Hydrocarbons Class 11 Important Extra Questions Chemistry Chapter 13

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Class 11 Chemistry Chapter 13 Important Extra Questions Hydrocarbons

Hydrocarbons Important Extra Questions Very Short Answer Type

Question 1.
Give different isomers of C4H10 with their I.U.P.A.C. names.
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 1

Question 2.
Give the I.U.P.A.C. name of the lowest molecular weight alkane that contains a quaternary carbon.
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 2
It is and its I.U.P.A.C. name is 2, 2-Dimethylpropane.

Question 3.
Which of the following has the highest boiling point?
(i) 2-methylpentane
(ii) 2, 3 – dimethylbutane
(iii) 2, 2-dimethylbutane.
Answer:
(i) 2—methyl pentane has the largest surface area and hence has the highest boiling point.

Alkene-reactions-cheat-sheet-summary-for-organic-chemistry-reactions.

Question 4.
Give the structure of the alkene (C4H8) which adds on HBr in the presence and in the absence of peroxide to give the same product C4H9Br.
Answer:
2-Butene with structure CH3 – CH = CH — CH3 being symmetrical gives the same product, i.e., 2-bromobutane CH3 CH (Br) CH2CH3.

Question 5.
How will you separate propene from propyne?
Answer:
Bypassing the mixture through ammoniacal AgNO3 solution when propyne reacts while propene passes over.

Question 6.
Name two reagents that can be used to distinguish \ between ethene and ethyne.
Answer:
Tollen’s reagent | Ammoniacal AgNO3 | and amm. CuCl solution.

Question 7.
How will you detect the presence of unsaturation in an organic compound?
Answer:
Either by Baeyer’s reagent
Hydrocarbons Class 11 Important Extra Questions Chemistry 3
or by Br, in CC14.

Question 8.
Arrange the following In order of increasing volatility: gasoline, kerosene, and diesel.
Answer:
Diesel, kerosene, gasoline.

Question 9.
Arrange the following: HCl, HBr, HI, HF in order of decreasing reactivity towards alkenes.
Answer:
HF, HCl, HBr, HI.

Question 10.
Out of ethylene and acetylene which is more acidic and why?
Answer:
Acetylene. Ethylene and acetylene have sp2, sp hybridized C atoms respectively. Due to the 50% S character of the C – H bond of acetylene rather than the 33% S-Character of the C – H bond in ethene, acetylene is more acidic.

Question 11.
Write the structure of the alkene which on reductive ozonolysis gives butanone and ethanal.
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 4

Question 12.
Write the I.U.P.A.C. names of
Hydrocarbons Class 11 Important Extra Questions Chemistry 5
Answer:
3-methylpent-l— en—4-yne.

(ii) CH2 = CH – CH (CH3) – CH = CH – CH – CH2,
Answer:
3-methylhept-1, 4, 6—triene.

Question 13.
Draw the structures of the following:
(i) Dicyclopropyl methane
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 6

(ii) 2-methyl-3—isopropyl heptane.
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 7

Question 14.
What effect the branching of an alkane has on its boiling point?
Answer:
Branching decreases the boiling point.

Question 15. Which of the following polymerizes most readily?
(i) Acetylene
(ii) Ethene
(iii) Buta —1, 3—diene
Answer:
(iii) Buta—1, 3—diene polymerizes most readily, being more reactive.

Question 16.
Arrange the following in increasing order of their release of energy on combustion
Hydrocarbons Class 11 Important Extra Questions Chemistry 8
Answer:
The more the number of C atoms having maximum hydrogen hydrogens, i.e., CH3 groups, the greater is the heat of combustion. Thus the increasing order of heat of combustion is (iii) < (iv) < (i) < (ii).

Question 17.
Arrangement of the following set of compounds in order of their decreasing relative reactivity with an electrophile E+.
(i) Chloro benzene, 2, 4 —dinitrochlorobenzene p—nitrochlorobenzene.
(ii) toluene, p – CH3 – C6H4 – CH3, p – CH3 – C6H4 – NO2, p – O2N – C6H4 – NO2
Answer:
Electron-donating groups increase the reactivity towards an electrophile E+, while electron-withdrawing groups decrease the reactivity. Thus
(i) Chlorobenzene > p — nitrochlorobenzene > 2. 4—di nitrochlorobenzene.
(ii) p – CH3 – C6H4 – CH3 > toluene > p – CH3 – C6H4 – NO2 > p – O2N – C6H4 – NO2.

Question 18.
What is the order of reactivity of halogen and alkyl groups in the dehydrohalogenation of alkyl halides to give alkenes?
Answer:

  • Halogens: Iodine > Bromine > Chlorine
  • Alkyl group: Tert > secondary > primary.

Question 19.
What products are formed when zinc reacts with
(i) vicinal C2H4Br2 and
Answer:
CH2Br – CH2Br + Zn → CH2 = CH2 + ZnBr2.

(ii) CH3CHBr – CH2Br.
Answer:
CH3 – CHBr – CH2Br + Zn → CH3 — CH = CH2 + ZnBr2.

Question 20.
What does L.P.G. stand for?
Answer:
L.P.G. stands for liquefied petroleum gas.

Question 21.
What do the terms (i) CNG and LPG stand for?
Answer:

  • CNG: Compressed natural gas
  • LPG: Liquefied Petroleum gas.

Question 22.
What are sources to obtain:
(i) LPG
Answer:
LPG is obtained by the fractional distillation of petroleum.

(ii) CNG?
Answer:
CNG is obtained by the fractional distillation of coal tar.

Question 23.
Write down the structures and names of all isomers with a molecular formula of C5H12.
Answer:

  1. (i) CH3 — CH2 — CH2 — CH2 — CH3 is n-Pentane or Pentane.
  2. Hydrocarbons Class 11 Important Extra Questions Chemistry 9
    is iso-pentane or 2-Methylbutane.
  3. Hydrocarbons Class 11 Important Extra Questions Chemistry 10
    is neo-pentane or 2, 2-Dimethylpropane.

Question 24.
The sodium salt of which acid will give ethane on heating with soda-lime? Give reaction.
Answer:
Propionic acid.
Hydrocarbons Class 11 Important Extra Questions Chemistry 11

Question 25.
What products are obtained by the acidic dehydration of
(i) Ethanol
Answer:
Ethene

(ii) Propan – 2—ol.
Answer:
Propene.

Question 26.
Out of cis-2-butene and trans-2-butene which is polar and which one is non-polar?
Answer:
Cis – 2 – butene is polar (μ = 0.33 D) and trans – 2 – butene is non-polar (μ = 0). .

Question 27.
Arrange the following in the decreasing order of acidic character.
(i) C2H4, C2H6, C2H2
Answer:
H – C ≡ C H > H2C = CH2 > H3C – CH3

(ii) CH3 – C = CH, C2H2, CH3 – C = C – CH3
Answer:
HC = CH > CH3 — C ≡ CH > > CH3 – C ≡ C – CH3.

Question 28.
Name two industrial sources of hydrocarbons.
Answer:

  1. Petroleum
  2. Coal.

Question 29.
Arrange the following in the increasing order of C – C bond length C2H6, C2H4, C2H22.
Answer:
C2H2 < C2H4 < C2H6.

Question 30.
What type of hydrocarbons is present in high octane gasoline?
Answer:
Branched-chain aliphatic and/or aromatic hydrocarbons.

Question 31.
What are the chief constituents of light oil fraction?
Answer:
Benzene, toluene, and xylenes.

Question 32.
Which of the following shows geometrical isomerism?
(i) CHCl = CHCl
(ii) CH2 = CCl2
(iii) CCl2 = l. Give the structures of cis-and transforms.
Answer:
(i) HC (Cl) = CH (Cl);
Hydrocarbons Class 11 Important Extra Questions Chemistry 12

Question 33.
Name the product formed when methyl bromide is treated with sodium and ether.
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 13

Question 34.
Which of the following shows geometrical isomerism?
But-1—ene or but—2—ene.
Answer:
But-2—ene CH3 – CH = CH – CH3.

Question 35.
Write the structure and I.U.P.A.C. name of Acetonitrile.
Answer:
Acetonitrile is CH33 CN. Its I.U.P.A.C. name is Ethane nitrile.

Question 36.
Why does carbon has a larger tendency of catenation than silicon although they have the same number of valance electrons?
Answer:
It is due to the smaller length of the C – C bond which is stronger (335 kJ mol-1) than the Si-Si bond (225.7 kJ mol-1).

Question 37.
Give name atm structure to the first organic compound synthesized in the laboratory.
Answer:
Urea
Hydrocarbons Class 11 Important Extra Questions Chemistry 14

Question 38.
Benzene is highly unsaturated, yet it does not give usual addition reactions readily. Why?
Answer:
Benzene is highly unsaturated, yet, resonance imparts extra-stability to benzene and it does not give additional reactions.
Hydrocarbons Class 11 Important Extra Questions Chemistry 15

Question 39.
What is Lindlar’s catalyst? What is it used for?
Answer:
Pd/BaSO4 poisoned with quinoline. It is used for the partial reduction of alkynes to cis-alkenes.

Question 40.
How will you detect the presence of unsaturation in an organic compound?
Answer:
Either by Baeyer’s reagent or by Br2 in CCl4. The color is discharged.

Question 41.
How can ethylene be converted to ethane?
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 16

Question 42.
Arrange the following in increasing order of C — C bond length C2H6, C2H4, C2H2.
Answer:C2H2 < C2H44 < C2H6.

Question 43.
What type of hybridization is involved in
(i) planar and
Answer:
sp2

(ii) linear molecules?
Answer:
sp.

Question 44.
Name the chain isomer of C55H12 which has a tertiary hydrogen atom.
Answer:
2-Methyl butane (CH3), CHCH2CH3.

Question 45.
What type of isomerism is shown by butane and isobutane.
Answer:
Chain or nuclear isomerism,

Question 46.
What do you mean by cracking?
Answer:
The thermal decomposition of higher hydrocarbons into lower hydrocarbons in the presence or absence of a catalyst is called cracking.

Question 47.
Complete the reaction HC ≡ CH
Hydrocarbons Class 11 Important Extra Questions Chemistry 17
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 18

Question 48.
Out of Octane and is heptane, which has a lower octane number?
Answer:
n-Octane.

Question 49.
What are the main components of LPG?
Answer:
Butane and isobutane.

Question 50.
Write a chemical reaction to illustrate the Saytzeffs rule.
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 19

Hydrocarbons Important Extra Questions Short Answer Type

Question 1.
The following organic compounds are known by their common names
(i) Neopentane
Answer:
Neopentane is
Hydrocarbons Class 11 Important Extra Questions Chemistry 20
& its h.U.P.A.C. name is 2, 2—dimethyl propane.

(ii) Acetone
Answer:
Acetone is
Hydrocarbons Class 11 Important Extra Questions Chemistry 87
and its I.U.P.A.C. name is Propanone.

(iii) Vinyl chloride
Answer:
Vinyl chloride is CPU = CH — Cl and its I.U.P.A.C. name are chloroethene.

(iv) Tert butyl alcohol. Write their structural formulae and I.U.P.A.C. names.
Answer:
Tert; butyl alcohol
Hydrocarbons Class 11 Important Extra Questions Chemistry 21
and its I.U.P.A.C. name is 2-methyl propan-2-ol.

Question 2.
What are the various products expected when propane reacts with fuming nitric acid?
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 22

Question 3.
How will you convert methane into
(i) Methanol
Answer:
Conversion of methane into methanol:
Hydrocarbons Class 11 Important Extra Questions Chemistry 23

(ii) Methanal.
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 24

Question 4.
What is aromatization? How will you convert ^hexane into benzene?
Answer:
Aromatization. It is the process that involves cyclization, isomerization, and dehydrogenation with the application of heat and catalyst to convert alkanes containing six or more carbon atoms into aromatic hydrocarbons.
Hydrocarbons Class 11 Important Extra Questions Chemistry 25

Question 5.
Give the different conformations of ethane with their
(i) Sawhorse representation and
(ii) Newmann Projection formulae.
Answer:
Sawhorse representation Newmann projection models
Hydrocarbons Class 11 Important Extra Questions Chemistry 26
Hydrocarbons Class 11 Important Extra Questions Chemistry 27

Question 6.
What are the relative stabilities of different conforma¬tions of ethane? Is it possible to isolate these at room temperature?
Answer:
The staggering form of ethane is more stable than the eclipsed form because the force of repulsion between hydrogen atoms on adjacent C atoms is minimum. The energy difference between the staggered form and eclipsed form of ethane is just 12.55 kJ mol-1. Therefore, it is not possible to separate these two forms of ethane at room temperature.

Question 7.
What is Saytzeff Rule? What are the expected products when 2-Bromobutane is dehydrohalogenation with ale. KOH?
Answer:
Saytzeff Rule. Whatever two alkenes are theoretically possible during a dehydrohalogenation reaction, it is always the more highly substituted alkene that predominates.
Hydrocarbons Class 11 Important Extra Questions Chemistry 28

Question 8.
What is the order of reactivity of H2C = CH2, (CH3)2, H2C = CH2, CH3 – CH = CH2, CH3 – CH = CH – CH3, (CH3)2 C = C (CH3)2, (CH3)2 C = CH CH3 towards electrophilic addition reactions?
Answer:
The order of reactivity of the above alkenes towards electrophilic addition reactions decreases in the order.
(CH3)2 C = C (CH3)2 > (CH33)2 C = CH CH3 > (CH3)2 C = CH2 > CH3 CH – CH – CH3 > CH3 – CH = CH, > CH2 = CH2.

Question 9.
Define Markownikov rule. Explain it with an example.
Answer:
Markownikov rule states. The negative part of the addendum adding to an unsymmetric alkene goes to that C atom of the double bond which is attached to a lesser number of C atoms.
Hydrocarbons Class 11 Important Extra Questions Chemistry 29

Question 10.
What is the Peroxide effect/Kharasch Effect? Illustrate with an example.
Answer:
In the presence of peroxides such as benzoyl peroxide, the addition of HBr (but not of HCl or HI) to an unsymmetrical alkene takes place contrary to the Markownikov rule. This is known as the peroxide/Kharsch effect.
Hydrocarbons Class 11 Important Extra Questions Chemistry 30

Question 11.
An alkene with the molecular formula C7H14 gives propanone and butanal on ozonolysis. Write down its structural formula and its I.U.P.A.C. name.
Answer:
The structures of the compounds on ozonolysis of C7H14 and
Hydrocarbons Class 11 Important Extra Questions Chemistry 31
Remove the oxygen atoms and connect them by a double bond, the structure of the alkene is
Hydrocarbons Class 11 Important Extra Questions Chemistry 32

Question 12.
How will you prepare propyne and I-Butyne from acetylene;?
Answer:
(i) Preparation of propyne from acetylene
Hydrocarbons Class 11 Important Extra Questions Chemistry 33

(ii) Preparation of 1-butyne from acetylene
Hydrocarbons Class 11 Important Extra Questions Chemistry 34

Question 13.
Describe a method to distinguish between ethane, ethene, ethyne.
Answer:
(i) Ethene (C2H4) and ethyne (C2H2) decolorize bromine in carbon tetrachloride whereas ethane (C2H6) does not.
Hydrocarbons Class 11 Important Extra Questions Chemistry 35

(ii) Ethyne (and not ethane, ethene) reacts with ammoniacal AgNO3 (Tollen’s reagent) to form white precipitates.
Hydrocarbons Class 11 Important Extra Questions Chemistry 36

Question 14.
Give the mechanism of an electrophilic addition of chlorine into propene.
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 37
Cl4 (chloronium ion) formed by the heterolytic fission of Cl2 in step (i) being an electrophile attacks propene in (ii) step, propene undergoes electrometric effect combined with + I effect to form carbocation which, is a slow step. In (iii) step Cl ion being a. nucleophile attacks carbocation and forms the product. It is a fast step.

Question 15.
Identify the correct order of reactivity in electrophilic substitution reactions of the following compounds: [I.I.T. 2002]
Hydrocarbons Class 11 Important Extra Questions Chemistry 38
Answer:
— NO2 group in structure IV is an electron attracting group, it deactivates the benzene ring largely towards electrophilic substitution reactions. Cl group in III is also a deactivating group, but its deactivation is lower too — NO2, where—CH3, a group in II is an electron-releasing group and so activates the benzene ring towards electrophilic substitution. Therefore, the order of reactivity is
Hydrocarbons Class 11 Important Extra Questions Chemistry 39

Question 16.
What is meant by (i) delocalization
Answer:
Delocalisation: Delocalisation means that pairs of 7t electrons extend over 3 or more atoms. They belong to the whole molecule. For example, 6n electrons present in benzene are delocalized and are spread on the whole of the ring and this imparts extra stability to the molecule.

(ii) resonance energy.
Answer:
Resonance energy: The difference between the energy of the most stable contributing/canonical structural and the energy of the resonance hybrid is known as resonance energy. In the case of benzene, the resonance hybrid has 147 kJ mol-1 than either A or B below. Thus resonance energy of benzene is 147 kJ mol-1.
Hydrocarbons Class 11 Important Extra Questions Chemistry 40

Question 17.
Describe Friedel’s craft reaction with suitable examples.
Answer:
When an alkyl or acid halide is treated with benzene or its derivative in the presence of anhydrous AlCl3 as a catalyst, we got alkyl or acyl benzene.
Hydrocarbons Class 11 Important Extra Questions Chemistry 41

Question 18.
Classify the following hydrocarbons into alkanes, alkenes, alkynes, and arenes.
(i) (CH3)4C
Answer:
Alkane

(ii) C2H2
Answer:
Alkyne

(iii) C3H6
Answer:
Alkene

Hydrocarbons Class 11 Important Extra Questions Chemistry 42
Answer:
Arene.

Question 19.
Write all possible structures of C5H8 and give their I.U.P.A.C. names.
Answer:
Structural isomers of C5H8 (Pentyne) are
(i) CH3 – CH2 – CH2 – C ≡ CH (Pent-l-yne)
(ii) CH3 – CH2 – C ≡ C – CH3 (Pent-2-yne)
(iii) Hydrocarbons Class 11 Important Extra Questions Chemistry 43
(3-Methylbut-l-yne)

Question 20.
How does ethylene undergo polymerization? What is the use to which the polymer obtained is put?
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 44
Low-density polythene [LDPE] and high-density polyethylene [HDPE] as used for the manufacture of plastic bags, squeeze bottles, refrigerator dishes, toys, pipes, radio, and T.V. cabinets, etc.

Question 21.
What is the action of water on
(i) Ethyne
(ii) Propyne? Name the end products obtained.
Answer:
(i) Action of water on ethyne: When ethyne is warmed with dilute H2SO4 at 333K and H8SO4 as catalyst ethanol is obtained.
Hydrocarbons Class 11 Important Extra Questions Chemistry 45
(ii) Action of water on Propyne CH3 — C ≡ CH + H2O
Hydrocarbons Class 11 Important Extra Questions Chemistry 46

Question 22.
Propene reacts with HBr to give Isopropyl bromide but does not give n-propyl bromide. Why?
Answer:
The addition of unsymmetrical addendum (HBr) to unsymmetrical olefines (CH3CH = CH2) takes place according to Markownikov rule, the negative part of reagent (i.e. Br-) adds on the carbon atom having a minimum number of hydrogen atoms. Hence Isopropyl bromide will be formed.
Hydrocarbons Class 11 Important Extra Questions Chemistry 47

Question 23.
An oxidizing agent is needed in the iodination of methane but not in the chlorination or bromination. Give reason.
Answer:
In the iodination of methane, H — I is also formed as the product with iodomethane, since it is a strong reducing agent, it reduces iodomethane back to methane & makes the reaction reversible. In order to destroy HI, an oxidizing agent like HIO3 (or HNO3) is needed. But HCl& HBr formed in the chlorination & bromination reactions of methane & not in a position to react with the monosubstituted products (CH3Cl & CH3Br) since they are comparatively weak reducing agents.

Therefore, no oxidizing agent is needed for these reactions.
CH4 + I2 ⇌ CH3I + HI
5HI + HIO3 → 3H2O + 3I2

Question 24.
The dipole moment of trans 1, 2-chloroethene is less than the cis isomer. Explain.
Answer:
The structure of the trans isomer is more symmetrical to the cis isomer. In the trans isomer, the dipole moments of the polar C—Cl bonds are likely to cancel with each other & the resultant dipole moment of the molecule is nearly zero. But in the cis isomer, these do not cancel. Therefore, the cis isomer has a specific moment but is zero in the case of the trans isomer.
Hydrocarbons Class 11 Important Extra Questions Chemistry 48

Question 25.
Write l.U.P.A.C. names of the products obtained by addition reactions of HBr to hex-l-ene.
(a) In the absence of Peroxide
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 49
(b) In the presence of Peroxide.
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 50

Question 26.
How will you distinguish between the following:
(a) Butyne-l & Butyne-2 ,
Answer:
Butyne-1 having an acetylene hydrogen atom will give white ppt. With ammoniacal silver nitrate & red ppt. with ammonical cuprous chloride. On the other hand, butyne-2-having no acetylenic hydrogen atom does not respond to either of the two reagents.

(b) Butene-1 & Butene-2
Answer:
Butene-1 & butene-2 can be distinguished either by ozonolysis or by oxidation with acidic KMnO2 solution with which they give different carbonyl compounds.
Hydrocarbons Class 11 Important Extra Questions Chemistry 51

Question 27.
Alkynes are less reactive than alkenes towards addition reaction even though they contain 2-7t bond. Give reason.
Answer:
This is due to

  1. greater electronegativity of sp-hybridized carbon of alkynes than sp2 hybridized carbon atoms of alkenes which holds the π-electrons of alkynes more tightly and
  2. greater delocalization of π-electrons in alkynes (because of the cylindrical nature of their n electron cloud) than in alkenes. As a result, n electrons of alkynes are less easily available for addition reactions than those of alkenes.

Consequently, alkynes are less reactive than alkenes towards addition reactions.

Question 28.
Why do addition reactions occur more readily with alkenes & alkynes than with aromatic hydrocarbons?
Answer:
The energy gained by forming two sigma bonds (of four sigma bonds) more than compensates for the loss of one or two n bonds when addition occurs to an alkene or alkyne. However, in aromatic hydrocarbons, the aromatic ring is specially stabilized by the delocalization of n electrons about the ring.

It, therefore, requires substantial activation energy to cause the loss of its aromatic character. The most usual reaction in arenes is thus substitution rather than addition, since substitution does not result in loss of aromatic character.

Question 29.
A Hydrocarbon A, adds one mole of hydrogen in presence of platinum catalyst from n-Hexane. When A is oxidized vigorously with KMnO4, a single carboxylic acid, containing three carbon atoms is isolated. Give the structure of A & explain.
Answer:

  1. Since hydrocarbon A adds one molecule of H2 in presence of platinum to form n-hexane. A must be a hexene.
  2. Since A on vigorous oxidation with KMnO4 gives a single carboxylic acid containing three carbon atoms, therefore, A must be asymmetrical hexene i.e. hex-3-ene.
    Hydrocarbons Class 11 Important Extra Questions Chemistry 52
    Thus, the given hydrocarbon A is hex-3-ene.

Question 30.
How would you carry out the following conversion?
Answer:
Propene to Ethyne
Hydrocarbons Class 11 Important Extra Questions Chemistry 53

Hydrocarbons Important Extra Questions Long Answer Type

Question 1.
How would you convert the following compounds to benzene?
(i) Acetylene
Answer:
Acetylene into benzene. Ethyne (Acetylene) in passing through a red hot iron tube at 873 K undergoes cyclic polymerization as shown below.
Hydrocarbons Class 11 Important Extra Questions Chemistry 54

(ii) Benzoic acid
Answer:
Benzoic acid into benzene
Hydrocarbons Class 11 Important Extra Questions Chemistry 55

(iii) Hexane
Answer:
Hexane into benzene
Hydrocarbons Class 11 Important Extra Questions Chemistry 56

(iv) Benzene diazonium chloride
Answer:
Benzene diazonium chloride into benzene
Hydrocarbons Class 11 Important Extra Questions Chemistry 57

Question 2.
How will you convert benzene into
(i) Nitrobenzene
Answer:
Benzene into Nitrobenzene
Hydrocarbons Class 11 Important Extra Questions Chemistry 58

(ii) Benzene sulphonic acid
Answer:
Benzene into benzene sulphonic acid
Hydrocarbons Class 11 Important Extra Questions Chemistry 59

(iii) Toluene
Answer:
Benzene into Toluene
Hydrocarbons Class 11 Important Extra Questions Chemistry 60

(iv) Acetophenone?
Answer:
Benzene into Acetophenone
Hydrocarbons Class 11 Important Extra Questions Chemistry 61

Question 3.
What is the mechanism of nitration of benzene?
Answer:
Nitration of benzene. It is carried out by treating benzene with a mixture of cones. HNO3+ Cone. H2SO4. The various steps involved are:
Step I: Generation of an electrophile, i.e., NOt (nitronium ion)
Hydrocarbons Class 11 Important Extra Questions Chemistry 62

Step II: Formation of complex or carbocation intermediate
Hydrocarbons Class 11 Important Extra Questions Chemistry 63
This step is slow and hence is the rate-determining step of the reaction.

Step III: Loss of a proton from the carbocation intermediate
Hydrocarbons Class 11 Important Extra Questions Chemistry 64
This step is fast and does not affect the rate of the reaction.

Question 4.
(a) What is the general formula of Alkynes?
Answer:
The general formula of alkynes ¡s CnH2n-2.

(b) Give the I.U.P.A.C. names and structure of all alkynes having the molecular formula C2H8.
Answer:
C5H8 has the following isomers.
(i) CH3 — CH2 — CH2 — C ≡ CH Pent-1-yne
(ii) CH3 CH, C ≡ C — CH3 Pent-2-yen
(iii) Hydrocarbons Class 11 Important Extra Questions Chemistry 65
3-Methyl but-1-yen

(c) Give any two methods for preparing acetylene.
Answer:
Acetylene can be prepared by the following two methods
Hydrocarbons Class 11 Important Extra Questions Chemistry 66
(ii) By dehydrohalogenation of dihaloalkanes
Hydrocarbons Class 11 Important Extra Questions Chemistry 67
(d) Discuss any three chemical properties of acetylene.
Answer:
(i) Addition of Hydrogen:
Hydrocarbons Class 11 Important Extra Questions Chemistry 68

(ii) On oxidation with alkaline KMnO4, it gives oxalic acid.
Hydrocarbons Class 11 Important Extra Questions Chemistry 69

(iii) When acetylene is passed through a red hot iron tube, it trickeries to give benzene
Hydrocarbons Class 11 Important Extra Questions Chemistry 70

Question 5.
(a) Write structures of different chain isomers of alkanes corresponding to the molecular formula C6H14. Also, write their I.U.P.A.C. names.
(b) Write structures of different isomeric alkyl groups corresponding to the molecular formula C5H11. Write IUPAC names of alcohols obtained by attachment of —OH groups at different carbons of the chain.
Answer:
Hydrocarbons Class 11 Important Extra Questions Chemistry 71
(b) Structures of – C5H11 group Corresponding alcohols
Hydrocarbons Class 11 Important Extra Questions Chemistry 72
Hydrocarbons Class 11 Important Extra Questions Chemistry 73

Question 6.
How will you prepare Alkanes by
(i) Wurtz reaction
Answer:
Methods of preparation of Alkanes
Wurtz reaction. Alkyl halides on treatment with sodium in dry ether give higher alkanes, preferably containing an even number of carbon atoms.
Hydrocarbons Class 11 Important Extra Questions Chemistry 74
(ii) Decarboxylation of sodium salts of fatty acids
Answer:
Sodium salts of fatty acids on heating with-soda lime (a mixture of NaOH and CaO) give alkanes containing one carbon atom less than the carboxylic acid. The process of elimination of carbon dioxide from a carboxylic acid is known as Decarboxylation
Hydrocarbons Class 11 Important Extra Questions Chemistry 75

(iii) Kolbe’s electrolytic method.
Answer:
Kolbe’s electrolytic method. An aqueous solution of sodium or potassium salt of a carboxylic acid on electrolysis gives alkanes containing an even number of carbon atoms at the anode.

The probable mechanism for the reaction is
Hydrocarbons Class 11 Important Extra Questions Chemistry 77

(b) How do alkanes udergo:
(i) substitution reactions with halogens
Answer:
Substitution reactions with Halogens. The order of reactivity of halogens is F2 > Cl2 > Br2 > I2
Hydrocarbons Class 11 Important Extra Questions Chemistry 78
Bromination is similar. With fluorine, the reaction is too violent to be controlled. Iodination is very slow and a reversible reaction.
It can proceed in the presence of oxidizing agents like HNO3, HIO3.
CH4 + I2 ⇌ CH3I + HI
HIO3 + 5HI → I3 + 3H2O

Halogenation of alkanes proceeds via a free-radical mechanism which consists of three steps.

  1. Chain initiation step
  2. Chain propagation step
  3. Chain termination step

1. Chain initiating step. Cl2 undergoes hemolysis in the presence of heat and light.
Hydrocarbons Class 11 Important Extra Questions Chemistry 79
2. Chain propagation step
Hydrocarbons Class 11 Important Extra Questions Chemistry 80
3. Chain termination step
Hydrocarbons Class 11 Important Extra Questions Chemistry 81

(ii) oxidation by combustion
Answer:
Oxidation of alkanes by combustion: Alkanes on heating in the presence of air or dioxygen are completely oxidized to carbon dioxide and water with the evolution of a large amounts of heat.
Hydrocarbons Class 11 Important Extra Questions Chemistry 82
In the presence of an insufficient amount of air or dioxygen, carbon black is formed.
Hydrocarbons Class 11 Important Extra Questions Chemistry 83

(iii) Isomerisation
Answer:
Isomerization reactions of alkanes, n-Alkanes on heating in the presence of anhydrous aluminum chloride and hydrogen chloride gas isomerize to branched-chain alkanes.
Hydrocarbons Class 11 Important Extra Questions Chemistry 84

(iv) Aromatisation.
Answer:
Aromatisation. n-Alkanes having six or more C atoms on heating to 773 K at 10-20 atmospheric pressure in. the presence of oxides of vanadium, molybdenum get dehydrogenated and cyclized. This reaction is called aromatization or reforming.
Hydrocarbons Class 11 Important Extra Questions Chemistry 85
Hydrocarbons Class 11 Important Extra Questions Chemistry 86

Equilibrium Class 11 Important Extra Questions Chemistry Chapter 7

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 7 Equilibrium. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 7 Important Extra Questions Equilibrium

Equilibrium Important Extra Questions Very Short Answer Type

Question 1.
Write the expression for the equilibrium constant Kp for the reaction 3Fe (s) + 4H2O (g) ⇌ Fe3O4 (s) + 4H2 (g)
Answer:
Kp = \(\frac{p_{\mathrm{H}_{2}}^{4}}{p_{\mathrm{H}_{2} \mathrm{O}}^{4}}=\frac{p_{\mathrm{H}_{2}}}{p_{\mathrm{H}_{2} \mathrm{O}}}\)

Question 2.
How are Kc and Kp related to each other in the reaction
N2 (g) + O2 (g) ⇌ 2NO (g)
Answer:
Kp = Kc.

Question 3.
What is the equilibrium constant expression for the reaction
Al (s) + 3H+ (aq) ⇌ Al3+ (aq) + \(\frac{3}{2}\)H2 (g)
Answer:
Kc = [Al3+ (aq)][H2 (g)3/2/[H+ (aq)]3.

Question 4.
What happens to the equilibrium
PCl5 (g) ⇌ PCl3 (g) + Cl2 (g) if nitrogen is added to it
(i) at constant volume
Answer:
The state of equilibrium remains unaffected.

(ii) at constant pressure?
Answer:
Dissociation increases, i.e., the equilibrium shifts forward.

Question 5.
What does the equilibrium K < I indicate? ,
Answer:
The reaction does not proceed much in the forward direction.

Question 6.
For an exothermic reaction, what happens to the equilibrium constant if the temperature is increased?
Answer:
K = K/Kb.
Kb increases much more than when the temperature is increased in an exothermic reaction. Hence K decreases.

How to find Kp from Best fit line.

Question 7.
Under what conditions, a reversible process becomes irreversible?
Answer:
If one of the products (gaseous) is allowed to escape out (i.e., in the open vessel).

Question 8.
What is the effect of increasing pressure on the equilibrium?
N2 (g) + 3H2 (g) ⇌ 2NH3 (g)?
Answer:
Equilibrium will shift in the forward direction forming more ammonia.

Question 9.
What is- the effect of increasing pressure on the equilibrium.
CaCO3 (s) ⇌ CaO (s) + CO2 (g)
Answer:
Kc = [CaO (s)][CO2 (g)J/[CaCO3 (s)]
Kc = [CO2 (g)]
Similarly, Kp = PCO2.

Question 10.
The equilibrium constant for the reaction SO3 (g) ⇌ SO2 (g) + \(\frac{1}{2}\) O2 (g) is 0.18 at 900 K. What will be the equilibrium
constant for the reaction SO2 (g) + \(\frac{1}{2}\) O2 (g) ⇌ SO3 (g)?
Answer:
K = \(\frac{1}{0.18}\) = 5.55.

350 Degrees F to C 176.667 °C The Celsius (ºC) degree or Fahrenheit (ºF) degree […..] by Helen C.

Question 11.
For which of the following cases does the reaction go farthest to completion: K = 1, K = 1010, K = 10-10.
Answer:
The reaction having K = 1010 will go farthest to completion because the ratio (product)/(reactants) is maximum in this case.

Question 12.
Under what conditions ice water system is in equilibrium?
(a) at 273 K
(b) below 273 K
(c) above 273 K.
Answer:
(a) At 273 K.

Question 13.
The equilibrium constant for the reaction
N2 (g) + 3H2 (g) ⇌ 2NH3-(g) is K. How is the equilibrium constant for the reaction
NH3 (g) ⇌ \(\frac{1}{2}\)N2 (g) + \(\frac{3}{2}\) H2 (g) related to K?
Answer:
If K’ is the equilibrium constant for the reaction
NH33(g) = \(\frac{1}{2}\)N2(g)+ \(\frac{3}{2}\)H2(g)
Then, K’ = \(\frac{1}{K}\).

Question 14.
Write the relation between Kc and Kp for the reaction
PCl5 (g) ⇌ PCl3 (g) + Cl2 (g)
Answer:
Δn (g) = 2 – 1 = 1
Hence Kp = Kc × (RT)Δn = K × RT.

Question 15.
The equilibrium constant of a reaction is 2 × 10-3 at 25°C and 2 × 10-2 at 50°C. Is the reaction exothermic or endothermic? ,
Answer:
It is endothermic as the equilibrium constant has increased with temperature.

Question 16.
What is the thermodynamic criterion for the state of equilibrium?
Answer:
At equilibrium, (ΔG)TP = 0.

Question 17.
Which measurable property becomes constant in water ⇌ water vapor equilibrium at constant temperature?
Answer:
Vapour pressure.

Question 18.
What happens to the dissociation of PCl5 in a closed vessel if helium gas is introduced into it at the same temperature?
Answer:
No effect.

Question 19.
What are the conditions for getting NH33 by Haber’s process?
N2 (g) + 3H2 (g) 2NH3 (g); ΔH = – Q.
Answer:
High concentration of N2 and H2, low temperature, and high pressure.

Question 20.
What happens if ferric salt is added to the equilibrium of the reaction between Fe3+ and SCN̅ ions?
Answer:
The red color deepens. Equilibrium shifts in the forward direction.

Question 21.
Does the value of equilibrium constant change on adding, a catalyst?
Answer:
No.

Question 22.
Name the factors which affect the equilibrium state.
Answer:
Temperature, pressure, and concentration.

Ionic Strength Calculator … Ionic strength is used to measure the concentration of the ions in a solution.

Question 23.
What happens to the ionic product of water if some acid is added into the water?
Answer:
It remains unchanged.

Question 24.
What is the pH of 0.1 M HCl?
Answer:
pH = – log [H+] = – log 10-1 = 1.

Question 25.
Which conjugate base is stronger: CN̅ or F̅?
Answer:
CN̅ is a stronger base than F̅

Question 26.
The dimethyl ammonium ion (CH3)2 NH2+ is a weak acid. What is its conjugate base?
Answer:
(CH3)2NH.

Question 27.
Write the solubility product expression for:
Ag2CrO4 (s) ⇌ 2Ag+ (aq) + CrO42- (aq)
Answer:
Ksp = [Ag+]2[CrO42-].

Question 28.
When does salt get precipitated in solution?
Answer:
When the Ionic product of its ions exceeds its solubility product.

Question 29.
The Ksp value of salt is high. What does it indicate?
Answer:
This means that the salt is highly soluble in water.

Question 30.
Select the Lewis acid and Lewis base in
SnCl4 + 2Cl → [SnCl66]2-.
Answer:
SnCl4 is acid and Cl is a base.

Question 31.
Out of pure water and 0.1. KCl solution in which AgCl will dissolve more?
Answer:
Pure water.

Question 32.
Write the conjugate acid and conjugate base of water?
Answer:
Conjugate acid H3O+ and conjugate base OH̅.

Question 33.
Select Lewis acids and Lewis bases from the following: Cu2+, H2O, BF3, OH.
Answer:
Lewis acids: Cu2+, BF3
Lewis base: H2O, OH.

Question 34.
Give two examples of cations that can act as Lewis acids.
Answer:
Ag+, H+.

Question 35.
What is the difference between a conjugate acid and a conjugate base?
Answer:
A conjugate acid and its conjugate base differ from each other by a proton.

Question 36.
What is the active mass of water?
Answer:
55.5 mol L-1

Question 37.
Whether the pH value of an aqueous solution of sodium acetate will be 7 or greater than 7?
Answer:
It is greater than 7.

Question 38.
An old sample of an aqueous solution of CuSO4 is acidic. Why?
Answer:
On keeping CuS04 solution undergoes hydrolysis to give H2SO4 a strong acid and a weak base Cu(OH)2

Question 39.
What happens to the pH of ammonium acetate solution when a few drops of acid are added to it?
Answer:
pH will remain unchanged as ammonium acetate (CH3COONH4) solution is a buffer.

Question 40.
What happens when HCl gas is passed through NaCl solution?
Answer:
NaCl will precipitate out.

Question 41.
What is the value of H3O+ ions and OH ions in water at 298 K?
Answer:
[H3O+] = [OH̅] 1.0 × 10-7 mL-1.

Question 42.
Will the ionic products of water increase or decrease on increasing the temperature?
Answer:
It increases with the increase in temperature.

Question 43.
Give one example of
(i) acidic buffer
Answer:
Acidic buffer: CH3COOH + CH3COONa

(ii) basic buffer.
Answer:
Basic buffer: NH4OH + NH4Cl.

Question 44.
Write down the conjugate base of [Al(H2O)6]3+.
Answer:
[Al(0H)(H2O)5]2+ is the conjugate base of [Al(H2O)6]3+.

Question 45.
The Ksp of CuS, Ag2S, HgS is 10-31, 10-44, 10-54 respectively. What is the order of the solubility of these sulfides?
Answer:
CuS > Ag2S > HgS.

Question 46.
Ksp for HgSO4 is 6.4 × 10-5. What is the solubility of the salt?
Answer:
S = \(\sqrt{\mathrm{K}_{s p}}=\sqrt{6.4 \times 10^{-5}}\)= 8 × 10-3 mol L-1.

Question 47.
The solubility product (K ) of silver chloride is 1.8 × 10-10 at 298 K. What is the solubility of AgO in 0.01 M HCl in mol L-1?
Answer:
[Ag+] = \(\frac{\mathrm{K}_{s p}}{\left|\mathrm{Cl}^{-}\right|}=\frac{1.8 \times 10^{-10}}{0.01}\) = 1.8 × 10-8 mol L-1
∴ Solubility of AgCl = 1.8 × 10-8 Mol L-1.

Question 48.
How does dilution with water affect the pH of a buffer solution?
Answer:
Dilution with water of a buffer solution has no effect on the pH of a buffer solution.

Question 49.
What is the pH of our blood? Why does it not change in spite of the variety of foods and Spices we eat?
Answer:
the pH of our blood is about 7.4. It remains constant because blood is a buffer.

Question 50.
Why does boric acid act as Lewis acid?
Answer:
Boric a.cid acts as a Lewis acid by accepting electrons from hydroxyl ions.
B(OH)3 + 2HOH → B(OH)4 + H3+O.

Question 51.
When a precipitate formed when solutions of BaCl2 and Na2SO4 are mixed?
Answer:
When in the final solution after mixing, the ionic product.
[Ba2+] [SO42- ] exceeds Ksp for BaSO4.

Question 52.
What is the common ion effect?
Answer:
The suppression of the dissociation of a weak electrolyte by the addition of a strong electrolyte having a common ion.

Question 53.
Arrange them in increasing order of the extent of hydrolysis.
CCl4, MgCl2, AlCl3 PCl5, SiCl4.
Answer:
The increasing order of the extent of hydrolysis is CCl4 < MgCl2 < AlCl3 < PCl5 < SiCl4.

Question 54.
Mention two different ways of drawing the following equilibrium towards the right
Equilibrium Class 11 Important Extra Questions Chemistry 1
[W.B. JEE 2003]
Answer:

  1. By adding more of CH3COOH or CH3CH2OH.
  2. By removing the ester or water formed.

Question 55.
Give one example of everyday life in which there is gas ⇌ solution equilibrium.
Answer:
Soda-water bottle.

Question 56.
What is the relationship between pKa and pKb values where Ka, Kb represent ionization constants of the acid and its conjugate base respectively?
Answer:
pKa + pKb = pKw =14.

Question 57.
What is the relationship between pH and pOH?
Answer:
pH + pOH= pKw = 14.

Question 58.
What is the function of adding NH4OH in group V?
Answer:
It converts any NH4HCO3 present into (NH4)2CO3.

Question 59.
What happens to the solubility of AgCl in water if NaCl solution is added to it?
Answer:
Solubility of AgCl decreases due to the common ion effect.

Question 60.
Write the expression for comparison of relative strengths of two weak acids in terms of their ionization constants.
Answer:
\(\frac{\text { Strength of acid }_{1}}{\text { Strength of acid }_{2}}=\sqrt{\frac{\mathrm{K}_{a_{1}}}{\mathrm{~K}_{a_{2}}}}\)

Equilibrium Important Extra Questions Short Answer Type

Question 1.
Justify the statement that water behaves like acid as well as a base on the basis of the protonic concept.
Answer:
Water ionizes as H2O + H2O ⇌ H3O+ + OH
With strong acids, water behaves as a base by accepting a proton from an acid.
HCl + H2O ⇌ H3O+ (aq) + Cl (aq)
While with bases, water behaves as an acid by liberating a proton
NH3 + H2O ⇌ NH4+ (aq) + OH (aq).

Question 2.
What is pOH? What is its value for pure water at 298 K?
Answer:
pOH = – log [OH]
pH + pOH = 14 for pure water at 298 K
pH = 7
or
pOH of water at 298 = 7.

Question 3.
Calculate the pH of a buffer solution containing 0.1 moles of acetic acid and 0.15 mole of sodium acetate. The ionization constant for acetic acid is 1.75 × 10-5.
Answer:
Equilibrium Class 11 Important Extra Questions Chemistry 2

Question 4.
An aqueous solution of CuSO4 is acidic while that of Na2SO4 is neutral. Explain.
Answer:
CuSO4 + 2H2O ⇌ Cu(OH)2 + H2SO4 (weak base strong acid)
CuS04 is the salt of weak base Cu(OH)2 and a strong acid H2SO4.
Thus the solution will have free H+ ions and will, therefore, be acidic.

Na2SO4, being the salt of a strong acid H2SO4 and a strong base.
NaOH does not undergo hydrolysis. The solution is, therefore, neutral.

Question 5.
The dissociation constants of HCN, CH3COOH, and HF are 7.2 × 10-10, 1.8 × 10-5, and 6.7 × 10-4 respectively. Arrange them in increasing order of acid strength.
Answer:
More the value of Ka, the stronger the acid
Their Ka1S are 6.7 × 10-4 > 1.8 × 10-5 > 7.2 × 10-10
∴ HCN < CH3COOH < HF.

Question 6.
The dissociation of PCl5 decreases in presence of Cl2. Why?
Answer:
For PCl5 ⇌ PCl3 + Cl2.
According to Le Chatelier’s principle, an increase in the concentration of Cl2 (one of the products) at equilibrium will favor the backward reaction, and thus the dissociation of PCl5 into PCl3 and Cl2 decreases.

Question 7.
The dissociation of HI is independent of pressure while dissociation of PCl5 depends upon the pressure applied. Why?•
Answer:
For 2HI ⇌ H2 + I2
Kc = \(\frac{x^{2}}{4(1-x)^{2}}\)
where x = degree of dissociation

For PCl5 ⇌ PCl3 + Cl2
K = \(\frac{x^{2}}{V(1-x)^{2}}\); V = Volume of container.

Kc for HI does not have a volume factor- and dissociation is independent of volume and hence pressure.
Kc for PCl5 has volume in the denominator and hence dissociation of PCl5 depends upon the volume and consequently pressure.

Question 8.
The reaction between ethyl acetate and water attains a state of equilibrium in an open vessel, but not the decomposition of CaCO3. Explain.
Answer:
CH2 COOC2H5 (l) + H2O (l) ⇌ CH3COOH (l) + C2H5OH (l)
Here both reactants and products are liquids and they will not escape from the vessel even if it is open. Therefore equilibrium is attained.
CaSO3 (s) ⇌ CaO (s) + CO2 (g)
Here CO2 is a gas. It will escape from the vessel if it is open and so backward reaction cannot take place. Therefore equilibrium is attained.

Question 9.
The degree of dissociation of N2O4 is α according to the reaction
N2O4 (g) ⇌ 2NO2 (g) at temperature T and total pressure P.
Find the expression for the equilibrium constant of this reaction.
Answer:
Equilibrium Class 11 Important Extra Questions Chemistry 3
Total moles at eqbm. = 1 – α + 2α = 1 + α
If P is total pressure
Equilibrium Class 11 Important Extra Questions Chemistry 4

Question 10.
Why NH4Cl is added in precipitating III group hydroxides before the addition of NH4OH?
Answer:
To prevent precipitation of IV group hydroxides (especially Mn) along with III group hydroxides. NH4Cl decreases dissociation of NH4OH and thus limited OH ions are present in solution to precipitate III group cations only
NH4OH ⇌ NH4 (aq) + OH (aq) to a small extent
NH4Cl (aq) ⇌ NH4 (aq) + Cl (aq) to a large extent
Due to common ion affect the degree of dissociation of NH4OH decreases leaving only a small no. of OH̅ ions.

Question 11.
If concentrations are expressed in moles L-1 and pressures in atmospheres. What is the ratio of Kp to Kc for the
2SO2 + O2 (g) ⇌ 2SO3 (g)
Answer:
Equilibrium Class 11 Important Extra Questions Chemistry 5

Question 12.
The equilibrium constants for the reactions
N2 + O2 ⇌ 2NOand
2NO + O2 ⇌ 2NO2 are K1 and K2 respectively, then what would be the equilibrium constant for the reactions.
N2 + 2O2 ⇌ 2NO2?
Answer:
Equilibrium Class 11 Important Extra Questions Chemistry 6

Question 13.
What qualitative information can be obtained from the magnitude of the equilibrium constant?
Answer:

  1. Large values of equilibrium constant (> 103) show that the forward direction is favored i.e. concentration of products is much larger than that of the reactants of equilibrium.
  2. Intermediate values of K (10-3 to 103) show that the concentrations of the reactants and products are comparable.
  3. The low value of K (< 10-3) shows that the backward reaction is favored, i.e., the concentration of reactants is much large than that of. products.

Question 14.
The following reaction has attained equilibrium
CO (g) + 2H2 (g) ⇌ CH3OH (g); ΔH° = – 92.0 kJ mol-1.
What will happen if
(i) the volume of the vessel is suddenly reduced to half?
Answer:
Kc = [CH3OH]/[CO][H2]2, Kp = PCH3OH/PCO × PH2
When the volume of the vessel is reduced to half, the concentration of each reactant or product becomes double. Thus
Qc = 2[CH3OH]/2[CO] × {2[H2]}2 = i K..
As Qc < Kp, equilibrium will shift in the forward direction.

(ii) The partial pressure of hydrogen is suddenly doubled (ii) an inert gas is added to the system.
Answer:
As volume remains constant, molar concentration will not change. Hence there is no effect on the state of equilibrium.

Question 15.
How does the degree of ionization of a weak electrolyte vary with concentration? Give exact relationship. What is this law called?
Answer:
α = \(\sqrt{\mathrm{K}_{i} / c}\) . It is called Ostwald’s dilution law.
(Ki is ionization constant and c is the molar concentration).

Question 16.
The ionization constant for formic acid and acetic acid is 17.7 × 10-5 and 1.77 × 10-5. Which acid is stronger and how many times the other if equimolar concentrations of the two are taken?
Answer:
Ka for HCOOH > Ka for CH3COOH. Hence formic acid is stronger.
Further \(\frac{\text { Strength of } \mathrm{HCOOH}}{\text { Strength of } \mathrm{CH}_{3} \mathrm{COOH}}=\sqrt{\frac{\mathrm{K}_{\mathrm{HCOOH}}}{\mathrm{K}_{\mathrm{CH}_{3} \mathrm{COOH}}}}\)
= \(\sqrt{10}\) = 3.16 times.

Question 17.
Out of CH3COO̅ and OH̅ which is the stronger base and why?
Answer:
OH̅ ions can combine with H+ ions more readily than CH3COO̅ ions can do. Hence OH̅ is a stronger base.

Question 18.
What is pKw? What is it? value at 25°C?
Answer:
pKw = – log Kw = – log 10-14 = 14.

Question 19.
What are pH and pOH values of a neutral solution at a temperature at which = 10-13?
Answer:
pKw = pH + pOH, But pKw =13
Hence pH = pOH = 6.5.

Question 20.
The ionization constants of HF = 6.8 × 10-4. Calculate the ionization constant of the corresponding conjugate base.
Answer:
Kb = \(\frac{\mathrm{K}_{w}}{\mathrm{~K}_{c}}=\frac{10^{-14}}{6.8 \times 10^{-4}}\) = 1.47 × 10-11.

Question 21.
What is the difference between an ionic product a rich solubility product?
Answer:
Solubility product is the product of the molar concentrations of the ions in a saturated solution, but the ionic product is for any solution.

Question 22.
Why common salt is added to precipitate soap from the solution during its manufacture?
Answer:
Soap is the sodium salt of higher fatty acid [RCOONa].
On adding common salt, Na+ ion concentration increases.
Hence the equilibrium RCOONa (s) ⇌ RCOO̅ + Na+ shifts in the backward direction, i.e., soap precipitates out.

Question 23.
Through a solution containing Cu2+ and Ni2+, H2S gas is passed after adding dil. HCl, which will precipitate out and why?
Answer:
Cu2+ ions will precipitate out because in the acidic medium only ionic product [Cu2+][S2-] exceeds the solubility product of CuS.

Question 24.
Why in Group V of qualitative analysis sufficient NH4OH solution should be added before adding (NH4)2CO3 solution?
Answer:
This is done to convert NH4HCO3 usually present in large amounts to (NH4)2CO3.
NH4HCO3 + NH4OH → (NH4)2CO3 + H2O.

Question 25.
The pH of an enzyme-catalyzed reaction has to be maintained between 7 and 8. What indicator should be used to monitor and control the pH?
Answer:
Bromothymol blue or phenol red or cresol red.

Question 26.
The ionization constant of formic acid is 1.8 × 10-4. Around what pH will its mixture with sodium formate give buffer solution of highest capacity?
Answer:
Buffer solution of highest capacity is formed at which
pH = PKa = – l0g 1.8 × 10-4 = 3.74.

Question 27.
Why PO4 ion is not amphiprotic?
Answer:
An amphiprotic ion is one that can donate proton as well as accept a proton. PO43- ion can accept a proton(s) but cannot donate any proton. Hence, PO43- is not amphiprotic.

Question 28.
In the reaction between BF3 and C2H5OC2H5 which one of them will act as an acid. Justify your answer.
Answer:
The reaction between BF3 and C2H5OC2H5 is
Equilibrium Class 11 Important Extra Questions Chemistry 7
As BF3 is electron-deficient and accepts a pair of electrons from C2H5OC2H5, hence BF3 is the Lewis acid.

Question 29.
Will the water be the same at 4°C and 25°C? Explain. [IIT 2003]
Answer:
No. the pH of water is not the same at 4°C and 25°C. This is because with an increase in temperature dissociation of H20 molecules increases. Hence concentration of [H+] ions will increase, i.e., pH will decrease. Thus pH of H2O at 4°C will be more than at 25°C.

Question 30.
What type of salts are Na2HPO3 and NaHS? [W.B. /EE 2003]
Answer:
Na2HPO3 is obtained by the reaction between NaOH and H3PO3 a dibasic acid.
Equilibrium Class 11 Important Extra Questions Chemistry 8
Both displaceable hydrogens are displaced by Na. No acidic hydrogen is left. Hence Na2HPO3 is a normal salt. NaHS is obtained by the replacement of one acidic hydrogen of H2S by Na (on reaction with NaOH)’. Hence NaHS is an acidic salt.

Question 31.
Explain why the pH of 0.1 molar solution of acetic acid will be higher than that of 0.1 molar solution of HCl.
Answer:
Acetic acid is a weak electrolyte. It is not completely ionized and hence gives less H+ ion concentration. HCl is a strong acid. It is completely ionized giving more H+ ions concentration. As pH = – log [H+], the lesser the no. of H+ ions, the more the value of pH. Therefore pH of 0.1 M acetic acid is more than that of 0.1 M HCl.

Question 32.
Benzoic acid is a monobasic acid. When 1.22 of its pure sample are dissolved in water and titrated against base 50 mL of 0.2 M NaOH are used up. Calculate the molar mass of benzoic acid.
Answer:
1000 ml of 1.0 M NaOH will neutralize acid
= \(\frac{1.22}{50 \times 0.2}\) × 1000 = 122 g.

But 1000 ml of 1.0 M NaOH contains 1 mole of NaOH and will neutralize 1 mole of monobasic acid. Hence the molar mass of benzoic acid is 122 g mol-1

Question 33.
Explain why ammonium chloride is acidic in liquid ammonia solvent.
Answer:
When NH4Cl is present in liquid ammonia, the following reaction takes place:
NH4 + NH3 ⇌ NH3 + NH4
Thus NH4Cl gives protons to liquid ammonia solvent.
Hence it is acidic.

Question 34.
Arrange the following in order of their increasing basicity.
H2O, OH̅, CH3OH, CH3 O̅.
Answer:
H2O < CH3OH < OH̅ < CH3 O̅.

Question 35.
The following can act both as Bronsted acid and Bronsted base. Write the formula in each case (of the product).
(i) HCO3-
Answer:
CO32-, H2CO3

(ii) H2PO4-
Answer:
HPO4, H3PO4

(iii) NH3
Answer:
NH̅2, NH4+

(iv) HS̅.
Answer:
S2-, H2S.

Question 36.
NaCl solution is added to a saturated solution of PbCl2. What will happen to the concentration of Pb2+ ions?
Answer:
Pb2+ ion concentration will decrease to keep K constant.

Question 37.
Which is a stronger base in each of the following pairs and why?
(i) H2O, Cl
Answer:
H2O is a stronger base being the conjugate base of a weak acid H3O+. Cl ion is the conjugate base of a strong acid and so is weak.

(ii) CH3COO, OH.
Answer:
OH is the stronger base, being the conjugate base of H2O a very weak acid.

Question 38.
For an aqueous solution of NH4Cl, prove that [H3O]+ = \(\sqrt{\mathbf{K}_{h} c}\) [CBSE PMT 2004]
Answer:
For salt of a strong acid weak base
Equilibrium Class 11 Important Extra Questions Chemistry 9

Question 39.
What are the conjugate bases of the following?
CH3OH, HN3, [Al(H2O)6]3+.
Answer:
CH3O (methoxide ion), N3 (azide ion), [Al(H2O)S(OH)]2+.

Question 40.
Write reaction for autoprotolysis of water. How is the ionic product of water-related to ionization constant of water? Derive the relationship.
Answer:
Autoprotolysis of H2O takes place as follows:
Equilibrium Class 11 Important Extra Questions Chemistry 10

Question 41.
Glycine is an a-aminoacid that exists in the form of Zwitter ion as NH3CH2OO̅. Write the formula of its conjugate acid and conjugate base.
Answer:
Equilibrium Class 11 Important Extra Questions Chemistry 11

Equilibrium Important Extra Questions Long Answer Type

Question 1.
Explain chemical equilibrium with th^ help of an example of formation and decomposition of hydrogen iodide.
Answer:
Consider the reaction between hydrogen and iodide at a constant temperature of 720 K in a closed vessel. The reaction involved is:
H2(g) + I2(g) → 2HI(g)

Accordingly, the effective collision amongst the reactant molecules will result in the production of HI. Since the product molecules are not permitted to leave the vessel (i.e., the reaction is carried out in a closed vessel), they will also collide amongst themselves leading to the formation of reactant molecules. Under these conditions, the reaction takes place in both directions. Hence, it is called a reversible reaction.
Equilibrium Class 11 Important Extra Questions Chemistry 12
Graphical representation of the change of reaction rates with time for the formation and decomposition of hydrogen iodide

Forward reaction: H2 (g) + I2 (g) → 2HI (g)
Backward reaction: 2HI (g) → H2 (g) + I2 (g)
Reversible reaction: H2 (g) + I2 (g) ⇌ 2HI (g).

To begin with, with the concentration of the reactants being higher in comparison to the product molecules, the rate of the forward reaction will be high as compared to the backward reaction. As the reaction proceeds further, the molar concentration of the reactants will gradually decrease while that of the product will gradually increase.

Apparently, the rate of forwarding reaction goes on decreasing while that of the backward reaction. This state is the reversible chemical reaction is called a chemical equilibrium state.

Question 2.
Name and explain the factors which influence the equilibrium state.
Answer:
The various factors which influence the equilibrium state are:
1. Concentration: Concentration change influences the equilibrium state. If the concentration of the reactants is increased, the equilibrium will shift in such a direction in which more to the products are formed and vice-versa.
On the other hand, if the concentration of the products is increased, the equilibrium will shift in such a direction in which more of the reactants are formed.

2. Temperature: Like concentration, the temperature change also affects the equilibrium state. An increase in temperature of the system will shift the equilibrium in such a direction in which heat is absorbed (i.e. rate of endothermic reaction will increase).

On the other hand, a decrease in temperature of the system will shift the equilibrium in such a direction in which heat is evolved (i.e., rate of exothermic reaction will increase).

3. Pressure: Like concentration and temperature, the pressure also influences the equilibrium state only when the reaction proceeds with a change in volume. An increase in pressure of the system will shift the equilibrium in such a direction in which the volume of the system decreases.

On the other hand, a decrease in pressure of the system will shift the equilibrium in such a direction in which the volume of the system increases.

To explain the effect of temperature, pressure, and concentration on the equilibrium state, consider the combination of N2 and H2 to form NH3
N2 (g) + 3H2 (g) ⇌ 2NH3 (g); ΔH = – 93.6 kJ

The reaction is reversible, exothermic, and accompanied by a decrease in volume.
Effect of temperature: According to Le-Chatelier’s principle, an increase in temperature shifts the equilibrium in the direction in which heat is absorbed, and a decrease in temperature shifts the equilibrium in the direction in which heat is evolved. Since the formation of ammonia is accompanied by the evolution of heat, it is favored by a decrease in temperature.

Effect of pressure: According to Le-Chatelier’s principle, an increase of pressure on a system in equilibrium, favors the direction which is accompanied by a decrease in volume and vice-versa. While going from, left to right in the above reaction, there is a decrease in the number of moles or say volume, the formation of ammonia is favored by an increase in pressure.

Effect of concentration: According to Le-Chatelier’s principle, an increase of concentration of any of the substances in the system shifts the equilibrium in the direction in which the concentration of that substance is reduced. Thus, the addition of N2 or H2 favors the formation of ammonia.

Question 3.
What is salt hydrolysis? Explain hydrolysis of salts of
(i) strong acids and strong bases
(ii) strong acids and weak bases
(iii) strong bases and weak acids
(iv) strong acids and weak bases.
Answer:
Salt hydrolysis: Hydrolysis is a process in which a salt reacts with water to form acid and base.
Salt + Water ⇌ Acid + Base
B A + H2O ⇌ HA + BOH

That is the interaction of the cations of the salt with OH ions furnished by water and anions of the salt with H+ ions furnished by water to form an acidic or basic solution is called salt hydrolysis.
(i) Salts of strong acids and strong bases like NaCl, KCl, KNO3 NaNO3, Na2SO4, K2SO4 do not undergo hydrolysis because the acids an.d bases furnished by them in aqueous solutions are strong acids and strong bases which are completely dissociated.
NaCl + H2O ⇌ NaOH + HCl
NaOH (aq) ⇌ Na+ + OH
HCl (Aq) ⇌ H+ + Cl

Since [H+] = [OH] the resulting solution is neutral and its pH = 7.

(iii) Hydrolysis of salts of strong acids and weak bases:
The salts belonging to this type are NH4NO3, NH4Cl, (NH4)2SO4, CuSO4, AlCl3, Ca(NO3)2, etc.

Let us take the case of NH4NO3
NH4NO3 + H2O ⇌ NH4OH + HNO3
NH4+ + NO3 + H2O ⇌ NH4OH + HNO3
or
NH4+ + H2O ⇌ NH4OH + H+

The resulting solution after hydrolysis is basic (pH > 7). Since only the anions of the salt have taken place in the hydrolysis, it is called anionic hydrolysis.

(iv) Hydrolysis of salts of weak acids and weak bases:
The salts belonging to this type are:
CH3COONH4, (NH4)2CO3, Ca3(PO4)2 etc.

Let us take the case of hydrolysis of CH3COONH4
CH3COONH4 + H3O ⇌ CH3COOH + NH4OH
or
CH3COO̅ + NH4 + H2O ⇌ CH3COOH + NH4OH

Since both the cations and anions of the salt have participated in the hydrolysis, it is known as cationic as well as anionic hydrolysis. The nature of the solution or pH depends upon the relative strengths of the acid and base that are formed on hydrolysis.

Equilibrium Important Extra Questions Numerical Problems

Question 1.
Calculate the pH of \(\frac{N}{1000}\) sodium hydroxide solution assuming complete ionisation (Kw = 1.0 × 10-14).
Answer:
Since NaOH is completely ionized
∴ [NaOH] = [OH] = 10-3 N = 10-3 M

Now [H2O+][OH] = Kw = 10-14 [Given]
∴ [H3O+] = \(\frac{\mathrm{K}_{w}}{\left[\mathrm{OH}^{-}\right]}=\frac{10^{-14}}{10^{-3}}\) = 10-11
pH = – log [H3O+] = – log 10-11 = 11.

Question 2.
Calculate the pH of a 0.01 N solution of acetic acid. Ka for acetic acid is 1.8 × 10-5 at 25°C.
Answer:
Equilibrium Class 11 Important Extra Questions Chemistry 13
Applying the law of chemical equilibrium
Ka = [CH3COO̅][H3O+]/[CH3COOH]
Equilibrium Class 11 Important Extra Questions Chemistry 14

Question 3.
Calculate the pH value of a solution of 0.1 M NH3 (Kb = 1.8 × 10-5). ,
Answer:
NH3 H2O ⇌ NH4+ + OH
Equilibrium Class 11 Important Extra Questions Chemistry 15

Question 4.
Equal volumes of solutions with pH = 4 and pH = 10 are mixed. Calculate the pH of the resulting solution?
Answer:
pH = 4
∴ [H3O+] = 10-4 M

pH = 10
∴ [H3O+] = 10-10 M
or [OH] = 10-4 M
Thus, they will exactly neutralize each other, and the pH of the resulting solution will be = 7.

Question 5.
An aqueous solution of 0.02 g of NaOH in 50 mL has been prepared. What is its pH and pOH.
Answer:
50 mL of NaOH contains 0.02 g of it
1000 mL of it contains = \(\frac{0.02}{50}\) × 1000 = 0.4 g

Strength of NaOH = 0.4 gL-1
Equilibrium Class 11 Important Extra Questions Chemistry 16

Question 6.
A reaction mixture containing N2 at 0.50 atm, H2 at 3.0 atm and NH3 at 0.50 atm is heated to 450°C, in which direction the reaction N2 (g) + 3H2 (g) ⇌ 2NH3 (g) will go if Kp = 4.28 × 10-5?
Answer:
Concentration quotient Q = \(\frac{p^{2} \mathrm{NH}_{3}}{p_{\mathrm{N}_{2}} \times p_{\mathrm{H}_{2}}^{3}}=\frac{(0.50)^{2}}{0.5 \times(3.0)^{3}}\) = 0.018

Kp = 4.28 × 10-5
As Q > > Kp reaction will go in the backward direction.

Question 7.
Under what pressure must an equimolar mixture of PCl3 and Cl2 be placed at 250°C in order to obtain PCl5 at 1 atm > Kp for dissociation of PCl5 = 1.78.
Answer:
Let partial pressure of PCl3 at eqbm. = p atm
Then partial pressure of Cl2 at eqbm. = p atm
Partial pressure of PCl5 at eqbm. = 1 atm

Then for PCl5 ⇌ PCl3 + Cl2
Equilibrium Class 11 Important Extra Questions Chemistry 17
Equilibrium Class 11 Important Extra Questions Chemistry 18
∴ Total initial pressure of PCl3 and Cl2 = 2.33 + 2.33 = 4.66 atm.

Question 8.
At 773 K, the equilibrium constant Kc for the reaction N2 (g) + 3H2 (g) ⇌ 2NH3 (g) is 6.02 × 10-2 L2 mol-2. Calculate the value of Kp at the same temperature.
Answer:
Equilibrium Class 11 Important Extra Questions Chemistry 19

Question 9.
Kp for the reaction N2 (g) + 3H2 (g) ⇌ 2NH3 (g) is 49 at a certain temperature. Calculate the value of Kp at the same
temperature for the reaction NH3 ⇌ \(\frac{1}{2}\)N2 + \(\frac{3}{2}\)H2 (g)
Answer:
Given for the reaction N2 (g) + 3H2 (g) ⇌ 2NH3 (g), Kp = 49
∴ for the reverse reaction 2NH3 (g) ⇌ N2 (g) + 3H2 (g)
Equilibrium Class 11 Important Extra Questions Chemistry 20

Question 10.
In the equilibrium CaCO3 (s) ⇌ CaO (s) + CO2 (g) at 1073 K, the pressure of CO2 is formed to be 2.5 × 104 Pa. What is the equilibrium constant for this reaction at 1073 K.
Answer:
With reference to the standard state pressure of 1 bar which is = 105 Pa
Equilibrium Class 11 Important Extra Questions Chemistry 21

Question 11.
Determine the concentration of CO2 which will be in equilibrium with 2.5 × 10-2, mol L-1 of CO at 100°C for the reaction FeO (s) + CO (g) ⇌ Fe (s) + CO2 (g); Kc = 5.0.
Answer:
Kc = \(\frac{\left[\mathrm{CO}_{2}\right]}{[\mathrm{CO}]}\)
i.e., 5 = \(\frac{\left[\mathrm{CO}_{2}\right]}{2.5 \times 10^{-2}}\)
or
[CO2] = 5 × 2.5 × 10-2 = 12.5 × 10-2 mol L-1

Question 12.
In a reaction between hydrogen and iodine, 6.34 moles of hydrogen and 4.02 moles of iodine are found to be in equilibrium with 42.85 moles of hydrogen iodide at 350°C. Calculate the equilibrium constant.
Answer:
Equilibrium Class 11 Important Extra Questions Chemistry 22

Question 13.
Prove that the pressure necessary to obtain 50% dissociation of PCl5 at 500 K is numerically equal to the three times the value of the equilibrium constant, Kp.
Answer:
Equilibrium Class 11 Important Extra Questions Chemistry 23
Total no. of moles = 1.5
If P is the total required pressure, then
Equilibrium Class 11 Important Extra Questions Chemistry 24

Question 14.
The equilibrium constant for the reaction A2 + B2 ⇌ 2AB is Kp. What will be the equilibrium constant for the reaction AB ⇌ \(\frac{1}{2}\)A2 + \(\frac{1}{2}\) B22? [A, B and AB are all gases] [WB JEE 2004]
Answer:
For A2 + B2 ⇌ 2AB, eqbm. constant = Kp
For the reverse reaction 2AB ⇌ A2 + B2 eqmb. constant = \(\frac{1}{\mathbf{K}_{p}}\)

On dividing by 2, AB ⇌ \(\frac{1}{2}\)A2 + \(\frac{1}{2}\)B2
eqbm. constant = \(\sqrt{\frac{1}{K_{p}}}\)

Question 15.
The concentration of hydronium ions in a cup of black coffee is 1.3 × 10-5 M. Find the pH of the coffee. Is this coffee acidic or alkaline?
Answer:
Here, given [H3O+] = 1.3 × 10-5 M
pH = – log [H3O+] = – log (1.3 × 10-5-5)
= 5 – log 1.3 = 5 – 0.1139 = 4.8861
As pH is less than 7, the black coffee is acidic.

Question 16.
A solution is found to contain 0.63 g of nitric acid in 100 mL of the solution. What is the pH of the solution.
Answer:
HNO3 → H+ + NO3 completely
∴ [HNO3] = [H+] as nitric acid is a strong acid.
Cone, of HNO3 = 0.63 g per 100 mL

Strength L-1 = 6.3
[HNO3] = \(\frac{6.3}{63}\) = 0.1 = 10-1 M
∴ [H+] = 10-1 M
∴ pH = – log 10-1 = 1.

Question 17.
The value of Kw is 9.55 × 10-14 at a certain temperature. Calculate the pH of water at this temperature.
Answer:
Kw = 9.55 × 10-14
Now for water [H3O+] = [OH]
∴ Kw = [H3O+] × [OH] = [H3O+]2
or
[H3O+] = \(\sqrt{\mathrm{K}_{w}}=\sqrt{9.55 \times 10^{-14}}\) = 3.09 × 10-7M
pH = – log (3.09 × 10-7) = – (- 7 – log 3.09) = 7 – log 3.09 = 7 – 0.49 = 6.51

Question 18.
Calculate the mass of HCl to be dissolved per litre of the solution so that its pH = 1.301.
pH = – log [H3O+]
[H3+O] = antulog (- 1.301) = antilog 2.699
= 5.0 × 10-2 × 36.5 gL-1 = 1.825 gL-1

Question 19.
Calculate the pH of 0 001 N H2SO4 solution.
Answer:
H2SO4 is completely ionised as
H2SO4 + 2H2O ⇌ 2H3O+ + SO42- as it is, a strong acid.
[H3O+] = 2[H2SO4] as one molecule of H2SO4 gives 2 H3O+ ions

But H2SO4 is given to be 0.001 N = 0.001 × 49 g L-1 [Eq. wt of H2SO4 = 49]
Equilibrium Class 11 Important Extra Questions Chemistry 25

Question 20.
What would be the pH of a solution obtained by mixing 100 mL of 0.1 NHCl and 9.9 mL of 1.0 N NaOH solution?
Answer:
100 mL of 0.1 N HCl = 100 × 0.1 = 10 millieq.
9.9 mL of 1 N NaOH = 9.9 × 1 = 9.9 millieq.
∴ HCl left unneutralized = 10 – 9.9 = 0.1 millieq
Volume of solution = 100 + 9.9 = 109.9 mL

Normality of resulting HCl solution
= \(\frac{0.1}{109.9}=\frac{0.1}{110}\) = 0.09 × 10-4 N

Molarity = 9.09 × 10-4 M = [H+] = [HCl]
pH = – log(9.09 × 10-4) = 3.05

Question 21.
Calculate the pH of 10-8 M HCl solution.
Answer:
From acid [H+] = 10-8 M as HCl is completely ionised
Now the concentration of [H+] = 10-7 M cannot be ignored as compared to [H+] = 10-8 M from HCl as HCl is very dilute solution.

∴ Total [H+] = 10-8 + 10-7 = 10-8(1 + 10)
= 11 × 10-8 M
pH = – log (11 × 10-8) = 8 – log 11
= 8 – 1.04 = 6.96.

Question 22.
Calculate the hydrolysis constant and degree of hydrolysis and pH of 0.10 M KCN solution at 25°C. Ka for HCN = 6.2 × 10-10.
Answer:
KCN is the salt of a weak acid HCN and strong base KOH
∴ Kh = Hydrolysis constant
Equilibrium Class 11 Important Extra Questions Chemistry 26
The hydrolysis reaction of KCN is
Equilibrium Class 11 Important Extra Questions Chemistry 27

Question 23.
Calculate the pH of 0.01 M solution of NH4CN. Given that the Ka for HCN = 6.2 × 10-10 and Kb for NH3 = 1.6 /10“5.
Answer:
Applying the formula
PH = 7+ [pKa – pKb] = 7 + [- log Ka + log Kb]
= 7 + \(\frac{1}{2}\)[- log (6.2 × 10-10) + log (1.6 × 10-5)]
= 7+ \(\frac{1}{2}\)[(10- 0.7924) + (5 – 0.2041)]
pH = 9,31

Question 24.
The solubility product of AgCl in water is 1.5 × 10-10. Calculate its solubility in 0.01 M NaCl solution.
Answer:
As NaCl dissociates completely
∴ [NaCl] = [Cl] = 0.01 M

If solubility of AgCl in 0.01 M is s mol L-1, then
[Ag+] = [Cl] = s mol L-1

∴ Total [Cl] = 0.01 + s ≈ 0.01 M
Ksp for AgCl = [Ag+][Cl- ] = s × 0.01 = 0.01 s

∴ 0.01 s = 1.5 × 10-10
s = 1.5 × 10-8 M.

Question 25.
Given that the solubility product of BaSO4 is 1 × 10-10. Will a precipitate form when
(i) Equal volumes of 2 × 10-3 M BaCl2 solution and 2 × 10-4 M NaSO4 solutions are mixed?
Answer:
BaCl2 ionizes completely in solution
Equilibrium Class 11 Important Extra Questions Chemistry 28
Ionic product of BaSO4 = [Ba++][SO4]
= 10-3 × 10-4 = 10-7

As Ionic product 10-7 is > Ksp 1 × 10-10
Hence a precipitate of BaSO4 will be formed.

(ii) Equal volumes of 2 × 10-8 M BaCl2 solution and × 2 × 10-3 M Na2SO4 solutions are mixed?
Answer:
Similarly [Ba++] = \(\frac{2 \times 10^{-8}}{2}\) = 10-8 M
[SO4] = \(\frac{2 \times 10^{-3}}{2}\) = 10-3

Ionic product of BaSO4 = 108 × 10 = 10-11
It is less than the solubility product Ksp × 1 × 10-10
Hence, no precipitate will be formed in this case.

NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules

NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules.

Question 1.
What are macromolecules? Give examples.
Solution:
Biomolecules i.e. chemical compounds found in living organisms are of two types. One, those which have molecular weights less than one thousand and are usually referred to as macromolecules or simply as biomolecules while those which are found in the acid-insoluble fraction are called macromolecules or as biomacromolecules.

The molecules in the insoluble fraction with the exception of lipids are polymeric substances. Then why do lipids, whose molecular weights do not exceed 800, come under acid-insoluble fractions i.e., macromolecular fractions?

Question 2.
Illustrate a glycosidic, peptide and a phospho-diester bond.
Solution:
(a) Glycosidic bond: It is a bond formed between two monosaccharide molecules in a polysaccharide. This bond is formed between two carbon atoms of two adjacent monosaccharides.

(b) Peptide bond: Amino acids are linked by a peptide bond which is between the carboxyl (- COOH) group of one amino acid and the amino (- NH2) group of the next amino acid which is formed by the dehydration process.

(c) Phosphodiester bond: This is the bond present between the phosphate and hydroxyl group of sugar which is called an ester bond. As this ester bond is present on either side, it is called a phosphodiester bond.

Question 3.
What is meant by the tertiary structure of proteins?
Solution:
Tertiary structure of protein : When the individual peptide chains of secondary structure of protein are further extensively coiled and folded into sphere-like shapes with the hydrogen bonds between the amino and carboxyl group and various other kinds of bonds cross-linking on-chain to another they form tertiary structure. The ability of proteins to carry out specific reactions is the result of their primary, secondary and tertiary structure.
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 1

protein molecular weight calculator. Molecular mass is the most fundamental characteristics for proteins and peptides.

Question 4.
Find and write down structures of 10 interesting small molecular weight biomolecules. Find if there is any industry which manufactures the compounds by isolation. Find out who are the buyers?
Solution:
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 2

NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 3
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 4
Fat is being manufactured by many companies in pharmaceuticals business as well as in food business. Vitamins come in many combination and are being used as supplementary medicines. Lactose is made by companies in manufacturing baby food. All of us are buyers of fat, protein and lactose.

Question 5.
Proteins have primary structures. If you are given a method to know which amino acid is at either of two termini (ends) of a protein, can you connect this information to purity or homogeneity of a protein?
Solution:
The primary structure of proteins is described as the type, number, and order of amino acids in the chain. A protein is imagined as a line whose left end represents the first and right end represents the last amino acid. But in fact, this is not so simple. Actually, the number of amino acids in between the two termini determines the purity or homogeneity of a protein.

Question 6.
Find out and make a list of proteins used as therapeutic agents. Find other applications of proteins (e.g., Cosmetics, etc.)
Solution:
Haemoglobin, Insulin, thyroxine, growth hormone, other hormones of the adenohypophysis, serum albumen, serum globulin, fibrinogen, etc. are used as the therapeutic agents. Proteins are also used for the synthesis of food supplements, film, paint, plastic, etc.

Question 7.
Explain the composition of triglyceride.
Solution:
Triglycerides are esters of three molecules of fatty acids and one molecule of glycerol.
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 5

Question 8.
Can you describe what happens when milk is converted into curd or yoghurt, from your understanding of proteins.
Solution:
Conversion of milk into curd is the digestion of milk protein casein. Semi digested milk is the curd. In the stomach, renin converts milk protein into paracasein which then reacts with Ca++ ion to form calcium paracaseinate which is called the curd or yoghurt.

Question 9.
Can you attempt building models of biomolecules using commercially available atomic models (Ball and Stick model)?
Solution:
Yes, the Three-dimensional structure of cellulose can be made using balls and sticks. Similarly, models of other bimolecular can be made
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 6

Question 10.
Attempt titrating an amino acid against a weak base and discover the number of dissociating (ionizable) functional groups in the amino acid.
Solution:
When an amino acid is titrated with weak base then its-COOH group also acts as weak acid. So it forms a salt with weak base then the pH of the resulting solution is near 7, so there is no sudden change. Number of dissociating functional groups are two, one is amino group (NH2) and another is carboxylic group ( – COOH). In the titration, amino acid acts as an indicator. Amino acids in solution acts as basic or acidic as situation demands. So these are also called amphipathic molecules.

Question 11.
Draw the structure of the amino acid, alanine.
Solution:
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 7

Question 12.
What are gums made of? Is fevicol different?
Solution:
Gums are categorized into secondary metabolites or biomolecules. Thousands of compounds one present in plant-fungal and microbial cells. They are derived from these things. But is different. Fevicol has not derived from paper written cells.

Question 13.
Find out a qualitative test for proteins, fats and oils, amino acid and test any fruit juice, saliva, sweat and urine for them.
Solution:
Qualitative Tests for proteins, amino acids, and fats:
Biuret Test: Biuret test for protein identifies the presence of protein by producing violet colour of solution. Biuret H2NCONHCONH2 reacts with copper ion in a basic solution and gives violet colour.
Liebermann-Burchard Test for cholesterol:
This is a mixture of acidic anhydride and sulphuric acid. This gives a green colour when mixed with cholesterol.
Grease Test for oil: Certain oils give a translucent stain on clothes. This tesi can be used to show presence of fat in vegetable oils. These tests can be performed to check presence of proteins and amino acids and fats in any of the fluid mentioned in the question.

Question 14.
Find out how much cellulose is made by all the plants in the biosphere and compare it with how much of paper is manufactured by man and hence what is the consumption of plant material by man annually. What a loss of vegetation?
Solution:
According to a 2006 report from the UN, forests store about 312 billion tons of carbon in their biomass alone. If you add to that the carbon in deadwood, litter, and forest soil, the figure increases to about 1.1 trillion tons! The UN assessment also shows that the destruction of forests adds almost 2.2 billion tons of carbon to the atmosphere each year, the equivalent of what the U.S. emits annually. Many climate experts believe that the preservation and restoration of forests offers one of the least expensive and best ways to fight against climate change.
Although it is difficult to get exact data about the quantum of cellulose produced by plants, but above information can give some idea. About 10% of cellulose is used in paper making. The percentage is less but wrong practice of cutting wood and re-plantation makes the problem complicated. Usually older trees are cut for large quantity of cellulose and re-plantation is limited to selected species of plants. Selected species disturb the biodiversity as it leads to monoculture.
Add to this the problem of effluents coming out of a paper factory and the problem further aggravates.

Question 15.
Describe the important properties of enzymes.
Solution:
Properties of enzymes

  • Enzyme catalysis hydrolysis of ester, ether, peptide, c-c, c-halids, or P-N bonds.
  • Enzymes catalysis removal of the group from the substrate by mechanisms other than hydrolysis of leaving double bonds.
  • Enzymes generally function in a narrow range of temperature and pH.
  • Activity declines both below and above optimum temperature and pH.
  • The higher the affinity of the enzyme for its substrate the greater is its catalytic activity.
  • The activity of an enzyme is also sensitive to the presence of specific chemicals that bind to the enzyme.
  • For eg: Inhibitors that shuts off enzyme activity and Co-factors that facilitate catalytic activity.
  • Enzymes retain their identity at the end of the reaction.

VERY SHORT ANSWER QUESTIONS

Question 1.
Which organic compound is commonly called animal starch?
Solution:
Glycogen

Question 2.
Name the biomolecules of life.
Solution:
Carbohydrates, Lipids, Proteins, Enzymes, and nucleic acids.

Question 3.
Name one basic amino acid.
Solution:
Lysine.

Question 4.
Name one heteropolysaccharide.
Solution:
Chitin

Question 5.
Name the biomolecules present in the acid-insoluble fraction.
Solution:
Protein, polysaccharide, nucleic acid, and lipids.

Question 6.
Name the bond formed between sugar molecules.
Solution:
Glycosidic bond.

Question 7.
Name three pyrimidines.
Solution:
Thymine, cytosine, and uracil

Question 8.
Which enzyme does catalyse covalent bonding between two molecules to form a large molecule?
Solution:
Ligases.

Question 9.
On reaction with iodine, starch turns blue-black, why?
Solution:
The appearance of blue colour with the addition of iodine is due to its reaction with amylose fraction of starch.

Question 10.
Which type of bonds are found in proteins and polysaccharides?
Solution:
Peptides bond in protein and glycosidic bonds in polysaccharides.

Question 11.
Name one neutral amino acid.
Solution:
Valine.

Question 12.
Where does histone occur?
Solution:
Chromosomes.

Question 13.
Name two different kinds of metabolism.
Solution:
Anabolism and catabolism.

SHORT ANSWER QUESTIONS

Question 1.
Which type of bonds are found in nucleic acids?
Solution:
Phosphodiester bond.

Question 2.
What are the monosaccharides present in DNA and RNA? (Chikmagalur 2004)
Solution:
Deoxyribose in DNA and Ribose in RNA.

Question 3.
What are fatty acids? Give two examples.
Solution:
Fatty acids are compounds which have a carboxyl group attached to an R-group, which could be a methyl (CH3), or ethyl (C2H5) group or a higher number of CH2 groups e.g., Linoleic acid, Palmitic acid.

Question 4.
What are co-enzymes? Give two examples.
Solution:
Coenzymes are the non-protein organic ^compounds bound to the apoenzyme in a conjugate enzyme, their association with the apoenzyme is only transient, e.g., Nicotinamide adenine dinucleotide (NAD). Flavin adenine dinucleotide (FAD), Nicotinamide adenine dinucleotide phosphate (NADP).

Question 5.
(i) What is meant by complementary base pairing?
(ii) What is the distance between two successive bases in a strand of DNA?
(iii) How many base pairs are present in one turn of the helix of a DNA strand?
Solution:
(i) Complementary base pairing is the type of
pairing in DNA, where a purine always pairs with a pyrimidine, i.e., adenine pairs with thymine (A=T) and guanine pairs with cytosine (G=C).
(ii) 0.34 nm or 34 A is the distance between two successive bases in the strand of DNA
(iii) 10 base pairs

Question 6.
Differentiate between DNA and RNA.
Solution:
The main differences between DNA add RNA are as following
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 8
Question 7.
What la a prosthetic group? Give an example.
Solution:
The non-protein part of a conjugated protein is called a prosthetic group. For example in a nucleoprotein (nucleic acid is the prosthetic group).

Question 8.
Differentiate between essential amino acids and non-essential amino acids.
Solution:
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 9

Question 9.
Differentiate between Structural Proteins and Functional Proteins.
Solution:
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 11

Question 10.
What is activation energy?
Solution:
Activation Energy: An energy barrier is required for the reactant molecules for their activation. So this energy with enzyme-substrate reaction is called Activation energy.
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 12

The activation energy is low for reactions with catalysts [enzymes] than those with Non enzymatic reactions.

Question 11.
What are the components of enzymes?
Solution:
Enzymes are made up of protein as well as non – protein parts. The protein part is called an apoenzyme and the non-protein part is a coenzyme. These two together are called a holoenzyme.

LONG ANSWER QUESTIONS

Question 1.
How many classes are enzymes divided into? Name all the classes.
Solution:
Enzymes are divided into 6 classes. Namely

  1. Oxidoreductases/dehydrogenases: Enzymes which catalyze oxidoreduction between two substrates
  2. Transferases: Enzymes catalyzing a transfer of group between a pair of substrates.
  3. Hydrolases: Enzymes catalyzing the hydrolysis of ester, ether, peptide, glycosidic, C-C-C-halide or P.N bonds.
  4. Lyases: Enzymes catalyze the removal of groups from – substrates by mechanisms other than hydrolysis leaving double bonds.
  5. Lyases: Enzymes catalyzing the interconversion of optical geometric or positional isomers.
  6. Ligases: Enzymes catalyzing the linking together of 2 compounds.

Question 2.
Distinguish between the primary, secondary, and tertiary structures of proteins.
Solution:
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 13

Question 3.
Explain the effect of the following factors on enzyme activity:
(i) Temperature
(ii) pH.
Solution:
Temperature: An enzyme is active within a narrow range of temperature. The temperature at which an enzyme shows its highest activity is called optimum temperature.

It generally corresponds to the body temperature of warm blood animals e.g., 37°C in human beings. Enzyme activity decreases above and below this temperature. Enzyme becomes inactive below minimum temperature and beyond maximum temperature.

Low temperature present inside cold storage prevents spoilage of food. High temperature destroys enzymes by causing their denaturation.
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 15

The relation between temperature and enzyme controlled reaction velocity

pH – Every enzyme has an optimum pH when it is most effective.

A rise or fall in pH reduces enzyme activity by changing the degree of ionisation of its side chains. A change in pH may also reverse the reaction.

Most of the intracellular enzymes function near-neutral pH with the exception of several digestive enzymes which work either in acidic range of pH or alkaline range of pH. pH for trypsin is 8.5.

Question 4.
Discuss the B-DNA helical structure with the help of a diagram.
Solution:

  • Watson & Crick suggested the double-helical structure of DNA in 1953.
  • The backbone of the DNA molecule is made up of deoxyribonucleotide units joined by a phosphodiester bond.
  • The DNA molecule consists of two chains wrapped around each other.
  • The two helical strands are bound to each other by Hydrogen Bonds.
  • Purines bind with pyrimidines A = T, C = G
  • The pairing is specific and the two chains are complementary.
  • One strand has the orientation 5’ → 3’ and other has 3’ → 5’.
  • Both polynucleotides strands remain separated with a 20A° distance.
  • The coiling is right-handed.

NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 16

Question 5.
What are different kinds of enzymes? Mention with enzyme examples.
Solution:
Enzymes with substrate bonds are broken and changed to different kinds as

  1. Oxidoreductases: eg Alcohol dehydrogenase, oxidation, Reduction occurs
  2. Transferases: transfer a particular group to another substrate, eg. transavninase
  3. Hydrolases: cleave their substrates by hydrolysis of a covalent bond e.g. Urease, amylase.
  4. Lyases: break the covalent bond eg. Deaminase
  5. Isomerase: by changing the bonds they make isomers. eg: Aldolase.
  6. Ligase: These bind two substrate molecules eg: DNA ligase, RNA ligase

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 9 Biomolecules, help you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 9 Biomolecules, drop a comment below and we will get back to you at the earliest.

Thermodynamics Class 11 Important Extra Questions Chemistry Chapter 6

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 6 Thermodynamics. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 6 Important Extra Questions Thermodynamics

Thermodynamics Important Extra Questions Very Short Answer Type

Question 1.
Under what conditions the heat evolved or absorbed is equal to the internal energy change?
Answer:
At constant volume.

Question 2.
What is the sign of AH for endothermic reactions and why?
Answer:
AH is positive as ΔH = Hp – Hr and Hr < Hp.

Question 3.
What is the relationship between the standard enthalpy of formation and the enthalpy of a compound?
Answer:
They are equal.

Question 4.
Why enthalpy of neutralization of HF is greater than 57.1 kJ mol-1?
Answer:
This is due to the high hydration energy of fluoride ions.

Question 5.
What are the specific heat capacity and molar heat capacity for water?
Answer:
Specific heat capacity for H2O = 4.18 JK-1 g-1
Molar heat capacity for H2O = 4.18 × 18 = 75.24 JK-1 mol-1.

Question 6.
Why enthalpy of neutralization is less if either the acid or the base or both are weak?
Answer:
A part of the heat is used up for dissociation of the weak acid or weak base or both.

Question 7.
What do you mean by a system?
Answer:
A specified part of the universe that is under thermodynamic observation is called a system.

Question 8.
Define a cyclic process.
Answer:
A process in which a system undergoes a series of changes and ultimately returns to the original state is called a cyclic process. For a cyclic process; ΔU = 0.

Question 9.
Why the entropy of a diamond is less than that of graphite?
Answer:
Diamond is more compact than graphite.

Question 10.
Under what conditions, ΔH of a process is equal to ΔU?
Answer:
At constant temperature and constant volume.

Question 11.
Is the enthalpy of neutralization of HCl is same as that of H2S04? If so, why?
Answer:
Yes. Because both are strong acids and ionized almost completely in aqueous solutions.

Question 12.
Which is a better fuel for the animal body: proteins or carbohydrates?
Answer:
Carbohydrates. They have high calorific value.

Question 13.
Is the experimental determination of enthalpy of formation of CH4 possible?
Answer:
No.

An online enthalpy calculator is specially designed to calculate exact amount of enthalpy generated in a thermodynamic system.

Question 14.
Can we calculate ΔH of every process from the bond energy data of reactants and products?
Answer:
No.

Question 15.
Define enthalpy of fusion.
Answer:
It is defined as the heat change when one mole of solid changes to its liquid form at its melting point. ‘

Question 16.
For the reaction NaCl (aq) + AgNO3 (aq) → AgCl(s) + NaNO3(aq), will ΔH be greater than, equal to or less than ΔE?
Answer:
ΔH will be ΔE.

Question 17.
Write the mathematical relationship between heat, internal energy, and work done on the system.
Answer:
ΔE = q + w.

Question 18.
What is the limitation of the I law of thermodynamics?
Answer:
It cannot tell us about the direction of the process

Question 19.
What is the relationship between ΔH and ΔE?
Answer:
ΔH = ΔE + PΔV = ΔE + ΔngRT.

Question 20.
State the I law of thermodynamics.
OR
State the Law of conservation of energy.
Answer:
The energy of an isolated system remains conserved.
OR
The energy can neither be created nor destroyed, though it can be converted from one form into another.

Question 21.
Which of the following is a state function?
(i) height of a hill
(ii) distance traveled in climbing the hill
(iii) energy consumed in climbing the hill.
Answer:
Energy consumed in climbing the hill.

Question 22.
What is the internal energy of one mole of a noble gas?
Answer:
U = \(\frac{3}{2}\)RT.

Question 23.
Which of the following are state functions?
(i) q
(ii) heat capacity
(iii) specific heat capacity
(iv) ΔH and ΔU.
Answer:
(iii) specific heat capacity and
(iv) ΔH and ΔU

Question 24.
Name the two most common modes by which a system and surroundings exchange their energy?
Answer:
Heat and work.

Question 25.
What will happen to internal energy if work is done by the system?
Answer:
The internal energy of the system will decrease.

Question 26.
How many times is the molar heat capacity greater than the specific heat capacity of water?
Answer:
18 times.

Question 27.
Neither q nor w is a state function but q + w is a state function. Why?
Answer:
q + w is equal to ΔU, which is a state function.

Question 28.
Name the state function Which remains constant during an isothermal change.
Answer:
Temperature.

Question 29.
What is the value of enthalpy of neutralization of a strong acid and a strong base?
Answer:
– 57.1 kJ/gm equivalent.

Question 30.
Out of 1 mole of H2O(g) and 1 mole of H2O(l) which one will have greater entropy?
Answer:
H2O(g) will have greater entropy.

Question 31.
When a substance is said to be in its standard state?
Answer:
When it is present at 298 K and under one atmospheric pressure.

Question 32.
What is the entropy of the formation of an element in its standard state?
Answer:
By convention, the enthalpies (heats) of the formation of all elements in their most stable form (standard state) is taken as zero.

Question 33.
What is the physical significance of free energy change of a system?
Answer:
The decrease in free energy (- ΔG) is a measure of useful work or network obtainable during the process at constant temp and pressure.

Question 34.
Why is it essential to mention the physical state of reactants and products in thermochemical reactions?
Answer:
It is because the physical state of reactants and products also contributes significantly to the value ΔU or ΔH.

Question 35.
Name two intensive and extensive properties of a system.
Answer:

  • Intensive properties: Viscosity, refractive index.
  • Extensive properties: Mass, volume, heat capacity, etc.

Question 36.
What is fuel value or calorific value?
Answer:
It is defined as the amount of heat released when 1 gm of fuel or food is burnt completely in air or oxygen.

Question 37.
Is the process of diffusion of gases enthalpy driven or entropy-driven?
Answer:
It is an entropy-driven process.

Question 38.
The value of ΔHsol of NaNO3 is positive, yet the dissolution is spontaneous, why?
Answer:
There is a large increase in entropy i.e., it is an entropy-driven process.

Question 39.
Write a thermochemical equation of enthalpy of combustion of methanol.
Answer:
2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(1); ΔH = – Q kJ
where Q kJ mol-1 is the enthalpy of combustion.

Question 40.
What is the enthalpy of the formation of Cl2?
Answer:
Enthalpy of formation of a homonuclear molecule like Cl2 is zero.

Question 41.
For a reaction also ΔH and ΔS are positive. What is the condition that this reaction occurs spontaneously?
Answer:
To make ΔG negative TΔS > ΔH.

Question 42.
Arrange the following fuels in order of increasing fuel efficiency; kerosene, diesel oil, wax, natural gas.
Answer:
Lower hydrocarbons have higher calorific value and thus are more efficient.
wax < diesel oil < kerosene < natural gas.

Question 43.
What is the sign of ΔS when N2 and H2 combine to form NH3?
Answer:
ΔS is negative as the number of molecules decreases.

Question 44.
How will compare the efficiency of the three given fuels?
Answer:
By comparing their calorific values. Larger the calorific value, the greater the fuel efficiency.

Question 45.
How is a non-spontaneous process made spontaneous?
Answer:
By continuously supplying energy to it from outside.

Question 46.
What is the expression for entropy change for a phase transition? ‘
Answer:
ΔS = \(\frac{q_{\mathrm{rev}}}{\mathrm{T}}\)

Question 47.
Give an example of an isolated system.
Answer:
Thermos.

Question 48.
Is the enthalpy of formation of SnCl2(s) the same as that of ZnCl2(s)?
Answer:
No. They are different.

Question 49.
Why we usually study enthalpy change and not internal energy change?
Answer:
Most of the processes including reactions are carried out in open vessels at constant pressure.

Question 50.
What is the enthalpy of one mole of a noble gas?
Answer:
H = U + PV = \(\frac{3}{2}\) RT + RT = \(\frac{5}{2}\) RT.

Question 51.
Which of the thermodynamic properties out of U, S, T, P, V, H, and G are intensive properties and why?
Answer:
T and P, because they depend only upon the nature of the substance.

Question 52.
What is the relationship between qp and qv?
Answer:
qp = qv + Δng RT, where Δn = np– nr (gaseous).

Question 53.
What is the Gibb’s Helmholtz equation?
Answer:
ΔG = ΔH – TΔS, where ΔG, ΔH, and ΔS are free energy change, enthalpy change, and entropy change respectively.

Question 54.
Comment on the bond energies of four C-H bonds present in CH4?
Answer:
The bond energies of 1st, 2nd, 3rd, and 4th C-H bonds are not equal and so average values are taken.

Question 55.
What is the main limitation of the first law of thermodynamics?
Answer:
It cannot predict the spontaneity of a process.

Question 56.
What is entropy?
Answer:
Entropy is a measure of the randomness/disorder of a system.

Question 57.
A reversible reaction has ΔG° negative for forwarding reaction. What will be the sign of ΔG° for the backward reaction?
Answer:
Negative.

Question 58.
What is the effect of increasing temperature on the entropy of a substance?
Answer:
It increases.

Question 59.
When is the entropy of a perfectly crystalline solid zero?
Answer:
At absolute zero (O.K).

Question 60.
What is an adiabatic process?
Answer:
In which no eat flow between the system and surroundings.

Thermodynamics Important Extra Questions Short Answer Type

Question 1.
Ice is lighter than water, but the entropy of ice is less than that of water. Explain.
Answer:
Water is the liquid form while ice is its solid form. Molecular motion in ice is restricted than in water, i.e., a disorder in ice is restricted than water, i.e., a disorder in ice is less than in water.

Question 2.
Define spontaneity or-feasibility of a process.
Answer:
Spontaneity or feasibility of a process means its inherent tendency to occur on its own in a particular direction under a given set of conditions.

Question 3.
Enthalpy of neutralization of CH3COOH and NaOH is 55.9 kJ. What is the value of ΔH for ionization of CH3COOH?
Answer:
The heat of neutralization of strong acid and strong base + ΔH of ionization of CH3COOH = Enthalpy of neutralization of CH3COOH and NaOH
∴ – 57.1 kJ + ΔH of ionisation of CH3COOH = – 55.9 kJ
∴ ΔH of ionisation of CH3COOH = (- 55.9 + 57.1) kJ
= 1.2 kJ.

Question 4.
When 1 gm of liquid naphthalene (C10H8) solidifies, 150 J of heat is evolved. What is the enthalpy of fusion of C10H8?
Answer:
ΔHsolidifcation = – 150 × 128 = – 19200 J = – 19.2 kJ
[∵ M.wt.of C10H8 = 128]

Question 5.
Why most of the exothermic processes (reactions) are spontaneous?
Answer:
ΔG = ΔH – TΔS; For exothermic reactions,
ΔH is -ve For a spontaneous process ΔG is to be -ve.

Thus decrease in enthalpy (- AH) contributes significantly to the driving force (To make ΔG negative).

Question 6.
What is meant by the term state function? Give examples.
Answer:
A state function is a thermodynamic property that depends upon the state of the system and is independent of the path followed to bring about the change. Internal energy change (ΔU), enthalpy change (ΔH) entropy change (ΔS), and free energy change (ΔG) are examples.

Question 7.
The enthalpy of combustion of sulfur is 297 kJ.
Write the thermochemical equation for the combustion of sulfur. What is the value of ΔfH of SO2?
Answer:
S(s) + O2(g) → SO2(g); ΔfH = – 297 kJ
∴ ΔH of SO2 = – 297 kJ mol-1

Question 8.
What would be the heat released when 0.35 mol of HC1 in solution is neutralized by 0.25 mol of NaOH solution?
Answer:
HCl and NaOH being strong acid and strong- base is completely ionization in dilute aqueous solutions. The net reaction is H + (0.25 mol) + OH (0.25. mol) → H2O (0.25 mol)

Now the heat of neutralization of 1 mole of a strong acid is – 57.1 kJ
∴ The heat released will be 57.1 × 0.25 kJ = 14.27 kJ.

Question 9.
Predict which of the following entropy increases/ decreases?
(i) A liquid crystallizes into a solid.
Answer:
After crystallization molecules attain an ordered state and therefore entropy decreases.

(ii) Temperature bf a crystalline solid is raised from 0 K to 115 K.
Answer:
When the temperature is raised, a disorder in molecules increases, and therefore entropy increases.

(iii) 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)
Answer:
The reactant is solid and hence has low entropy. Among the products there are two gases and be solid, therefore products represent a condition of higher entropy.

(iv) H2(g) → 2H(g).
Answer:
Here 2 moles of H atoms have higher entropy than one mole of the hydrogen molecule.

Question 10.
Discuss the effect of temperature on the spontaneity of reactions.
Answer:
Effect of temperature on the spontaneity of reactions:
Thermodynamics Class 11 Important Extra Questions Chemistry 1
The terms low temperature and high temperature are relative. For a particular reaction, the high temperature could even mean room temperature.

Question 11.
What is the most important condition for a process to be reversible in thermodynamics?
Answer:
The process should be carried out infinitesimally slowly or the driving force should be infinitesimally greater than the opposing force.

Question 12.
Why heat is not a state function?
Answer:
According to first law of Thermodynamics; ΔU = q + w or q – = ΔU – w. As ΔU is a state function, but w is not a state function, therefore q is also not a state function.

Question 13.
Why the absolute value of enthalpy cannot be determined?
Answer:
As H = U + PV
The absolute value of U – the internal energy cannot be determined as it depends upon various factors whose value, cannot be determined.
∴ The absolute value of H cannot be determined.

Question 14.
Which of the following is/are exothermic and which are endothermic?
(i) Ca(g) → Ca2+(g) + 2e
Answer:
Endothermic (Ionisation enthalpy is required)

(ii) O(g) + e → O(g) .
Answer:
Exothermic (first electron affinity-energy is released)

(iii) N2-(g) + e- → N3-(g).
Answer:
Endothermic (higher electron affinities are required).

Question 15.
Calculate Δr Gθ for the conversion of oxygen to ozone, \(\frac{3}{2}\)O2(g) → O3(g) at 298 K, if Kp for this conversion is 2.47 × 10-29
Answer:
Δr Gθ = – 2.303RT log Kp
and R = 8.314 JK-1 mol-1
∴ Δr Gθ = – 2.303 × 8.314 × 298 log 2.47 × 10-29
= 163000 j mol-1 = 163 kJ mol-1.

Question 16.
Find the value of the equilibrium constant for the following conversion reaction at 298 K.
Thermodynamics Class 11 Important Extra Questions Chemistry 2
Δr Gθ = – 13.6 kJ mol-1.
Answer:
r Gθ = 2.303 RT log K
Thermodynamics Class 11 Important Extra Questions Chemistry 3

Question 17.
At 60°C, N2O4 is 50% dissociated. Calculate Δr Gθ this temperature and at one atmospheric pressure.
Answer:
N2O4(g) ⇌ 2NO2(g)
If N2O4 is 50% dissociated, the mole fraction of both the substances are given by
Thermodynamics Class 11 Important Extra Questions Chemistry 4

Question 18.
Calculate the heat released when 0.5 moles of the nitric acid solution is mixed with 0.2 moles of potassium hydroxide solution.
Answer:
HNO3 is a strong acid and KOH is a strong base. Therefore, both are completely ionized in water. The net reaction is
Thermodynamics Class 11 Important Extra Questions Chemistry 5
∴ Heat evolved = – 57.1 × 0.2 = – 11.42 kJ.

Question 19.
Define Hess’s Law of Constant Heat Summation.
Answer:
The enthalpy change for a reaction remains the same whether it proceeds in one step or in series of steps all’ measurements being done under similar conditions of temperature.

This law is a corollary from I Law of Thermodynamics.

Question 20.
What are the applications of Hess’s Law of constant heat summation?
Answer:

  1. It helps to calculate the enthalpies of formation of those compounds which Cannot be determined experimentally.
  2. It helps to determine the enthalpy of allotropic transformations like C(graphite) → C (diamond).
  3. It helps to calculate the enthalpy of hydration.

Question 21.
Calculate the maximum work obtained when 0.75 mol of an ideal gas expands isothermally and reversible at 27°C from a volume of 15 L to 25 L.
Answer:
For an isothermal reversible expansion of an ideal gas
w = – nRT log \(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\) = – 2.303 nRT log \(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\)

Putting n = 0.75 mol; V1 = 15 L; V2 = 25 L, T = 27 + 273 = 300 K R = 8.314 JK-1 mol-1.
w = – 2.303 × 0.75 × 8.314 × 300 log \(\frac{25}{15}\)
= – 955.5 J.

Question 22.
What are heat capacities at constant volume and constant pressure? What is the relationship between them?
Answer:
Heat capacity at constant volume (Cv): Heat supplied to a system to raise its temperature through 1°C keeping the volume of the system constant is called its heat capacity at constant volume (Cv).

Heat capacity at constant pressure (Cp): Heat supplied to a system to raise its temperature through 1°C keeping the external pressure constant is called its heat capacity at constant pressure (Cp).

Relationship between Cp and Cv: Cp – Cv = R.

Question 23.
Explain what do you mean by a reversible process.
Answer:
A process or a change is said to be reversible if it can be reversed at any moment by an infinitesimal change. It proceeds infinitesimally slowly by a series of equilibrium states such that the system and surroundings are always in near equilibrium with each other.

Question 24.
Two liters of an ideal gas at a pressure of 10 atm expands isothermally into a vacuum until its. total volume is 10 liters. How much heat is absorbed and how much work is done in the expansion?
Answer:
We have q = – w: pext(10 – 2) = 0 × 8 = 0
No. work is done; No heat is absorbed.

Question 25.
Define (i) Molar enthalpy of fusion
Answer:
The enthalpy change that accompanies the melting of one mole of a solid substance at its melting point is called the articular enthalpy of fusion.

(ii) Molar enthalpy of vaporisation.
Answer:
The amount of heat required to vaporize one mole of a liquid at constant temperature and under standard pressure (1 bar) is called molar enthalpy of vaporization.

Question 26.
How will you arrive at the relationship qp = qv + ΔngRT?
Answer:
Enthalpy change ΔH = qp; where qp = heat change at constant pressure,
Internal energy change ΔU = qv; where qv = heat change at. constant volume.

Now ΔH = ΔU + PΔV
For ideal gases PV = nRT
∴ ΔH = ΔU + (PV2 – PV1)
= ΔU + P(V2 – V1) = ΔU + (n2RT – n1RT)
= ΔU + RT(n2 – n1) = ΔU + ΔngRT
or
qp = qv + ΔngRT

Question 27.
Define
(i) Specific heat capacity
Answer:
Specific heat capacity: It is the amount of heat required to raise the temperature of 1 gram of the substance through 1°C.

(ii) molar heat capacity.
Answer:
Molar heat capacity: It is the amount of heat required to raise the temperature of one mole of the substance through 1 °C.

Question 28.
Derive the relationship Cp – Cv = R.
Answer:
Thermodynamics Class 11 Important Extra Questions Chemistry 6

Question 28.
Derive the expression -ΔG = w non-expansion
Answer:
From I law of thermodynamics
Thermodynamics Class 11 Important Extra Questions Chemistry 7

Question 29.
How does the sign of G help in predicting the spontaneity/non-spontaneity of a process?
Answer:

  1. IF ΔG is negative, the process is spontaneous.
  2. If ΔG = O, the process does not occur and the system is in equilibrium.
  3. If ΔG is positive, the process does not proceed in the forward direction.

Question 30.
An exothermic reaction A B is spontaneous in the backward direction. What will be the sign of ΔS for the forward direction?
Answer:
The backward reaction will be endothermic. Thus, the energy factor opposes the backward reaction., As the backward reaction is spontaneous, the randomness factor must favor i.e., ΔS will be positive for the backward reaction or it will be negative for the forward direction.

Thermodynamics Important Extra Questions Long Answer Type

Question 1.
Define
(i) Standard enthalpy of formation.
Answer:
Standard enthalpy of formation: The heat change accompanying the formation of 1 mole df a substance from its elements in their most stable state of aggregation is called its standard enthalpy of formation.
H2(g) + \(\frac{1}{2}\)O2(g) H2O(l); Δf He = 285.8 kJ mol-1

(ii) Standard enthalpy of combustion
Answer:
Standard enthalpy of combustion: It is the heat change accompanying the complete combustion or burning of one mole of a substance in its standard state in excess of air or oxygen.
C4H10(g) + \(\frac{13}{2}\)O2(g) → 4CO2(g) + 5H2O(1); ΔHθ = – 2658.0 kJ mol-1.

(iii) Enthalpy of atomization
Answer:
Enthalpy of atomization: It is defined as the enthalpy change accompanying the breaking of one mole of a substance completely into its atoms in the gas phase.
H2(g) → 2H(g) ΔcHe = 435.0 kJ mol-1

(iv) Enthalpy of solution
Answer:
Enthalpy of solution: It is defined as the heat change when one mole of a substance dissolves in a specified amount of the solvent. The enthalpy of solution at infinite dilution is the enthalpy change observed on dissolving 2 moles of the substance in an infinite amount of the solvent.

(v) Lattice enthalpy
Answer:
Lattice Enthalpy: The lattice enthalpy of an ionic compound is the enthalpy change that occurs when one mole of an ionic compound dissociates into its ions in a gaseous state.
Thermodynamics Class 11 Important Extra Questions Chemistry 8

(vi) Thermochemical equation.
Answer:
Thermochemical Equation: A balanced chemical equation together with the value of its A^H is called a thermochemical equation.
Thermodynamics Class 11 Important Extra Questions Chemistry 9
The above equation describes the combustion of liquid ethanol. The negative sign indicates that tills are an exothermic reaction. We specify the physical state along with the allotropic state of the substance in a thermochemical equation.

Thermodynamics Important Extra Questions Numerical Problems

Question 1.
For the equilibrium PCl5(g) ⇌ PCl3(g) + Cl2(g) at 298 K, Kc = 1.8 × 10-7. What is ΔG° for the reaction? (R = 8.314 JK-1 mol-1).
Answer:
Thermodynamics Class 11 Important Extra Questions Chemistry 10

Question 2.
Calculate the equilibrium constant, K, for the following reaction at 400 K?
2NOCl(g) ⇌ 2NO(g) + Cl2(g)
Given that ΔrH° = 80.0 kJ mol-1 and ΔrS° = 120 JK-1 mol-1.
Answer:
Thermodynamics Class 11 Important Extra Questions Chemistry 11

Question 3.
Calculate the standard entropy change for the reaction X ⇌ Y if the value of ΔH° = 28.40 kJ and equilibrium constant is 1.8 × 10-7 at 298 K and ΔrG° = 38.484 kJ.
Answer:
Thermodynamics Class 11 Important Extra Questions Chemistry 12
= -33.8 JK-1 mol-1

Question 4.
Calculate the enthalpy of formation of methane, given that the enthalpies of combustion of methane, graphite, and hydrogen are 890.2 kJ, 393.4 kJ, and 285.7 kJ mol-1 respectively.
Answer:
Thermodynamics Class 11 Important Extra Questions Chemistry 13
Multiply equation (iii) by 2, add it to equation (ii) and subtract equation (i) from their sum
C + 2H2 → CH4 .
ΔH = – 393.4 + 2(-285.7) – (-890.2)
= – 74.6 kJ mol-1
Hence the heat of formation of methane (CH4) is
ΔfH = – 74.6 kJ mol-1.

Question 5.
CO is allowed to expand isothermally and reversibly from 10 m3 to 20 m3 at 300 K and work obtained is 4.75 KJ. Calculate the number of moles of carbon monoxide (CO). R = 8.314 JK-1 mol-1.
Answer:
Thermodynamics Class 11 Important Extra Questions Chemistry 14

Question 6.
Two moles of an ideal gas initially at 27°C and one atm pressure are compressed isothermally and reversible till the final pressure of the gas is 10 atm. Calculate q, w, and AU for the process.
Answer:
Here n = 2; p1 = 1 atm; P2 = 10 atm; T = 300 K
w = 2.303 nRT log \(\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}\)
= 2.303 × 2 × 8.314 JK mol-1 × 300 K × log \(\frac{10}{1}\)
= 11488 J .

For isothermal compression of ideal gas
ΔU = 0
Further, ΔU = q + w
∴ q = -w = 11488 J.

Question 7.
The heat of combustion of benzene in a bomb calorimeter (i.e. at constant volume) was found to be 3263.9 kJ mol-1 at 25°C. Calculate the heat of combustion of benzene at constant pressure.
Answer:
The given reaction is
C6H6(l) + \(\frac{15}{2}\) O2(g) → 6CO2(g) + 3H2O(l)
C6H6 is a liquid and H20 is a liquid at 25°C.
Thermodynamics Class 11 Important Extra Questions Chemistry 15

Question 8.
When 0.532 g of benzene (C6H6) with boiling point 353 K is burnt with an excess of O2 in a calorimeter, 22.3 kJ of heat is given out. Calculate ΔH for the combustion process (R = 8.31 JK-1 mol-1)
Answer:
Thermodynamics Class 11 Important Extra Questions Chemistry 16

Question 9.
Specific heat of an elementary gas is found to be 0.313 Jat constant volume. If the molar mass of the gas is 40 g mol-1, what is the atomicity of the gas? R = 8.31 JK-1 mol-1.
Answer:
Thermodynamics Class 11 Important Extra Questions Chemistry 17

Question 10.
Calculate the energy needed to raise the temperature of 10.0 g of iron from 25° C to 500°C if the specific heat of iron is 0.45 J (0°C)-1g-1.
Answer:
Energy needed (q) = m × C × ΔT
= 10.0 × 0.45 × (500 – 25) J
= 2137.5 J

Question 11.
Calculate the enthalpy of formation of carbon disulfide given that the enthalpy of combustion of it is 110.2 kJ mol-1 and those of sulfur and carbon are 297.4 kJ and 394.5 kJ/g atoms respectively.
Answer:
Our aim is C(s) + 2S(s) → CS2(l); ΔH =?
Given (i) CS2(l) + 3O2(g) → CO2(g) + 2SO2(g); ΔH = – 110.2 kJ mol-1
(ii) S(s) + O2(g) → SO2(g); ΔH = – 297.4 kj mol-1
(iii) C(s) + O2(g) → CO2(g); ΔH= – 394.5 kj mol-1

Add (iii) + 2(ii) and subtract (i), it gives, on rearranging
C(s) + 2S(s) → CS2(l);
ΔH = (- 394.5) +- 2(- 297.4) – (- 110.2)
= -879.1 kj mol-1

Thus the enthalpy of formation of CS2 = – 879.1 kJ mol-1

Question 12.
There are two crystalline forms of PbO; one is yellow and the other is red. The standard enthalpies of formation of these two forms are – 217.3 and – 219.0 kJ mol-1 respectively. Calculate the enthalpy change for the solid-solid phase transition:
PbO (yellow) → PbO(red)
Answer:
Our aim is PbO (yellow) → PbO (rpd); ΔH =?
Given (i) PbO(s) + \(\frac{1}{2}\)02(g) ) → PbO (yellow); ΔH = – 217.3 kJ mol-1
(ii) Pb(s) + \(\frac{1}{2}\)O2(g) ) → PbO(red); ΔH = —219.0 kJ mol-1

Subtracting (1) from (ii), we get
PbO (yellow) → PbO(red); ΔH =- 219.0 – (- 2173) = – 1.7 kJ mol-1.

Question 13.
The thermite reaction used for welding of metals involves the reaction 2Al (s) + Fe2O3(s) → Al2O3(s) + 2Fe(s)
What is Δr, H° at 25°C for this reaction? Given that the standard heats of formation of Al2O3 and Fe2O3 are – 1675.7 k J and – 828.4 kJ mol-1 respectively.
Answer:
Our aim is 2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s); ΔrH° =?
Thermodynamics Class 11 Important Extra Questions Chemistry 18

Question 14.
Calculate the bond enthalpy of HCI. Given that the bond enthalpies of H2 and Cl are 430 and 242 kJ mol-1 respectively and for HCI is -91 kJ mol-1.
Answer:
Thermodynamics Class 11 Important Extra Questions Chemistry 19

Question 15.
CaIculat the enthalpy of hydrogenation of C2H2(g) to C2H4(g). Given bond energies: C—H = 414.50 kJ mol-1; C≡C is 827.6 kJ mol-1, C=C is 606.0 kJ mol-1; H—H = 430.5 kJ mol-1.
Answer:
Thermodynamics Class 11 Important Extra Questions Chemistry 20
= [827.6 + 2 × 414.0 + 430.5] – [606.0 + 4 × 414.0]
= 175.9 kJ mol-1.

Question 16.
The entropy change for the vaporization of water is 109 JK-1 mol-1. Calculate the enthalpy change for the vaporization of water at 373 K.
Answer:
Δvap H = TΔvap S
= 373 × 109 = 40657 Jmol-1
= 40.657 kJ mol-1

Question 17.
Enthalpy and entropy changes of a reaction are 40.63 kJ mol-1 and 108.8 JK-1 mol-1 respectively. Predict the feasibility of the reaction at 27°C.
Answer:
ΔH = 40.63 kJ mol-1 = 40630 Jmol-1
ΔS = 108.8 JK-1 mol-1
T = 27°C =27 + 273 = 300 K
Now ΔG = ΔH – TΔS
= 40630 – 3 × 108.8 = 7990 J mol-1

Since AG comes out to be positive (i.e., ΔG > 0), the reaction is not feasible at 27°C in the forward direction.

Question 18.
At 0°C ice and water are in equilibrium and ×H = 6.0 kJ mol-1 for the process H2O(s) → H2O(l). What will be ΔS and
ΔG for the conversion of ice to liquid water.?
Answer:
Since the given process is in equilibrium
ΔG = 0
∴ from the equation ΔG = ΔH – TΔS
ΔH becomes = TΔS

∴ ΔS = \(\frac{6000 \mathrm{~J} \mathrm{~mol}^{-1}}{273 \mathrm{~K}}\) = 21.98 JK-1 mol-1.

Question 19.
Calculate the standard- free energy for the reaction
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)
Given that the standard free energies of formation (ΔfG°) for NH3(g), NO(g), H2O(l) are – 16.8, + 86.7, and – 237.2 kJ mol-1 respectively. Predict the feasibility of the above reaction at the standard state.
Answer:
Given ΔfG°(NH3) = – 16.8 kJ mol-1, ΔfG°(NO) = + 86.7 kj mol-1; ΔfG°(H2O) = – 237.2 kJ mol-1
Thermodynamics Class 11 Important Extra Questions Chemistry 21
= – [4 × ΔfG°(NH3) + 5 × ΔfG°(O2)]
= [4 × 86.7 + 6(- 237.2)] – [4 × (- 16.8) + 5 × 0]
= – 1009.2 kJ
Since ΔrG° is negative, the reaction is feasible.

Question 20.
The value of K for the water gas reaction
CO + H2O ⇌ CO2 + H2 is 1.06 × 105 at 25°C. Calculate the standard free energy change for the reaction at 25°C.
Answer:
ΔrG° = – 2.303 RT log K
= – 2.303 × 8.314 × 298 × log (1.06 × 105)
On solving, ΔrG° = – 28.38 KJ mol-1.

Question 21.
Calculate ΔrG° for the conversion of oxygen to ozone. \(\frac{3}{2}\)O2(g) → O3(g) at 298 K. If Kp for the reaction is 2.47 × 10-29
Answer:
ΔrG° = – 2.303 RT log K
R = 8.314 JK-1 mol-1
∴ ΔfG° = – 2.303 (8-314 JK-1 mol-1) × 298 × (log 2.47 × 10-29)

On solving
ΔrG° = 163000 J mol-1 = 163 kJ mol-1

Question 22.
Find out the value of the equilibrium constant for the following reaction at 298 K. Standard Gibbs energy change, ΔrG° at the given temperature is – 13.6 kJ mol-1.
Answer:
rG° = 2.303 RT log K
Thermodynamics Class 11 Important Extra Questions Chemistry 22

Question 23.
For oxidation of iron, 4Fe(s) + 3O2(g) → 2Fe2O3 (s) entropy change is – 549.4 JK-1 mol-1 at 298 K. Inspite .of negative entropy change of this reaction, why is the reaction spontaneous? (ΔrH° = -1648 × 103 J mol-1).
Answer:
The spontaneity of a reaction is determined from ΔStotal
(which is = Δsys + Δsutu)

Now ΔSsurr = – \(\frac{\Delta_{r} \mathrm{H}^{0}}{\mathrm{~T}}\) at constant pressure
= \(\frac{-1648 \times 10^{3}}{298 \mathrm{~K}}\) J mol-1 = 5530 JK-1 mol-1

Thus total entropy change for the reaction is
Δr Stotal = 5530 + (- 549.4) = 4980.6 JK-1 mol-1.
Since the total entropy change is positive, therefore, the above reaction is spontaneous.

Question 24.
A swimmer coming from a pool is covered with a film of water weighing about 18 g. How much heat must be supplied to evaporate this water at 298 K? Calculate the internal energy of vaporization at 100°C. AvapH° for water.
Answer:
Thermodynamics Class 11 Important Extra Questions Chemistry 23

No. of moles in 18 g H2O(l) = \(\frac{18 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 1 mol.
Δvap U = Δvap H° – pΔV = ΔvapH – Δng RT (assuming steam behaving as an ideal gas)
Δvap H° – Δng RT = 40.66 kJ mol-1 – (1)(8.314 JK-1 mol-1)(373 K) (10-3 kJ) ,
∴ Δvap U° = 40.66 kJ mol-1 – 3.10 kJ mol-1 = 37.56 kJ mol-1

Question 25.
1 g or graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atm pressure according to the reaction
C(graphite) + O2(g) → CO2(g). During the reaction, the temperature rises from 298 to 299 K. If the heat, the capacity of the bomb calorimeter is 20.7 kJ K-1 what is the enthalpy change for the above reaction at 298 K and 1. atm?
Answer:
q = Heat change = Cv × ΔT, where q is the heat absorbed by the calorimeter.
The quantity of heat from the reaction will have the same magnitude, but opposite sign, because heat lost by the system (reaction mixture) is equal to the heat gained by the calorimeter.
∴ q = – Cv × ΔT = – 20.7 kJ K-1 × (299 – 298) K
= -20.7kJ.

Here, negative sign indicates the exothermic nature of the reaction.
Thus, ΔU for the combustion of 1 g of graphite = – 20.7 kJ K-1 For combustion of 1 mol of graphite
= – \(\frac{12.0 \times(-20.7)}{1}\) = – 2.48 × 102 kJ mol-1

Since Δngg = 0
∴ ΔH = ΔU = – 2.48 × 102 kJ mol-1.

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry.

Question 1.
Calculate the molecular mass of the following :
(i) H2O (ii) CO2 (iii) CH4
Answer:
(i) Molecular mass of H2O : 2 × 1 + 1 × 16 = 18u
(ii) Molecular mass of CO2 : 1 × 12 + 2× 16 = 44 u
(iii) Molecular mass of CH4 : 12 + 4 × 1 = 16 u

The theoretical yield calculator will tell you how many grams of product each reagent can produce, if fully consumed with no byproducts.

Question 2.
Calculate the mass percent of different elements present in sodium sulphate (Na2SO4).
Answer:
Molecular mass of Na2SO4 = 2 × Atomic mass of Na + Atomic mass of S + 4 × Atomic mass of O
= 2 × 23 + 32 + 4 × 16 = 46 + 32 + 64 = 142 u.
The percentage of different elements present can be calculated as :
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 1

Question 3.
Determine the empirical formula of an oxide of iron which has 69-9% iron and 30-1% oxygen by mass.
Answer:
Step I. Calculation of simplest whole number ratios of the elements
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 2
The simplest whole number ratios of the different elements are : Fe : O : : 2 : 3
Step II. Writing the empirical formula of the compound.
The empirical formula of the compound = Fe2CO3.

Illustration about Set line Chemical formula Calculator, Electrical panel and Light bulb with concept of idea. Color circle button.

Question 4.
Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.
Answer:
The chemical equation for the combustion of carbon in dioxygen present in air is :
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 3
(i) When 1 mole of carbon is burnt in air
1 mole of carbon will form C02 = 1 mol = 44 g
(ii) When 1 mole of carbon is burnt in 16 g of dioxygen
For 1 mole of carbon, dioxygen required = 32 g = 1 mol
But the mass of dioxygen available = 16 g = 1/2 mol .
This means that dioxygen is in limited amount or-it is the limiting reactant.
Since dioxygen and carbon react in the same ratio, therefore mass of CO2 formed = \(\frac { 1 }{ 2 } \) mol = 22 g
(iii) When 2 moles of carbon are burnt in 16 g of dioxygen
For 2 moles of carbon, dioxygen required = 64 g = 2 mol
But the mass of dioxygen available = 16 g =  \(\frac { 1 }{ 2 } \) mol
This means that dioxygen is in limited amount or it is the limiting reactant.
∴ Mass of C02 formed =\(\frac { 1 }{ 2 } \) mol = 22 g

Question 5.
Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0-375 molar aqueous solution. Molar mass of sodium acetate is 82.0 g mol-1.
Answer:
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 4

Question 6.
Calculate the concentration of nitric acid in moles per liter in a sample which has a density of 1.41 g mL-1 and the mass percent of nitric acid in it being 69%.
Answer:
Mass percent 69 means that 69 g of HNo3 are dissolved in 100 g of the solution.
Mass of solution = 100 g Density of solution = 1.41 g mL-1
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 5

Question 7.
How much copper can be obtained from 100 g of copper sulphate (CuSO4)?
Answer:
The molecular mass of CuSO4 = Atomic mass of Cu + Atomic mass of S + 4 x Atomic mass of O
= 63.5 + 32 + 4 × 16
= 159.5 u
Gram molecular mass of CuSO4 = 159.5 g
Now, 159-5 g of CuSO4 have Cu = 63.5 g
∴ 100 g of CuSO4 have Cu = (63.5 g) × \(\frac { (100 g) }{ (159.5 g) } \) = 39.81 g

Question 8.
Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69.9 and 30.1 respectively.
Answer:
The empirical formula of the oxide of iron = Fe2O3
(For details, refer to No. 3)
Molecular formula of the oxide of iron = n × Empirical formula
= 1 × (Fe2O3)
= Fe2O3
(Since there is no common factor in Fe2O3, therefore n = 1).

Question 9.
Calculate the average atomic mass of chlorine from the following data: Isotope % Natural Abundance Atomic mass
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 6

Answer:
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 7

Question 10.
In three moles of ethane (C2H6), calculate the following:
(i) No. of moles of carbon atoms
(ii) No. of moles of hydrogen atoms No. of molecules of ethane.
Answer:
(i) 1 mole of C2H6 has moles of carbon atoms= 2 moles
3 moles of C2H6 have moles of carbon atoms= 2 × 3 = 6 moles
(ii) 1 mole of C2H6 has moles of hydrogen atoms = 6 moles
3 moles of C2H6 have moles of hydrogen atoms = 6 × 3 = 18 moles
(iii) 1 mole of C2H6 has molecules = 6.022 × 1023
3 moles of C2H6 have molecules = 6.022 × 1023 x 3 = 1.81 x 1024

Question 11.
What is the concentration of sugar (C12H22O11) in mol L-1 of it are dissolved in enough water to make final volume upto 2 L?
Answer:
The concentration in mol L-1 means molarity (M).
From the available data, it can be calculated as:
Mass of sugar = 20 g
Molar mass of sugar (C12H22o11) = 12 × 12 + 22 × 1 + 11 × 16 = 342 g mol-1
Volume of solution in litre = 2 L
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 8

Question 12.
If the density of methanol is 0.793 kg L-1, what is its volume needed for making 2.5 L of its 0.25 M solution?
Answer:
Step I. Calculation of the mass of methanol (CH3OH)
Molar mass of methanol (CH3OH) =12 + 4×1 + 16 = 32 g mol-1
Molarity of solution = 0.25 M = 0.25 mol L-1
Volume of solution = 2.5 L
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 9
Step II. Calculation of volume of methanol
Mass of methanol = 20 g = 0-002 kg
Density of methanol = 0-793 kg L-1
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 10

Question 13.
The pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below:
IPa = 1 Nm-2
If the mass of air at sea level is 1034 g, calculate the pressure in pascal.
Answer:
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 11

Question 14.
What is the SI unit of mass? How is it defined?
Answer:
Kilogram. It is equal to the mass of the prototype of the kilogram. It is in fact, the mass of a platinum block stored at the International Bureau of Weights and Measurements in France.

Question 15.
Match the following prefixes with their multiples
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 13
Answer:
After matching:
micro = 10-6
deca = 10
mega = 106
giga = 109
femto = 10-15

Question 16.
What do you understand by significant figures?
Answer:
We have seen that every measurement done in the laboratory involves the same error or uncertainty depending upon the limitation of the measuring instrument. In order to report scientific data, the term ‘significant figures’ has been used. According to this, all digits repotted in a given data are certain except the last one which is uncertain or doubtful. For example, let us suppose that the reading as reported by a measuring scale is 11-64. It has four digits in all. Out of the 1, 1, and 6 are certain digits while the last digit ‘4’ is uncertain. Thus, the number may be reported as follows :
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 14
Thus, the significant figures in any number are all certain digits plus one doubtful digit.
It may be noted all digits reported in a number are significant. However, only the last digit is uncertain while the rest are certain. Thus, the number 11-64 has all four digits as significant figures. Out of the 1, 1 and 6 are certain while 4 has some uncertainty about it.

Question 17.
A sample of drinking water was found to be severely contaminated with chloroform CHCI3, supposed to be a carcinogen. The level of contamination was 15 ppm (by mass)
(i) Express this in percent by mass.
(ii) Determine the molality of chloroform in the water sample.
Answer:
(i) Calculation of percent by mass
15 ppm level of contamination means that 15 parts or 15 g of chloroform (CHCI3) are present in 106 parts or 106 g of the sample i.e., water.
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 15
(ii) Calculation of molality of the solution
Mass of chloroform = 1.5 x 10-3 g
Molar mass of chloroform (CHCI3) = 12 + 1 + (3 × 35.5) = 119.5 g mol-1
Mass of sample i.e., water = 100 g
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 16

Question 18.
following in the scientific notation:
(i) 0.0048
(ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012

Answer:
(i) 4.8 x 10-3
(ii) 2.34 × 105
(iii) 8.008 × 103
(iv) 5.000 × 102
(v) 6.0012 × 100

Question 19.
How many significant figures are present in the following?
(i) 0.0025
(ii) 208
(iii) 5005
(iv) 126,000
(iv) 500.0.
(v) 2.0034

Answer:
(i) 2
(ii) 3
(iii) 4
(iv) 6
(v) 4
(vi) 5

Question 20.
Round up the following upto three significant figures :
(i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808

Answer:
(i) 34.2
(ii) 10.4
(iii) 0.0460
(iv) 281

Question 21a.
The following data is obtained when dinitrogen and dioxygen react together to form different compounds :
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 17

Answer:
By keeping 14 g as the fixed mass of dinitrogen (N2), the ratios by mass of dioxygen (O2) combining with 14 g dinitrogen are : 16 : 32 : 16 : 40 or 2 : 4 : 2 : 5. Since this ratio is simple whole number, the data obeys the Law Multiple Proportions.

Question 21b.
Fill in the blanks in the following conversions :
(i) 1 km = ……… mm = ……….. pm
(ii) 1 mg = …….. kg = …………. ng
(iii) 1 mL = ……. L = ………… dm3.
Answer:
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 18

Question 22.
If the speed of light is 3.0 x 108 m s-1, calculate the distance covered by light in 2.00 ns.
Answer:
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 20

Question 23.
Identify the limiting reactant if any in the following reaction mixtures ?
A + B → AB2
(i) 300 atoms of A + 200 molecules of B2
(ii) 100 atoms of A + 100 molecules of B2
(iii) 5 moles of A + 2.5 moles of B2
(iv) 2.5 moles of A + 5 moles of B2
(v) 2 moles of A + 3 moles of B2.
Answer:
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 21

In the light of the above information, let us find the limiting reactant if any in all the cases
(i) 1 atom of A will react with molecules of B2 =1
300 atoms of A will react with molecules of B2 = 300
But the molecules of B2 actually available = 200
∴B2 is the limiting reactant.

(ii) 1 atom of A will react with molecules of B2 = 1
100 atoms of A will react with molecules of B2 = 100
The molecules of B2 actually available = 100
∴There is no limiting reactant in this case.

(iii) 1 mole of A will react with moles of B2 =1
5 moles of A will react with moles of B2 =5
But the moles of B2 actually available = 2.5
∴B2 is the limiting reactant.

(iv) 1 mole of A will react with moles of B2 =1
2.5 moles of A will react with moles of B2 = 2.5
But moles of B2 actually available = 5
This shows that 5 moles of A can react whereas only 2.5 moles of A are actually available.
∴A is the limiting reactant.

(v) 1 mole of A will react with moles of B2 = 1
2 moles of A will react with moles of B = 2
But the moles of B2 actually available = 3
This shows that 3 moles of A can react whereas only 2 moles of A are actually available.
∴ A is the limiting reactant.

Question 24.
Nitrogen and hydrogen react to form ammonia according to the reaction
N2 (g) + 3H2 (g) →2NH3(g)
If 1000 g of H2 react with 2000 g of N2,
(i) will any of the two reactants remain unreacted ? If yes, which one and what would be its mass ?
(ii) Calculate the mass of ammonia (NH3) which will be formed.
Answer:
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 22

According to available data,
28 g of N2 require H2 = 6 g
2000 g of N2 require H2 = (6g) x \(\frac { 2000 g }{ 28 g }\)=428.6 g
But H2 actually available = 1000 g
This means that H2 is in excess and will remain unreacted.
(i) Mass of H2 that remains unreacted = 1000 – 428.6 = 571.4 g
(ii) Mass of NH3 formed may be calculated as follows :
6 g of H2 will form NH3 = 34 g
428.6 g H2 will form NH3 = (34g) x \(\frac { 428.6 g }{ 6.0 g }\) = 2428.8 g

Question 25.
How are 0.50 mol Na2Co3 and 0.50 M Na2Co3 different ?
Answer:
0.50 mol Na2CO3 represent concentration in moles.
0.50 M Na2CO3 represent concentration in moles/litre (molarity).

Question 26.
If 10 volumes of dihydrogen react with five volumes of dioxygen gas, how many volumes of water will be produced ?
Answer:
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 23

10 volumes of water vapours will be produced.

Question 27.
Convert the following into basic units
(i) 28.7 pm
(ii) 15.15 \xs
(iii) 25365 mg.

Answer:
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 24

Question 28.
Which of the following has largest number of atoms ?
(i) 1 g of Au
(ii) lg of Na
(iii) 1 g of Li
(iv) lg of Cl2

Answer:
(i) 197 g of Au have atoms = 6.022 x 1023
∴ l g of Au has atoms = 6.022 × 1023 × \(\frac { 1 g }{ 197g }\) = 3.06 x 1021
(ii) 23 g of Na have atoms = 6.022 × 1023
1 g of Na has atoms = 6 -022 × 1023 × \(\frac { 1 g }{ 23g }\) = 2.62 x 1022 atoms
(iii) 71 g of Cl2 have molecules = 6.022 × 1023
71 g of Cl2 have atoms = 2 × 6.022 x 1023
1 g of Cl2 has atoms = 2 × 6.022 × 1023 x \(\frac { 1 g }{ 71g }\) = 1.67 × 1022 atoms
Thus, 1 g of lithium (Li) has the largest number of atoms.

Question 29.
Calculate the molarity of a solution of ethanol in water in which mole fraction of ethanol is 0.04.
Answer:
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 25

2.31 moles of ethanol are dissolved in 1000 g (or 1000 mL) of water or 1000 mL of the solution. In this case, the volume of solution is considered to be the same as that of the solvent i.e., water. In other words, the solution is regarded as dilute solution,
∴Molarity of solution = 2.31 M

Question 30.
What will be mass of one 12C in g ?
Answer:
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 26

Question 31.
How many significant figures should be present in the answer of the following calculations ?
(i) \(\frac { 0.2856 x 298.15 x 0.112 }{ 0.5785 }\)
(ii) 5 x 5.364
(iii) 0.0125 + 0-7864 4- 0.0215.

Answer:
(i) The least precise figure (0.112) has 3 significant figures. Therefore, the answer should have three significant figures.
(ii) The second figure (5.364) has 4 significant figures. Therefore, the answer should be reported upto four significant figures. The exact figure (5) is not considered in this case.
(iii) In this case, the least precise figures (0.0125 and 0.0215) have 3 significant figures. Therefore, the answer should be reported upto three significant figures.

Question 32.
Use the data given in the following table to calculate the molar mass of naturally occurring argon.
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 27

Answer:
Molar mass of argon is the average molar mass and may be calculated as :
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 28

Question 33.
Calculate the number of atoms present in : (i) 52 moles of He (ii) 52 u of He (iii) 52 g of He.
Answer:
(i) 1 mole of He contains atoms = 6.022 × 1023
52 moles of He contain atoms = 6.022 x 1023 x 52 = 3-13 × 1025 atoms
(ii) Atomic mass of He = 4 u ; 4 u is the mass of He atoms = 1
52 u is the mass of He atoms = \(\frac { 1 }{ 4 }\) x 52 = 13 atoms
(iii) Gram atomic mass of He = 4 g ; 4 g of He contain atoms = 6.022 × 1023
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 29

Question 34.
A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g water and no other products. A volume of 10.0 L (measured at NTP) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula (ii) molar mass and (iii) molecular formula of the gas.
Answer:
Step I. Calculation of mass percent of carbon and hydrogen.
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 30

Step II. Determination of empirical formula of fuel gas.NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 31
Empirical formula of the fuel gas = CH.

Step III. Calculation of molecular mass of fuel gas.
10.0 L of the fuel gas at N.T.P. weigh = 11.6 g
22.4 L of the fuel gas at N.T.P. weigh =\(\frac { 11.6 }{ 10.0}\) × 24.4 =25.98 g
Molecular mass of the fuel gas = 25.98 g \(\approx \) 26.0 g = 26 u

Step IV. Calculation of molecular formula of the gas.
Empirical formula mass = 12 + 1 = 13 u
Molecular mass = 26 u
n = \(\frac { Molecular mass }{ Empirical formula mass }\) × \(\frac { 126 }{ 13}\) = 2
∴ Molecular formula = n × Empirical formula = 2 × CH = C2H2
The molecular formula of fuel gas is C2H2 and it is acetylene.

Question .35.
Calcium carbonate reacts with aqueous HCl to give CaCl2 and Co2 according to the reaction :
CaCo3 (s) + 2HCl (aq) + CaCl2 (aq) + Co2 (g) + H2o (l)
What mass of CaCo3 is required to react completely with 25 mL of 0.75 M HCl ?
Answer:

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 32

Question 36.
Chlorine is prepared in the laboratory by treating manganese dioxide (Mno2) with aqueous hydrochloric acid according to the reaction :
4HCl(aq) + MnO2(s) → MnCl2(aq) + Cl2(g) + 2H2O(l)
How many grams of HCl react with 5.0 g of manganese dioxide ?
Answer:
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry 33

We hope the NCERT Solutions for Class 11 Chemistry at Work Chapter 1 Some Basic Concepts of Chemistry, help you. If you have any query regarding NCERT Solutions for Class 11 Chemistry at Work Chapter 1 Some Basic Concepts of Chemistry, drop a comment below and we will get back to you at the earliest.

Kinetic Theory Class 11 Important Extra Questions Physics Chapter 13

Here we are providing Class 11 Physics Important Extra Questions and Answers Chapter 13 Kinetic Theory. Important Questions for Class 11 Physics with Answers are the best resource for students which helps in Class 11 board exams.

Class 11 Physics Chapter 13 Important Extra Questions Kinetic Theory

Kinetic Theory Important Extra Questions Very Short Answer Type

Question 1.
What does gas constant R signify? What is its value?
Answer:
The universal gas constant (R) signifies the work done by (or on) a gas per mole per kelvin. Its value is 8.31 J mol-1 K

Question 2.
What is the nature of the curve obtained when:
(a) Pressure versus reciprocal volume is plotted for an ideal gas at a constant temperature.
Answer:
It is a straight line.

(b) Volume of an ideal gas is plotted against its absolute temperature at constant pressure.
Answer:
It is a straight line.

Question 3.
The graph shows the variation of the product of PV with the pressure of the constant mass of three gases A, B and C. If all the changes are at a constant temperature, then which of the three gases is an ideal gas? Why?
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 1
Answer:
A is an ideal gas because PV is constant at constant temperature for an ideal gas.

Question 4.
On the basis of Charle’s law, what is the minimum possible temperature?
Answer:
– 273.15°C.

Question 5.
What would be the ratio of initial and final pressures if the masses of all the molecules of a gas are halved and their speeds are doubled?
Answer:
1: 2 (∵ P = \(\frac{1}{3} \frac{\mathrm{mn}}{\mathrm{V}}\)C2)

Question 6.
Water solidifies into ice at 273 K. What happens to the K.E. of water molecules?
Answer:
It is partly converted into the binding energy of ice.

Question 7.
Name three gas laws that can be obtained from the gas equation.
Answer:

  1. Boyle’s law
  2. Charle’s law
  3. Gay Lussac’s law.

Question 8.
What is the average velocity of the molecules of a gas in equilibrium?
Answer:
Zero.

An online Charles law calculator determines the value of initial temperature, final temperature, initial volume, final volume, pressure, ext…

Question 9.
A vessel is filled with a mixture of two different gases. Will the mean kinetic energies per molecule of both gases be equal? Why?
Answer:
Yes. This is because the mean K.E. per molecule i.e. \(\frac{3}{2}\) kT depends only upon the temperature.

Question 10.
Four molecules of a gas are having speeds, v1, v2, v3 and v4.
(a) What is their average speed?
Answer:
Vav = \(\frac{v_{1}+v_{2}+v_{3}+v_{4}}{4}\)

(b) What is the r.m.s. speed?
Answer:
Vrms = \(\sqrt{\frac{v_{1}^{2}+v_{2}^{2}+v_{3}^{2}+v_{4}^{2}}{4}}\)

Question 11.
The density of a gas is doubled, keeping all other factors unchanged. What will be the effect on the pressure of the gas?
Answer:
It will be doubled. (∵ P ∝ ρ if other factors are constant).

Question 12.
What is the average translational K.E. of an ideal gas molecule at a temperature T?
Answer:
\(\frac{3}{2}\) kT, where k is Boltzmann Constant.

Question 13.
Define the mean free path of a molecule.
Answer:
It is defined as the average distance travelled by a molecule between two successive collisions.

Question 14.
At what temperature, Charle’s law breaks down?
Answer:
At very low temperature, Charle’s law breaks down.

Question 15.
A container has an equal number of molecules of hydrogen and carbon dioxide. If a fine hole is made in the container, then which of the two gases shall leak out rapidly?
Answer:
Hydrogen would leak faster as r.m.s. speed of hydrogen is greater than the r.m.s. speed of CO2.

Question 16.
Two different gases have the same temperature. Can we conclude that the r.m.s? velocities of the gas molecules are also the same? Why?
Answer:
No. If temperature is same, then \(\frac{3}{2}\) kT is same. Also \(\frac{1}{2}\) mC2 is same. But m is different for different gases. C will be different.

Question 17.
A gas enclosed in a container is heated up. What is the effect on pressure?
Answer:
The pressure of the gas increases.

Question 18.
What is an ideal gas?
Answer:
It is a gas in which intermolecular forces are absent and it obeys gas laws.

Question 19.
Define absolute zero.
Answer:
It is defined as the temperature at which all molecular motions cease.

Question 20.
What do you understand by the term ‘Collision frequency’?
Answer:
It is the number of collisions suffered by a molecule in one second.

Question 21.
What do you understand by the term ‘mean free path’ of a molecule?
Answer:
It is the average distance travelled by the molecule between two successive collisions.

Question 22.
Mention two conditions when real gases obey the ideal gas equation PV = RT?
Answer:
Low pressure and high temperature.

Question 23.
Comment on the use of water as a coolant.
Answer:
Since water has a high value of specific heat so it can be used as a coolant.

Question 24.
Do Water and ice have the same specific heats? Why water bottles are used for fomentation?
Answer:
No. For water C = 1 cal g-1 °C-1 for ice, C = 0.5 cal g-1°C-1. It is because water has a high value of specific heat.

Question 25.
(a) What is the value of γ for a monoatomic and a diatomic gas?
Answer:
γ is 1.67 and 1.4 for monoatomic and diatomic gas respectively.

(b) Does the value of y depend upon the atomicity of the gas?
Answer:
Yes.

Question 26.
Which of the two has larger specific heat-monoatomic or diatomic gas at room temperature?
Answer:
Diatomic gas has more specific heat than a monoatomic gas e.g. Molar specific heat at constant volume is \(\frac{5}{2}\)R for monoatomic gas and \(\frac{5}{2}\)R for diatomic gas.

Question 27.
What is the effect on the pressure of an if it is compressed at constant temperature?
Answer:
Applying Boyle’s law, we find that the pressure increases.

Question 28.
What is the volume of a gas at absolute zero of temperature?
Answer:
Zero.

Question 29.
Why the pressure of a gas enclosed in a container increases on heating?
Answer:
This is because the pressure of a gas is proportional to the absolute ‘ temperature of the gas if V is constant.

Question 30.
The number of molecules in a container is doubled. What will be the effect on the r.m.s? speed of Slit molecules?
Answer:
No effect.

Question 31.
What will be the effect on K.E. and pressure of the gas in the above quation?
Answer:
Both will be doubled.

Question 32.
Does real gases obey the gas equation, PV = nRT.
Answer:
No.

Question 33.
On what factors does the internal energy of a real gas depend?
Answer:
It depends upon the temperature, pressure and volume of the gas.

Question 34.
What is the pressure of an ideal gas at absolute zero i.e. 0 K or – 273°C.
Answer:
Zero.

Question 35.
What do NTP and STP mean?
Answer:
They refer to a temperature of 273 K or 0°C and 1 atmospheric pressure.

Question 36.
What is the internal energy or molecular energy of an ideal gas at absolute zero?
Answer:
Zero.

Question 37.
Name the temperature at which all real gases get liquified?
Answer:
All real gases get liquified before reaching absolute zero.

Question 38.
Is the internal energy of the real gases sera at the absolute temperature?
Answer:
No.

Kinetic Theory Important Extra Questions Short Answer Type

Question 1.
Why cooling is caused by evaporation?
Answer:
Evaporation occurs on account of faster molecules escaping from the surface of the liquid. The liquid is therefore left with molecules having lower speeds. The decrease in the average speed of molecules results in lowering the temperature and hence cooling is caused.

Question 2.
On reducing the volume of the gas at a constant temperature, the pressure of the gas increases. Explain on the basis of the kinetic theory of gases.
Answer:
On reducing the volume, the space for the given number of molecules of the gas decreases i.e. no. of molecules per unit volume increases. As a result of which more molecules collide with the walls of the vessel per second and hence a larger momentum is transferred to the walls per second. Due to which the pressure of gas increases.

Question 3.
Why temperature less than absolute zero is not possible?
Answer:
According to the kinetic interpretation of temperature, absolute temperature means the kinetic energy of molecules.

As heat is taken out, the temperature falls and hence velocity decreases. At absolute zero, the velocity of the molecules becomes zero i.e. kinetic energy becomes zero. So no more decrease in K.E. is possible, hence temperature cannot fall further.

Question 4.
There are n molecules of a gas in a container. If the number of molecules is increased to 2n, what will be:
(a) the pressure of the gas.
(b) the total energy of the gas.
(c) r.m.s. speed of the gas molecules.
Answer:
(a) We know that
P = \(\frac{1}{3}\)mnC2.
where n = no. of molecules per unit volume.
Thus when no. of molecules is increased from n to 2n, no. of molecules per unit volume (n) will increase from n 2n
\(\frac{n}{V}\) to \(\frac{2n}{V}\), hence pressure will become double.

(b) The K.E. of a gas molecule is,
\(\frac{1}{2}\)mC2 = \(\frac{3}{2}\)kT
If the no. of molecules is increased from n to 2n. There is no effect on the average K.E. of a gas molecule, but the total energy is doubled.

r.m.s speed of gas is Crms = \(\sqrt{\frac{3 \mathrm{P}}{\rho}}=\sqrt{\frac{3 \mathrm{P}}{\mathrm{mn}}}\)

When n ¡s increased from n to 2n. both n and P become double and the ratio \(\frac{P}{n}\) remains unchanged. So there will be no effect of increasing the number of molecule from n to 2n on r.m.s. speed of gas molecule.

Question 5.
Equal masses of O2 and He gases are supplied equal amounts of heat. Which gas will undergo a greater temperature rise and why?
Answer:
Helium is monoatomic while O2 is diatomic. In the case of helium, the supplied heat has to increase only the translational K.E. of the gas molecules.

On the other hand, in the case of oxygen, the supplied heat has to increase the translations, vibrational and rotational K.E. of gas molecules. Thus helium would undergo a greater temperature rise.

Question 6.
Two bodies of specific heats S1 and S2 having the same heat capacities are combined to form a single composite body. What is the specific heat of the composite body?
Answer:
Let m1 and m2 be the masses of two bodies having heat capacities S1 and S respectively.
∴ (m1 + m2)S = m1S1 + m2S2 = m1S1 + m1S1 = 2m1S1

S = \(\frac{2 m_{1} S_{1}}{m_{1}+m_{2}}\).

Also, m2S2 = m1S1
( ∵ Heat capacities of two bodies are same.)
or
m2 = \(\frac{\mathrm{m}_{1} \mathrm{~S}_{1}}{\mathrm{~S}_{2}}\)

∴ S = \(\frac{2 \mathrm{~m}_{1} \mathrm{~S}_{1}}{\mathrm{~m}_{1}+\frac{\mathrm{m}_{1} \mathrm{~S}_{1}}{\mathrm{~S}_{2}}}=\frac{2 \mathrm{~S}_{1} \mathrm{~S}_{2}}{\mathrm{~S}_{1}+\mathrm{S}_{2}}\)

Question 7.
Tell the degree of freedom of:
(a) Monoatomic gas moles.
Answer:
A monoatomic gas possesses 3 translational degrees of freedom for each molecule.

(b) Diatomic gas moles.
Answer:
A diatomic gas molecule has 5 degrees of freedom including 3 translational and 2 rotational degrees of freedom.

(c) Polyatomic gas moles.
Answer:
The polyatomic gas molecule has 6 degrees of freedom (3 translational and 3 rotational).

Question 8.
State law of equipartition of energy.
Answer:
It states that in equilibrium, the total energy of the system is divided equally in all possible energy modes with each mode i.e. degree of freedom having an average energy equal to \(\frac{1}{2}\) kBT.

Question 9.
Explain why it is not possible to increase the temperature of gas while keeping its volume and pressure constant?
Answer:
It is not possible to increase the temperature of a gas keeping volume and pressure constant can be explained as follows:

According to the Kinetic Theory of gases,
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 2
( ∵ C2 = kT, when k is a constant)
T ∝ PV

Now as T is directly proportional to th^ product of P and V. If P and V are constant, then T is also constant.

Question 10.
A glass of water is stirred and then allowed to stand until the water stops moving. What has happened to the K.E. of the moving water?
Answer:
The K.E. of moving water is dissipated into internal energy. The temperature of water thus increases.

Question 11.
Why the pressure of a gas increases when it is heated up?
Answer:
This is due to the two reasons:

  1. The gas molecules move faster than before on heating and so strike the container walls more often.
  2. Each impact yields greater momentum to the walls.

Question 12.
R.m.s. velocities of gas molecules are comparable to those of a single bullet, yet a gas takes several seconds to diffuse through a room. Explain why?
Answer:
Gas molecules collide with one another with a very high frequency. Therefore, a molecule moves along a random and long path to go from one point to another. Hence gas takes several seconds to go from one comer of the room to the other.

Question 13.
Calculate the value of the universal gas constant (R).
Answer:
We know that R is given by
R = \(\frac{PV}{T}\)

Now one mole of all gases at S.T.P. occupy 22.4 litrês.
P = 0.76 m of Hg
= 0.76 × 13.6 × 103 × 9.8
= 1.013 × 105 Nm-2

V = 22.4 litre
= 22.4 × 10 3m3

T = 273 K .
n = 1

∴ R = \(\frac{1.013 \times 10^{5} \times 22.4}{273}\) × 10-3
= 8.31 J mol-1 k-1

Question 14.
Define and derive an expression for the mean free path.
Answer:
It is defined as the average distance travelled by a gas molecule between two successive collisions. It is denoted by X.

Derivation of Expression – Let us assume that only one molecule is in motion and all other molecules are at rest. ,
Let d = diameter of each molecule.
l = distance travelled by the moving molecule.

The moving molecule will collide with all those molecules whose centres lie inside a volume πd2l.
Let n = no. of molecules per unit volume in the gas.
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 3
Now λ = \(\frac{\text { distance travelled }}{\text { no. of collisions }}=\frac{l}{\mathrm{n} \pi \mathrm{d}^{2} l}\)
or
λ = \(\frac{1}{n \pi d^{2}}\) …(1)

In this derivation, we have assumed that all but one molecules are at rest. But this assumption is not correct. All the molecules are in random motion. So the chances of a collision by a molecule are greater.

Thus taking it into account, the mean free path can be shown to be \(\sqrt{2}\) times less than that in equation (1),
∴ λ = \(\frac{1}{\sqrt{2} n \pi d^{2}}\)
which is the required expression.

Question 15.
On what parameters does the λ (mean free path) depends?
Answer:
we know that λ = \(\frac{\mathrm{k} \mathrm{T}}{\sqrt{2} \pi \mathrm{d}^{2} \rho}=\frac{\mathrm{m}}{\sqrt{2} \pi \mathrm{d}^{2} \rho}\)

= \(\frac{1}{\sqrt{2} \pi \mathrm{nd}^{2}}\)
∴ λ depends upon:

  1. diameter (d) of the molecule, smaller the ‘d’, larger is the mean free path λ,.
  2. λ ∝ T i.e. higher the temperature, larger is the λ.
  3. λ ∝ \(\frac{1}{P}\) i.e. smaller the pressure, larger is the λ.
  4. λ ∝ \(\frac{1}{ρ}\) i.e. smaller the density (ρ), larger will he the X.
  5. λ ∝ \(\frac{1}{n}\) i.e. smaller the number of molecules per unit volume of the gas, larger is the λ.

Question 16.
What causes the Maxwellian distribution of molecular speed?
Answer:
Maxwellian distribution of molecular speed is a statistical phenomenon due to the intermolecular collisions in which the system tends to acquire equilibrium.

Question 17.
Why the pressure of a gas increases on increasing the temperature at constant volume?
Answer:
We know that Crms ∝ \(\sqrt{T}\).

Thus when T is increased, the root means square velocity of gas molecules also increases, thus they move faster and the number of collisions per second with the walls of the container increase and thus pressure increases.

Question 18.
Explain why the temperature of a gas rises when it is compressed?
Answer:
The work is done against pressure during the compression and the velocity of the individual molecules increases, so their K..E. is increased and thus the temperature of the gas is increased.

Question 19.
What determines the average speeds of the molecules of the gases?
Answer:
The average speed of the molecules of the gases depends upon their mass and temperature.

Question 20.
What is the average velocity of the molecules of an ideal gas? Why?
Answer:
The average velocity of the molecules of an ideal gas is zero because the molecules possess all sorts of velocities in all possible directions. Thus their vector sum and hence average value is zero.

Question 21.
A person putting on wet clothes may catch cold-Why?
Answer:
The water in the clothes evaporates. The heat required for evaporation is taken from the body of the person wearing wet clothes. So due to the cooling of the body, he may catch a cold.

Question 22.
At what temperature the molecular speed of the gas molecules should reduce to zero? Does it really happen? Why?
Answer:
The molecular speed should reduce to zero at absolute zero. But such a situation never arises because all gases liquefy before that temperature is attained.

Question 23.
The temperature of the gas in Kelvin is made 9 times. How does it affect the total K.E., average K.E., r.m.s? velocity and pressure?
Answer:
Total K.E., average K.E. and pressure become 9 times, but the RMS velocity is tripled.

Question 24.
Do diatomic molecules have all types of motions? Explain.
Answer:
No. At very low temperature, they have only translatory r motion. At moderate temperature, they possess both translatory and rotatory motion and at very high temperature, all three types of motions are possible.

Question 25.
Why the molecules of an ideal monoatomic gas have only three degrees of freedom?
Answer:
It is so because the molecules of an ideal gas are point masses, so rotational motions are not significant. Thus it can have only three degrees of freedom corresponding to the translatory motion.

Question 26.
The adiabatic expansion causes a lowering of the temperature of the gas. Why?
Answer:
As the gas expands work needs to be done by the gas. The process being adiabatic, no heat is absorbed by the gas from outside, so the energy for doing work is obtained from the gas itself and hence its temperature falls.

Question 27.
What are the different ways of increasing the number of molecular collisions per unit time against the walls of the vessel containing a gas?
Answer:
The number of collisions per unit time .an be increased in the following ways:

  1. By increasing the temperature of the gas.
  2. By increasing the number of molecules.
  3. By decreasing the volume of the gas,

Question 28.
Two identical cylinders contain helium at 2 atmospheres and argon at 1 atmosphere respectively. If both the gases are filled in one of the cylinders, then:
(a) What would be the pressure?
Answer:
(2 + 1) = 3 atmosphere.

(b) Will the average translational K.E. per molecule of both gases be equal?
Answer:
Yes, because the average translational K.E./molecule (\(\frac{3}{2}\)kT) depends only upon the temperature.

(c) Will the r.m.s. velocities are different?
Answer:
Yes, because of the r.m.s. velocity depends not only upon temperature but also upon the mass.

Question 29.
Why hydrogen escapes more rapidly than oxygen from the earth’s surface?
Answer:
We know that Crms ∝ \(\frac{1}{\sqrt{\rho}}\)

Also ρ0 = 16 ρH. So Crms of hydrogen is four times that of oxygen at a given temperature. So the number of hydrogen molecules whose velocity exceeds the escape velocity from earth (11.2 km s-1) is greater than the no. of oxygen molecules. Thus hydrogen escapes from the earth’s surface more rapidly than oxygen.

Question 30.
When a gas is heated, its molecules move apart. Does it increase the P.E. or K.E. of the molecules? Explain.
Answer:
It increases the K.E. of the molecules. Because on heating, the temperature increases and hence the average velocity of the molecules also increases which increases the K.E.

Question 31.
Distinguish between the terms evaporation, boiling and vaporisation.
Answer:
Evaporation: It is defined as the process of conversion of the liquid to a vapour state at all temperatures and occurs only at the surface of the liquid.

Boiling: It is the process of rapid conversion of the liquid to a vapour state at a definite temperature and occurs throughout the liquid.

Vaporisation: It is the general term for the conversion of liquid to vapour state. It includes both evaporation and boiling.

Kinetic Theory Important Extra Questions Long Answer Type

Question 1.
Derive gas laws from the kinetic theory of gases.
Answer:
(a) Boyle’s law: It states that P ∝ \(\frac{1}{V}\) if T = constant.
Derivation: We know from the kinetic theory of gases that
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 4
Here R = constant
If T = constant, then PV = constant
or
P ∝ \(\frac{1}{V}\).

(b) Charles’ law: It states that for a given mass of a gas, the volume of the gas is directly proportional to the absolute temperature of the gas if pressure is constant
i. e. V ∝ T.

Derivation: We know that
PV = \(\frac{1}{3}\)MC2= \(\frac{1}{3}\)mNC2
where N = Avogadro’s no.

Also, we know that mean K..E. of a molecule is
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 5
If P = constant, then V ∝ T. Hence proved.

(c) Avogadro’s Hypothesis: It states that equal volumes of all gases contain equal no. molecules if T and P are the same.

Derivation: Consider two gases A and B having n, and n2 as the no. of molecules, C1 and C2 are the r.m.s. velocities of these molecules respectively.

According to the kinetic theory of gases,
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 6
Also, we know that
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 7
Hence proved.

(d) Graham’s law of diffusion of gases: It states that the rate of diffusion of a gas is inversely proportional to the square root of the density of the gas.
Derivation: We know that
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 8
Also, we know that r.m.s. velocity is directly proportional to the rate of diffusion (r) of the gas, i.e.
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 9

Numerical Problems:

Question 1.
Calculate r.m.s. the velocity of hydrogen at N.T.P. Given the density of hydrogen = 0.09 kg m4.
Answer:
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 10

Question 2.
Calculate the temperature at which r.m.s. the velocity of the gas molecule is double its value at 27°C, the pressure of the gas remaining the same.
Answer:
Let t be the required temperature = ? and Ct, C27 be the r.m.s. velocities of the gas molecules at t°C and 27°C respectively.
\(\frac{\mathrm{C}_{\mathrm{t}}}{\mathrm{C}_{27}}\) = 2 (given)
Also let M = molecular weight of the gas
Now T = t + 273
and T27 = 27 + 273 = 300 K

∴ Using the relation
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 11
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 12

Question 3.
Calculate the K.E./mole of a gas at N.T.P. Density of gas at N.T.P. = 0.178 g dm-3 and molecular weight = 4.
Answer:
Here, ρ = 0.178 g dm-3
= 0.178 × 10-3 g cm-3 (∵ 1 dm3 = 10-3 cm3)
= 178 × 10-6 g cm-3

Volume of 1 mole of gas i.e. 4 g of gas = \(\frac{\text { Mass }}{\text { Density }}\)
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 13

Question 4.
Calculate the diameter of a molecule if n = 2.79 × 1025 molecules per m3 and mean free path = 2.2 × 10-8 m.
Answer:
Here, n = 2.79 × 1025 molecules m-3
λ = 2.2 × 10-8 m
d = ?

Using the relation.
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 14

Question 5.
Calculate the number of molecules in 1 cm3 of a perfect gas at 27°C and at a pressure of 10 mm ofHg. Mean K.E. of a molecule at 27°C = 4 × 1025 J. ρHg = 13.6 × 103 kg m-3.
Answer:
Here, K..E. per molecule at 27°C = 4 × 10-11 J
Let μ = number of molecules in 1 cm3 or 10-6 m3
∴ Mean K.E. per cm3 = μ × 4 × 1011 J ….(i)
Now K.E. per gram molecule = \(\frac{3}{2}\) RT
for a perfect gas, PV = RT

∴ K.E, per gram molecule = \(\frac{3}{2}\) PV
or
K.E. per cm3 of gas = \(\frac{3}{2}\) PV
P = 10 mm of Hg = 10-2 m of Hg
= 10-2 × 13.6 × 103 × 9.8
= 136 × 9.8 Nm-2 V
= 1 cm3
= 10-6 m3

∴ K.E per cm3 of gas = \(\frac{3}{2}\) × 136 × 9.8 × 106
= 1.969 × 10-3 J ….(ii)

∴ from (i) and (ii) we get
μ × 4 × 10-11 = 1.969 × 10-3
or
μ = \(\frac{1.969 \times 10^{-3}}{4 \times 10^{11}}\)
= 4.92 × 107 molecules

Question 6.
Gas at 27°C ¡n a cylinder has a volume of 4 litres and pressure 1oo Nm2.
(a) Gas is first compressed at a constant temperature so that the pressure is 150 Nm2. Calculate the change in volume.
Answer:
Here, V1 = 4 litres = 4 × 10-3 m3
P1 = 100 Nm-2
P2 = 150 Nm-2
V2 =?
T1 = 273 + 27 = 300 K

∴ According to Boyle’s Law
P1V1 = P2V2
or
V2 = \(\frac{P_{1} V_{1}}{P_{2}}=\frac{100 \times 4}{150}\) × 10-3
= 2.667 × 10-3 m3
= 2.667 litre.

∴ Change in volume = V1 – V2
= 4 – 2.667 = 1.333 litre.

(b) It is then heated at a constant volume so that temperature becomes 127°C. Calculate the new pressure.
Answer:
T1 = 300 K
T2 = 273 + 127 = 400 K
P1 = 150 Nm-2
P2 = ?

At constant volume, according to Gay Lussac’s law
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 15

Question 7.
The temperature of a gas is – 68°C. To what temperature should it be heated so that
(a) the average K.E. of the molecules be doubled.
(b) the root mean square velocity of the molecules to be doubled?
Answer:
(a) Let θ°C be the temperature of gas up to which it is heated.
∴ T2 = 273 + θ.
and T1 = 273 + (- 68)
Let E2 and E1 be the average K.E. of the molecules at T2 and T1 respectively.

According to the given condition
\(\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}\) = 2

Using the relation,
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 16

(b) Let t°C be the temperature up to which the gas is heated.
∴ T3 = 273 + t
and T4 = 273 – 68 = 205 K

Let C1 and C2 be the respective r.m.s. velocities of the molecules.
∴ \(\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}\) = 2 (given)

Now using the relation
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 17

Question 8.
A balloon contains 500 m3 of He at 27°C and I atm pressure. Find the volume of the helium at 3°C and 0.5 atm pressure?
Answer:
Here, P1 = 1 atm
T1 = 27°C = 273 + 27 = 300 K
V1 = 500 m3
P2 = 0.5 atm
V2 = ?
T2 = – 3°C = 273 – 3 = 270

Using ideal gas equation,
PV = RT, we get
P1V1 = RT1 …(i)
and P2V2 = RT2 …(ii)

Dividing (i) by (ii), we get
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 18

Question 9.
At what temperature will the average velocity of O2 molecules be sufficient so as to escape from earth? Ve = 11.0 km s-1 and mass of one molecule of O2 is 5.34 × 10-26 kg, k = 1.38 × 10-23 JK-1.
Answer:
Here, escape velocity from earth surface,
Ve = 11.0 kms-1
= 11 × 10-3 ms-1
k = 1.38 × 10-23 JK-1

Mass of one O2 molecule, M = 5.34 × 10-26 kg
T = ?
We know that K.E. per molecule is given by
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 19

Question 10.
The volume of the air bubble increases 15 times when it rises from the bottom to the top of a lake. Calculate the depth of the lake if the density of lake water is 1.02 × 103 kg m-3 and atmospheric pressure is 75 cm of Hg.
Answer:
Here, P1 = 75 cm of Hg
= 0.75 × 13.6 × 103 × 9.8 Nm-2
= 99.96 × 103 Nm-2

Let V2 = volume of bubble at depth h = x
i.e. V1 = x + 15x = 16x
P2 = 0.75 m of Hg = hρwater g
= 99.96 × 103 + h × 103 × 9.8

Using Byole’s law,
P1V1 = P2V2, we get
99.96 × 103 × 16x = (99.96 × 103 + h × 103 × 9.8)x
or
h = \(\frac{15 \times 99.96 \times 10^{3}}{9.8 \times 10^{3}}\) = 153 m.

Question 11.
Two glass bulbs of volumes 500 cm3 and 100 cm3 are connected by a narrow tube of negligible volume. When the apparatus is sealed off, the pressure of the air inside is 70 cm of Hg and temperature 20°C. What does the pressure become if a 100 cm3 bulb is kept at 20°C and the other bulb is heated to 100°C?
Answer:
Here, V2 = 500 cm3
V1 = 100 cm3
P1 = P2 = P = ?
T1 = 20°C = 293 K
T2 = 100°C = 373 K
T = 20°C = 293 K
P’ = 70 cm of Hg

For a given mass of the gas,
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 20
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 21

Question 12.
0.014 kg of nitrogen is enclosed in a vessel at a temperature of 27°C. How much heat has to be transferred to the gas to double the r.m.s. the velocity of its molecules?
Answer:
Vrms = \(\sqrt{\frac{3 R T}{M}}\)
Here, T1 = 273 + 27 = 300 K .

Now to double the r.m.s. velocity, the temperature should be raised to four times the initial temperature.
i.e. T2 = 4T, = 4 × 300 K = 1200 K
∴ ΔT = rise in temperature = T2 – T1
= 1200 – 300 – 900 K

k = 1.38 × 10-23 Jk-1
Cv = \(\frac{5}{2}\) R = \(\frac{5}{2}\)(kN)
= \(\frac{5}{2}\) × 1.38 × 10-23 × 6.023 × 1023
= 20.8 JK-1 mol-1 …(i)

∴ If Δθ be the amount of heat required,
Then Δθ = n Cv ΔT
where n = number of moles of nitrogen in 0.014 kg
= \(\frac{1}{0.028}\) × 0.014
= \(\frac{1}{2}\) (∵ 1 mole of N2 = 0.028 Kg)

∴ Δθ = \(\frac{1}{2}\) × 20.8 × 900
= 9360 J.

Question 13.
Four molecules of a gas have speeds 4,6,8 and 10 Km s-1 respectively. Calculate their rms speed.
Answer:
Here, C1 = 4 km s-1
C2 = 6 km s-1
C3 = 8 km s-1
C4 = 10 km s-1
C = rms speed = ?
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 22

Question 14.
At what temperature, pressure remaining constant, will the RMS velocity of gas on behalf of its value at 0°C?
Answer:
Here, T1 = 0°C 273 + 0 = 273 K
Let C1 = rmsvelocityat0°C
Let T2 be the temperature (= ?) at which rms velocity becomes half
i.e. C2 = \(\frac{\mathrm{C}_{\mathrm{I}}}{2}\)

Now using the relation, C2 ∝ T. we get
\(\frac{C_{2}^{2}}{C_{1}^{2}}=\frac{T_{2}}{T_{1}}\)
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 23

Question 15.
If the density of nitrogen at S.T.P. be 0.00125 g cm3. What is the velocity of its molecules? g = 980 cm
Answer:
Here, P = 1 atm = 76 cm of Hg
= 76 × 13.6 × 980 dyne cm-2
ρ = density of nitrogen
= 125 × 10-5 g cm-3
Crms = ?

Using the relation,
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 24

Question 16.
At a certain pressure and 127°C, the average K.E. of a hydrogen molecule is 8 × 10-27 J. At the same pressure, determine:
(a) ruts velocity of hydrogen molecules at 27°C.
(b) average K.E. of nitrogen molecules at 127°C.
Mass of hydrogen atom = 1.7 × 10-27 kg.
Answer:
(a) Here, T1 = 127°C = 127 + 273 = 400 K
E1 = 8 × 10-27 J
m = mass of hydrogen molecule
= 2 × 1.7 × 10-27 kg
= 3.4 × 10-27 kg

T2 = 27°C = 27 + 273 = 300 K
E2 = ?

r.m.s velocity at 27°C, Crms = ?
Using the relation,
K.E = E ∝ T, we get
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 25

Also, we know that
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 26
(b) As the average K.E. of a molecule of any gas at the same temperature is the same for all gases, so the average K.E. of nitrogen molecule at 127°C is = 8 × 10-27 J

Question 17.
Calculate the total random K.E. of one gram of nitrogen at 600 K.
Answer:
Here, T = 600 K
R = 8.31 J/mol/K
M = 28 g = molecular weight of nitrogen

∴ Total random K.E. for 1 g molecule of nitrogen is
E’ = \(\frac{3}{2}\)RT

∴ Total random K.E. for one gram of nitrogen
= \(\frac{3}{2} \frac{\mathrm{R}}{\mathrm{M}}\)T
= \(\frac{3}{2}\) × \(\frac{8.31}{28}\) × 600
= 266.8 J.

Question 18.
Find the temperature at which the RMS velocity of oxygen molecules in the earth’s atmosphere equals the velocity of escape from the earth’s gravitational field.
N = 6.023 × 1023
R = 6400 km = radius of earth
k = 1.38 × 10-23 JK-1
Answer:
Here, N = 6.023 × 1023
R = 6400 km = 6400 × 103m
k = 1.38 × 10-23JK-1

Ve = escape velocity from earth’s surface
= \(\sqrt{2 \mathrm{gR}_{\mathrm{e}}}\) ….(i)
T = temperature = ?
Let Crms = rms velocity of oxygen molecules at temp. T0

∴ Using the relation,
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 27

Where m = mass of one molecule = \(\frac{M}{N}\)
k = \(\frac{R}{N}\)

∴ According to the statement,
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 28

Question 19.
Calculate the temperature at which r.m.s. the velocity of a gas molecule is the same as that of a molecule of another gas at 27°C. The molecular weights of the two gases are 64 and 32 respectively.
Answer:
Here, T1 = ? :
T2 = 27°C = 273 + 27 = 300K
M1 = 64 .
M2 = 32
C1 = C2

∴ Using the relation,

\(\frac{1}{2}\)MC2 = \(\frac{3}{2}\)RT,weget
\(\frac{1}{3}\) M1C12= RT1 for gas of mass M1
and \(\frac{1}{3}\)M2C22 = RT2 for gas of mass M2
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 29

Question 20.
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 125.0 m3 at a temperature of 127°C and 2 atm pressure, k = 1.38 × 10-23 JK-1.
Answer:
Here, T = 127°C + 273 = 400 K
k = 1.38 × 10-23 JK-1 P = 2 atmosphere
= 2 × 1.01 × 105 Nm-2
= 2.02 × 105 Nm-2

V = volume of room = 125 m3
N’ = no. of molecules in the room =?
∴ R = Nk = 6.023 × 1023 × 1.38 × 10-23
= 8.31 JK-1 mol-1

Let n = no. of moles of the air in the given volume.
∴ Using gas equation,
PV = nRT, we get
n = \(\frac{\mathrm{PV}}{\mathrm{RT}}=\frac{2.02 \times 10^{5} \times 125}{8.31 \times 400}\)
= 7.60 × 103 moles

∴ N’ = Nn = 6.023 × 1023 × 7.60 × 103
= 45.77 × 1026.

Question 21.
Calculate the temperature at which the oxygen molecules will have the same r.m.s. velocity as the hydrogen molecules at 150°C. The molecular weight of oxygen is 32 and that of hydrogen is 2.
Answer:
Here, Molecular weight of oxygen, M0 = 32
Molecular weight of hydrogen. MH = 2

Let T0 = temp. of oxygen = ?
TH = temp. of hydrogen
= 150°C = 150 + 273 = 423 K
C0 = CH
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 30

Question 22.
Calculate the r.m.s. the velocity of molecules of gas for which the specific heat at constant pressure is 6.84 cal per g mol per °C. The velocity of sound in the gas being 1300 ms-1. R = 8.31 × 107 erg per g mol per °C. J = 4.2 × 107 erg cal-1.
Answer:
Here, Cp = 6.84 cal/g mol/°C
R = 8.31 × 10 erg/g mol/°C
J = 4.2 × 10 erg/cal
v = velocity = 1300 ms-1
= 1300 × 100 cm s-1
Crms =?

Using the relation,
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 31
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 32
Now using the relation,
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 33

Question 23.
Calculate the molecular K.E. of I g of an oxygen molecule at 127°C. Given R = 8.31 JK-1 mol-1. The molecular weight of oxygen = 32.
Answer:
Here, M = 32 g
T = 127 + 273 = 400 K

∴ Molecular K.E. of oxygen is given by
\(\frac{1}{2}\) MC2 = \(\frac{3}{2}\) RT
Now K.E. of 32 g of O2 RT = \(\frac{3}{2}\)RT

∴ K.E.of 1 g of O2 = \(\frac{3}{2} \cdot \frac{\mathrm{RT}}{32}\)
or
E = \(\frac{3}{64}\) × 8.31 × 400 J
= 155.81 J.

Question 24.
Calculate the intermolecular B.E. in eV of water molecules from the following data:
N = 6 × 1023 per mole
1 eV= 1.6 × 10-19 J
L = latent heat of vaporisation of water = 22.6 × 105 J/kg.
Answer:
Here, molecular weight of water, M = 2 + 16 = 18g
∴ No. of molecules in 1 kg of water = \(\frac{6 \times 10^{23}}{18}\) × 1000 = \(\frac{10^{26}}{3}\)

L = 22.6 × 105 J kg
∴ B.E .per molecule = 22.6 × 105 J = B.E of \(\frac{6 \times 10^{23}}{18}\) molecule

Thus B.E. per molecule
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 34

Question 25.
Two perfect gases at absolute temperatures T1 and T2 are mixed. There is no loss of energy. Find the temperature of mixture if masses of molecules are m1 and m2 and the no. of molecules in the gases are n1 and n2 respectively.
Answer:
Let E1 and E2 be the K.E. of the two gases,
∴ E1 = \(\frac{3}{2}\) kT1 × n1
and E2 = \(\frac{3}{2}\) kT2 × n2

Let E be the total energy of the two gases before mixing
∴ E = E1 + E2 = \(\frac{3}{2}\)K(n1T1 + n2T2) ….(1)

After mixing the gases, let T be the temperature of the mixture of the two gases
∴ E’ = \(\frac{3}{2}\)kT(n1 + n2) …(2)

As there is no loss of energy,
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 35

Value-Based Type:

Question 1.
Ram has to attend an interview. He was not well. He took the help of his friend Raman. On the way office, Ram felt giddy, He vomited on his dress. Raman washed his shirt. He made Ram drink enough amount of water. In spite of doing, a foul smell was coming from the shirt. Then Raman purchased a scent bottle from the nearby cosmetics shop and applied to Ram. Ram attended the interview, Performed well. Finally, he was selected.
(a) What values do you find in Raman?
Answer:
He has the presence of mind, serves others in need.

(b) The velocity of air is nearly 500m/s. But the smell of scent spreads very slowly, Why?
Answer:
This is because the air molecules can travel only along a zig-zag path due to frequent collisions. Consequently, the displacement per unit time is considerably small.

Question 2.
One day Shikha got up 7a.ni and saw that the rays of sunlight coming through a narrow hole contains some dust particles which is moving randomly. She kept it her mind and when she reached her school the next day, she first asked her physics teacher the reason behind it. The teacher explained that gas consists of rapidly moving atoms or molecules. The particles may also collide with each other when they come together. She becomes happy to hear the reason. ‘
(i) What values are exhibited by Shikha?
Answer:
Creative, Awareness, willing to know the scientific reasons.

(ii) What is r.m.s velocity?
Answer:
The root mean square speed (r.m.s) of gas molecules is defined as the square root of the mean of squares of the speeds of gas molecules.
i.e Vrms = \(\sqrt{\frac{V_{1}^{2}+V_{2}^{2}+V_{3}^{2}+\ldots \ldots \ldots \ldots \ldots .+V_{n}^{2}}{n}}\)
= \(\sqrt{\frac{3 \mathrm{~K}_{\mathrm{B}} \mathrm{T}}{\mathrm{m}}}\)

(iii) Calculate r.m.s velocity of one gram molecule of hydrogen at S.T.P. [density of hydrogen at S.T.P = 0.09 kg m-3]
Answer:
According to kinetic theory .of gases:
P = \(\frac{1}{3}\)ρC2
or
C = \(\sqrt{\frac{3 \mathrm{P}}{\rho}}\)

Here, ρ = 0.09 kg m-3, P = 1.01 × 105 Pa.
∴ C = \(\sqrt{\frac{3 \times 1.01 \times 10^{5}}{0.09}}\)
= 1837.5 ms-1

Question 3.
During the lecture of physics period, Madan’s teacher Mr Suresh has given a question to the whole class. The question was as under:
“A vessel contains two non-reactive gases: neon (monoatomic) and oxygen (diatomic). The ratio of their partial pressure is 3:2.
Estimate the ratio of
(a) Number of molecules
(b) Mass density
Madan raised his hand to answer the above two questions and gave a satisfactory answer to the question. .
(i) What value is displayed by Madan?
Answer:
He is intelligent, Creative, Sharp minded.

(ii) What explanations would have been given by Madan?
Answer:
(a) Since V and T are common to the two gases
Also, P1V = μ1 RT ….(i)
P2V = μ2 RT….(ii)

[Using ideal gas equation]
Here, 1 and 2 refer to neon and oxygen respectively
i.e \(\frac{P_{1}}{P_{2}}=\frac{\mu_{1}}{\mu_{2}}\)
⇒ \(\frac{3}{2}=\frac{\mu_{1}}{\mu_{2}}\) {∵ \(\frac{P_{1}}{P_{2}}=\frac{3}{2}\)(given)}

(b) By definition μ1 =\(\frac{\mathrm{N}_{1}}{\mathrm{~N}_{\mathrm{A}}}\) and \(\frac{\mathrm{N}_{2}}{\mathrm{~N}_{\mathrm{A}}}\) where N1 and N2 are the number of molecules of 1 and 2 and NA is the Avogadro’s number.
∴ \(\frac{N_{1}}{N_{2}}=\frac{\mu_{1}}{\mu_{2}}=\frac{3}{2}\)

Question 4.
A quiz contest was organized by a public school. They asked rapid-fire questions and decided to give lst, 2nd and 3rd prizes. The entire class was divided into ten groups and each group has S students.
The questions in the final round were as under:
(i) What is the ideal gas equation?
Answer:
The relationship between Pressure P, Volume V and absolute temperature T of a gas is called its equation of state. The equation of the state of an ideal gas is
PV = μRT

(ii) What would be the effect on the RMS velocity of gas molecules if the temperature of the gas is increased by a factor of 4?
Answer:
Class 11 Physics Important Questions Chapter 13 Kinetic Theory 36
Clearly, ‘C’ will be doubled.

(iii) Which values are being depicted here by the school by ask¬ing the above questions?
Answer:
Values are:
(a) To develop group activity.
(b) To teach in an interesting way.
(c) Award and prizes to motivate them.
(d) To develop leadership quality.

States of Matter Class 11 Important Extra Questions Chemistry Chapter 5

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 5 States of Matter. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 5 Important Extra Questions States of Matter

States of Matter Important Extra Questions Very Short Answer Type

Question 1.
What is the pressure of a gas? What is its S.I. unit?
Answer:
The force exerted by the gas molecules per unit area on the walls of the container is equal to its pressure SI. unit of pressure is the pascal (Pa).
1 Pa = Nm2- =1 kg m-1s-2.

Question 2.
What does the temperature of gas represent?
Answer:
The temperature of the gas represents the average kinetic energy of the gas molecules as we know K.E. ∝ \(\sqrt{T}\).

Question 3.
Why is it not possible to cool gas to o K?
Answer:
This is because all gases condense to liquids or solids before this temperature is reached.

Question 4.
What property of molecules of real gases is indicated by van der Waal’s constant ‘a’?
Answer:
Intermolecular attraction.

Question 5.
What do you understand by standard temperature?
Answer:
The standard temperature is 0°C or 273 K.

Question 6.
What is the effect of temperature on the vapour pressure of a liquid?
Answer:
Vapour pressure increases with rising in temperature.

Question 7.
What is an ideal gas equation?
Answer:
For given ideal gas PV = nRT
where P = gas pressure;
V = volume,
R = gas constant
T = Temp. (Kelvin scale);
n – no. of moles of gas.

Combined Gas Law Calculator … Combined gas law is the combination of Charles’s law, Boyle’s law, and Gay-Lussac’s law.

Question 8.
Which state of matter has a definite volume, but no definite shape?
Answer:
Liquid.

Question 9.
What is the value of the gas constant in S.I. units?
Answer:
8.314 JK-1 mol-1.

Question 10.
What are the S.I. units of surface tension?
Answer:
Nm-1

Question 11.
What is the compressibility factor?
Answer:
Z = \(\frac{PV}{nRT}\).

Question 12.
What is the equation of state for real gases?
Answer:
van der Waal’s equation(P + \(\frac{a}{\mathrm{~V}^{2}}\))(v – b) = RT for 1 mole.

Question 13.
How is the pressure of a gas related to its density at a particular temperature?
Answer:
d = \(\frac{MP}{RT}\)

Question 14.
How is the partial pressure of a gas in a mixture related to the total pressure of the gaseous mixture?
Answer:
The partial pressure of a gas = Mole fraction of that gas × total pressure.

Question 15.
What is the relationship between average kinetic energy , and the temperature of a gas?
Answer:
K.E. = \(\frac{3}{2}\)kT
where k is Boltzmann constant = \(\frac{\mathrm{R}}{\mathrm{N}_{0}}\).

Question 16.
What is the significance of van der Waal’s constant ‘a’ and ‘b’?
Answer:
‘a’ is a measure of the magnitude of the intermolecular forces of attraction while ‘b’ is a measure of the effective size of the as molecules.

Question 17.
Arrange the solid, liquid and gas in order of energy-giving reasons.
Answer:
Solids < liquid < gas. This is because a solid absorbs energy to change into a liquid which further absorbs energy to change into a gas.

Question 18.
What is the effect of pressure on the boiling point of a liquid?
Answer:
The boiling point increases as the prevailing pressure increases.

Question 19.
Why are vegetables cooked with difficulty at a hill station?
Answer:
The atmospheric pressure decreases as we go up. Therefore at the hills due to the lowering of atmospheric pressure, boiling points lowered.

Question 20.
The size of the weather balloon keeps on becoming larger as it rises to high altitude. Explain why?
Answer:
At higher altitudes, the external pressure on the balloon decreases and therefore, its size increases.

Question 21.
How is the pressure of a given sample of a gas related to the temperature at a constant volume?
Answer:
P ∝ T, i.e., Pressure varies as absolute temperature.

Question 22.
How is the pressure of a gas related to the number of molecules of a gas at constant volume and temperature?
Answer:
P ∝ N; N = No. of molecules of a gas.

Question 23.
What type of graph do we get when we plot a graph PV against P? What is shown by this graph?
Answer:
It is a straight line parallel to the r-axis. It shows that PV is constant at different pressures. It shows Boyle’s Law.

Question 24.
At what temperature will oxygen molecules have the same kinetic energy as ozone molecules at 30°C?
Answer:
AT 30°C. Kinetic energy depends only on the absolute temperature and not on the identity of the gas.

Question 25.
At a particular temperature why vapours pressure of acetone is less than that of ether?
Answer:
This is because intermolecular forces in acetone are more than those present in ether.

Question 26.
Out of NH3 and N2 which will have
(i) a larger value of V and
Answer:
NH3 will have a large value of ‘a’ because of hydrogen bonding.

(ii) larger value of ‘b’?
Answer:
N2 should have a large value of ‘b’ because of its larger molecular size.

Question 27.
Why are the gases helium and hydrogen not liquefied at room temperature by applying very high pressure?
Answer:
Because their critical temperature is lower than room temperature. Gases cannot be liquefied above the critical temperature by applying even very high pressure.

Question 28.
What will boil at a higher temperature at sea level or at the top of mountains?
Answer:
Water will boil at a higher temperature at sea level.

Question 29.
Under what conditions do real gases tend to show ideal gas behaviour?
Answer:
When the pressure of the gas is very low and the temperature is high.

Question 30.
Both propane (C3Hg) and carbon dioxide (CO2) diffuse at the same rate under identical conditions of temperature and pressure. Why?
Answer:
Both propane (C3Hg) and carbon dioxide (CO2) have the same molar mass (44 gm).

Question 31.
If the number of moles of a gas is doubled by keeping the temperature and pressure constant, what will happen to the volume?
Answer:
The volume will also double as V ∝ n according to Avogadro’s Law.

Question 32.
What is a Triple point?
Answer:
The temperature at which solid, liquid and vapour; i.e., all the three states of the substance exist together is called the triple point.

Question 33.
Is Dalton’s law of partial pressures valid for a mixture of SO2 and O2?
Answer:
No, the law holds good only for those gases which do not react with each other.

Question 34.
Under what conditions a gas deviates from ideal gas behaviour?
Answer:
It deviates at low temperature and high pressure.

Question 35.
Molecule A is twice as heavy as molecule B. Which of these has higher kinetic energy at any temperature?
Answer:
K.E. of a molecule is directly proportional to temperature and is independent of its mass. So both the molecule A and B at any temperature will have equal kinetic energy.

Question 36.
How will you define pascal?
Answer:
It is defined as the pressure exerted by a force of one newton on an area of one meter2.
Pa = 1 Nm-2.

Question 37.
How will you define London or dispersion forces?
Answer:
The forces of attraction between the induced momentary dipoles are called London or dispersion forces.

Question 38.
What is Absolute Zero?
Answer:
The lowest possible hypothetical temperature of – 273°C at which gas is supposed to have zero volume is called Absolute zero.

Question 39.
What is the nature of the gas constant R?
Answer:
States of Matter Class 11 Important Extra Questions Chemistry 1
= work done per degree per mole.

Question 40.
What is the value of R in SI units?
Answer:
R = 8.314 JK-1 mol-1 = 8.314 k Pa dm3 K-1 mol-1

Question 41.
What is meant from Boyle point or Boyle temperature?
Answer:
The temperature at which a real gas behaves like an ideal gas over an appreciable pressure range is called Boyle point or Boyle temperature.

Question 42.
How will you define viscosity?
Answer:
The internal resistance to the flow of a fluid is called its viscosity.

Question 43.
What is the effect of temperature on the viscosity of a liquid?
Answer:
The viscosity of a liquid decreases with an increase in temperature.

Question 44.
How will you convert pressure in atmospheres into SI units?
Answer:
1 atm = 1,01,325 Pa or Nm-2 or 1 bar = 105 Pa.

Question 45.
What do you understand from surface tension?
Answer:
The surface tension of a liquid is defined as the force acting at right angles to the surface along a one-centimetre length of the. liquid.

Question 46.
What is the effect of the increase in temperature on the surface tension of a liquid?
Answer:
Surface tension decreases in increasing the temperature.

Question 47.
What is the difference between normal boiling point and standard boiling point?
Answer:
When the external pressure is equal to one atmospheric pressure the boiling point is referred to as a normal boiling point, when it is one bar, the boiling point is called its standard boiling point.

Question 48.
What is the difference between vapour and gas?
Answer:
When gas is below its critical temperature, it is called vapour.

Question 49.
What happens if a liquid is heated to the critical temperature of its vapours?
Answer:
The meniscus between the liquid and the vapour disappears, the surface tension of the liquid becomes zero.

Question 50.
Why falling liquid drops are spherical?
Answer:
The surface area of a sphere is minimum. In order to have minimum surface area drops of the liquid become spherical.

States of Matter Important Extra Questions Short Answer Type

Question 1.
Ammonia and Sulphur dioxide gases are prepared in two comers of a laboratory. Which gas will be detected first by a student working in the middle of the laboratory and why?
Answer:
Molecular mass of NH3 = 17 Molecular mass of SO2 = 64
NH3 is a lighter gas and diffuses at a faster speed than SO2
∴ NH3 gas will be detected first.
[∵ Acc. to Graham’s law of diffusion \(\frac{r_{1}}{r_{2}}=\sqrt{\frac{d_{2}}{d_{1}}}\)

Question 2.
What is the effect of hydrogen bonding on the viscosity of a liquid?
Answer:
Hydrogen bonding leads to an increase in the effective size of the moving unit in the liquid. Due to an increase in the size and mass of the molecule, there is greater interval resistance of the flow of the liquid. As a result, the viscosity of the liquid rises.

Question 3.
Which are the two faulty assumptions in the kinetic theory of gases.
Answer:

  1. There is no force of attraction between the molecules of the gas.
  2. The volume of the molecules of the gas is negligibly small as compared to the total space (volume) occupied by the gas.

Question 4.
What is the relationship between the density and molar mass of a gaseous substance? Derive it.
Answer:
Ideal gas equation is PV = nRT
or \(\frac{n}{V}=\frac{p}{R T}\)
Replacing n by \(\frac{m}{M}\), we get
States of Matter Class 11 Important Extra Questions Chemistry 2

Question 5.
What is meant by the term: Non-ideal or real gas?
Answer:
The gas which does not obey the Cas law:
Boyle’s law, Charles’ law, Avogadro^ law at all temperatures and pressures is a called-non-ideal or real gas. Most of the real gases show ideal behaviour at low pressure and high temperature.

Question 6.
Derive the ideal gas equation PV = nRT.
Answer:
According to Boyle’s law V ∝ \(\frac{1}{P}\) if n and T are constant.
According to Charles’ law V ∝ T at constant P.and n
According to Avogadro’s law V ∝ n at constant T and P
Combining the three laws
In
V ∝ \(\frac{Tn}{P}\)
or
PV ∝ nT
or
PV = nRT where R is a constant of proportionality called universal gas constant.

Question 7.
Why liquids have a definite volume, but no definite shape?
Answer:
It is due to the fact that in liquids intermolecular forces are strong enough to hold the molecules together, but these are strong enough to hold the molecules together, but these forces are not strong enough to fix them into definite or concrete positions as in solids. Hence they possess fluidity but no definite shape.

Question 8.
How do the real gases deviate from ideality above and below the Boyle point?
Answer:
Above their Boyle point, real gases show positive deviations from ideality and the values of Z are greater than one. The forces of attraction between the molecules are very feeble. Below the Boyle point, real gases first show a decrease, in Z value with increasing pressure, the value Of Z increases continuously.

Question 9.
Write down the van der Waals, equation for n moles of a real gas. What do the constants ‘a’ and ‘b’ stand for?
Answer:
(P + \(\frac{a n^{2}}{V^{2}}\)) (V – nb) = nRT
where p, Vindicate gas pressure and its volume. T is the Kelvin temperature of the gas. R is gas constant. Value of ‘a’ is a measure of the magnitude of intermolecular attractive forces within the molecule and is independent of temperature and pressure, ‘rib’ is approximately the total volume occupied by the molecules themselves. ‘a’ and ‘b’ are called van der Waals constants and their value depends upon the nature of the gas.

Question 10.
How is compressibility factor Z related to the real and ideal volume of the gas?
Answer:
Z = \(\frac{P \mathrm{~V}_{\text {real }}}{n \mathrm{RT}}\) …(1)

If the gas shows ideal behaviour, then
Videal = \(\frac{nRT}{P}\) ….(2)

On putting the titis value of nRT in equation (1), we get
Z = \(\frac{\mathrm{V}_{\text {real }}}{\mathrm{V}_{\text {ideal }}}\)

Thus the compressibility factor Z is the ratio of the actual molar volume of a gas to the molar volume of it. if it were an ideal gas at that temperature and pressure.

Question 11.
What do the critical temperature, critical pressure, and critical volume for a gas stand for?
Answer:
The critical temperature is the temperature above which a gas cannot be liquified however large may be the pressure applied on it. The pressure sufficient to liquiíy a gas at its critical temperature is called its critical pressure. The volume of the gas at its crìtical temperature is called its critical volume’.

Question 12.
What do the absolute zero and absolute scale of temperature stand for?
Answer:
The lowest possible hypothetical temperature of – 273°C at which gas is supposed to have zero volume is called Absolute! zero.

Lord Kelvin suggested a new scale of temperature starting with – 273°C and it’s zero. This scale of temperature is called the absolute scale of temperature.

Absolute or Kelvin temperature is given by T, where
T = t°C + 273
t°C stands for the centigrade scale of temperature.

Question 13.
Give one application of Dalton’s law of partial pressures.
Answer:
One application of Dalton’s law of partial pressures is in determining the/pressure of dry gas. Gases are generally collected over water and, so they contain water vapours. The pressure exerted by the water Vapours at-a particular temperature is called the Aqueous Tension. By subtracting the aqueous tension from the ‘ vapour pressure of the moist gas, the pressure of the dry gas can be calculated.
Pdry gas= Pmoist gas – Aqueous Tension (at t°C).

Question 14.
How will you calculate the partial pressure of gas using Dalton’s Law of partial pressures?
Answer:
In a mixture of non-reacting gases A, B, C etc., if each gas is considered to be an ideal gas, then
PA = nA \(\frac{RT}{V}\)
PB = nB \(\frac{RT}{V}\)
PC = nC \(\frac{RT}{V}\)
where nA, nB nC stand for their respective moles in the same vessel .
[V = constant, keeping T constant]
Then according to Dalton’s law ‘
States of Matter Class 11 Important Extra Questions Chemistry 3
Thus, particle pressure of a. gas – Mole fraction of A × Total pressure.

Question 15.
Is there any effect of the nature of a liquid and temperature on the surface tension of a liquid?
Answer:

  1. Effect of nature of the liquid: Surface tension is a property that arises due to the intermolecular forces of attraction among the molecules of the liquid.
  2. The surface tension of the liquids generally decreases with an increase in temperature and becomes zero at the critical temperature.

Question 16.
Leaving electrostatic forces that exist between oppositely charged ions, which other intermolecular attractive forces exist between atoms or ions. Mention them.
Answer:

  1. van der Waals forces
  2. London Forces or Dispersion Forces
  3. Dipole-dipole forces
  4. Dipole-induced dipole forces
  5. Hydrogen bond.

Question 17.
What is the relationship between intermolecular forces and thermal interactions?
Answer:
Intermolecular forces tend to keep the molecules together but the thermal energy of the molecules tends to keep them apart. Three states of matter are the result of a balance between intermolecular forces and the thermal energy of the molecules.
Gas → Liquid → Solid
intermolecular forces increases →
Gas ← Liquid ← Solid
increase in Thermal energy ←

Question 18.
What are the main characteristics associated with gases?
Answer:

  1. Gases are highly compressible.
  2. Gases exert pressure equally in all directions.
  3. Gases have a much lower density than solids or liquids.
  4. Gases do not have definite shape and volume.
  5. Gases intermix freely and completely in all proportions.

Question 19.
What do you understand by the term? Isotherms and Isobars?
Answer:
Isotherms are the curves plotted by varying pressure against volume keeping temperature constant.

Each line obtained by plotting volume against temperature keeping, pressure constant is called an Isobar.

Question 20.
What are the forces which are responsible for the viscosity of a liquid? Why is glycerol more viscous than water?
Answer:
The forces responsible for the viscosity of the liquids are

  • Hydrogen bonding
  • van der Waals torches.

Glycerol possesses greater viscosity than water because glycerol has extensive hydrogen bonding in its molecules due to the pressure of three-Oil groups in it as compared to the hydrogen bonding in water molecules.

States of Matter Important Extra Questions Long Answer Type

Question 1.
Derive the Ideal gas equation PV = nRT.
Answer:
Let a certain mass of a gas at pressure P1 and temperature
T1 occupy a volume V1. On changing the pressure and temperature respectively to P2 and T2, let the volume of the gas change to V2.
(i) Let us first change the pressure P1 to P2 at constant temperature T1. Then, according to Boyle’s law, volume V is given by,
P2V = P1V1
or
V = \(\frac{P_{1} V_{1}}{P_{2}}\) …(1)

(ii) Now keeping the pressure constant at P2, heat the gas from temperature T1 to T2, the volume changes from V to V2.
By Charle’s Law, we must have,
States of Matter Class 11 Important Extra Questions Chemistry 4
The numerical value of k depends upon the amount of gas and the units in which P and V are expressed. By Avogadro’s law, for one mole of any gas, the value of \(\frac{PV}{T}\) will be the same and is represented
\(\frac{PV}{T}\) = R
or
PV = RT.
for n moles of an ideal gas
PV = nRT

Question 2.
State the fundamental assumptions of the kinetic theory of gases
Or
What are the postulates of the kinetic theory of gases?
Or
Write a short note on “Microscopic Model of a gas”.
Answer:
To explain the general properties of gases to provide some theoretical explanation for various gas laws (stated on basis of experimental studies), various, scientists Bernoulli, Clausius, Maxwells, Boltzmann and others, gave ‘Kinetic Theory of Gases’ which explains qualitatively as well as quantitatively the various aspects of gas behaviour.

The important postulates of the theory are:

  1. All gases are made up of a very large number of minutes particles called molecules.
  2. The molecules are separated from one another by large distances. The empty spaces among the molecules are so large that the actual volume of the molecules is negligible as compared to the total volume of the gas.
  3. The molecules are in a state of constant rapid motion in all directions. During their motion, they keep on colliding with one another and. also with the walls of the container and, thus, change their directions.
  4. Molecular collisions are perfectly elastic, no loss of energy occurs when the molecules collide with one another or with the walls of the container. However, there may be. redistribution of energy during the collisions.
  5. There are no forces of interaction (attractive or repulsive) between molecules. They move completely independent of one another.
  6. The pressure exerted by the gas is due to the bombardment of its molecules on the walls of the container per unit area.
  7. The average kinetic energy of the gas molecules is directly proportional to the absolute temperature.

Question 3.
Explain the following observations.
(a) Aerated water bottles are kept underwater during summer.
Answer:
Aerated water contains CO2 gas dissolved in an aqueous solution under pressure and the bottles are well stoppered. As in summer temperature increases and we know that the solubility of the gases decreases with increase in temperature and as a result move of gas is expected to be generated may be large in quantity and hence pressure exerted by the gas may be very high and the bottle may explode. So, to decrease the temperature and hence to avoid explosion of the bottles.

(b) Liquid ammonia bottle is cooled before opening the seal.
Answer:
Liquid ammonia bottle contains the gas under very high pressure. If the bottle is opened as such, then the sudden decrease in pressure will lead to a large increase in the volume of the gas. As a result, the gas will come out of the bottle with force. This will lead to the breakage of the bottle. Cooling under tap water will result in a decrease in volume. It reduces the chances of an accident.

(c) The tyre of an automobile is inflated at lesser pressure in summer than in winter.
Answer:
The pressure of the air is directly proportional to the temperature. During summer due to high temperature, the pressure in the tyre will be high as compared to that in water. The tube may burst under high pressure in summer. Therefore, it is advisable to inflate the tyre to lesser pressure in summer than in winter.

(d) The size of the weather balloon becomes larger and larger as it ascends up to higher altitudes.
Answer:
Answered somewhere else in the book.

States of Matter Important Extra Questions Numerical Problems

Question 1.
In a hospital, an oxygen cylinder holds 10 L of oxygen at 200 atm pressure. If a patient breathes in 0.50 ml of oxygen at 1.0 atm with each breath, for how many breaths the cylinder will be sufficient. Assume that all the data is at 37°C.
Answer:
10 L at 200 atm =? L at 1 atm. [Temp, is constant at 37°C]
∴ Boyle’s law P1V1 = P2V2 can be applied
200 × 10 = 1 × V2
or
V2 = 2000 L

Number of breaths = \(\frac{\text { Total volume }}{\text { Volume inhaled per breath }}\)
= \(\frac{2000 \mathrm{~L}}{0.5 \times 10^{-3} \mathrm{~L}}\)
= 4 × 106

Question 2.
Calculate the pressure of 1 × 1022 molecules of oxygen gas when enclosed ¡n a vessel of 5-litre capacity at a temperature of 27°C.
Answer:
Number f moles of O2 = \(\frac{1 \times 10^{22}}{6.02 \times 10^{23}}\)

= 0.166 × 10-1 mol
= 1.66 × 102 mol
The pressure exerted by O2 molecules can be calculated by the equation
PV = nRT [Ideal gas equation]
States of Matter Class 11 Important Extra Questions Chemistry 5

Question 3.
When a ship is sailing in the Pacific Ocean where the temperature is 23.4°C, a balloon is filled with 2 L air. What will be the volume of the balloon when the ship reaches the Indian Ocean, where the temperature is 26.1°C.
Answer:
V1 = 2 L
T1 = 23.4 + 273 = 296.4K
T2 = 26.1 + 273 = 299.1 K
From Charle’s Law
States of Matter Class 11 Important Extra Questions Chemistry 6

Question 4.
At what temperature centigrade, will the volume of a gas at 0°C double itself, pressure remaining constant?
Answer:
Let the volume of the gas at 0°C = V1 mL
Final Volume = V2 = 2V1 mL [Given]
T1 = OC + 273 = 273 K, T2 =?

Since Pressure remains constant
∴ Charles law can be applied
States of Matter Class 11 Important Extra Questions Chemistry 7

Question 5.
A bulb ‘B’ of the unknown volume containing gas at one atmospheric pressure is connected to an evacuated bulb of 0.5-litre capacity through a stop-cock. On opening the stop-cock, the final pressure was found to be 570 mm at the same temperature. ‘What is the volume of bulb ‘B’?
Answer:
Let the volume of the bulb = V1; P, = 1 atm
Evacuated Bulb has volume = 0.5 L;
Total volume of both bulbs after the opening of stop-cock = (V1 + 0.5) L i.e. V2 = (V1 + 0.5):
P2 = 570 mm = \(\frac{570}{760}\) atm

Apply Boyle’s law P1V1 = P2V2
1 × V1 = \(\frac{570}{760}\) × (V1 + 0.5)

On solving the above equation
V1 = 1.5 L
∴ Volume of the unknown bulb ‘B’ = 1.5 L

Question 6.
1 mole of SO2 occupies 1.5 L at 25°C. Calculate the pressure exerted by gas assuming that gas does not obey the ideal gas equation. (Given a = 3.6 atm L2mol-2, b = 0.04 L mol-1) where a, b are van der Waal’s constants.
Answer:
States of Matter Class 11 Important Extra Questions Chemistry 8
States of Matter Class 11 Important Extra Questions Chemistry 9

Question 7.
A gas occupies a volume of 2.5 L at 9 × 105 Nm-2. Calculate the additional pressure required to decrease the volume of the gas to 1.5 L, keeping the temperature constant.
Answer:
V1 = 2.5 L; P1 = 9 × 105 Nm-2
V2 = 1.5 L; P2 =?

Since temperature is constant, Boyle’s law is applied.
States of Matter Class 11 Important Extra Questions Chemistry 10

Question 8.
What volume of air will be expelled from a vessel containing 400 cm3 at 7° C when it is heated to 27° C at the same pressure?
Answer:
As the pressure is constant, Charles law is applied
States of Matter Class 11 Important Extra Questions Chemistry 11
428.6 cm3 is the volume after expansion
∴ Volume expelled = (428.6 – 400) cm3 = 28.6 cm3.

Question 9.
A 10.0 L container is filled with a gas to a pressure of 2.0 atm at 0°C. At what temperature will the pressure of the gas inside the container be 2.50 atm.
Answer:
As the volume of the container remains constant, Gay- Lussac’s Law/Amonton’s Law is applicable
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)
\(\frac{2}{273}=\frac{2.50}{\mathrm{~T}_{2}}\)
T2 = 341 K
or
t2 = (341 – 273)°C = 68°C.

Question 10.
One litre flask containing vapours of methyl alcohol (Mol. mass = 32) at a pressure of 1 atm and 25°C was evacuated till the final pressure was 10-3 mm. How many molecules of methyl alcohol were left in the flask?
Answer:
P1 = 10-3 mm; V1 = 1000 cm3; T1 = 298 K
Converting this volume to volume, at STP, where T2 = 2/3 K and P2 = 760 mm
States of Matter Class 11 Important Extra Questions Chemistry 12
Now, 22400 cm3 at STP contains 6.02 × 1023 molecules

∴ 1.205 × 10-3 cm3 at STP contains
= \(\frac{6.02 \times 10^{23}}{22400}\) × 1.205 × 10-3 molecules
= 3.24 × 1016 molecules.

Question 11.
A sealed tube that can withstand a pressure of 3 atm is filled with air at 27°C and 760 mm pressure. Find the temperature above which it will burst?
Answer:
Applying gas equation \(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\) as volume of the sealed tube remains constant .
\(\frac{1}{300}=\frac{3}{\mathrm{~T}_{2}}\)
T2 = 900 K
or
t2 = 900 – 273 = 627°C.

Question 12.
The temperature at the foot of a mountain is 30°C and pressure is 760 mm. Whereas at the top of the mountain these are 0°C and 710 mm. Compare the densities of the air at the foot and at the top.
Answer:
States of Matter Class 11 Important Extra Questions Chemistry 13
∴ the ratio of densities of air at the foot and the top of the mountain = 0.964: 1.

Question 13.
10.0 g of O2 were introduced into an evacuated vessel of 5-litre capacity maintained at 27°C. Calculate the pressure of the gas in the atmosphere in the container.
Answer:
Volume of the gas = V = 5 L;
Wt. of O2 = 10.0 g
Molar mass of O2 = 32.0
∴ No. of moles = \(\frac{10}{32}\)

T = 27 + 273 = 300 K; R = 0.0821 L atm K-1 mol-1
From ideal gas equation,
PV = nRT, we get
P = \(\frac{10}{32}\) × 0.0821 × 300 = 1.54 atm.

Question 14.
8.0 g of methane is placed in a 5 L container at 27°C. Find Boyle constant.
Answer:
Pressure × Volume is called Boyle’s constant
PV = nRT = \(\frac{W}{M}\) RT ,
= \(\frac{8}{16}\) × 0.0821 × 300
= 12.315 L atm

Question 15.
The density of A gas is 3.80 g L-1 at STP. Calculate its density at 27°C and 700 torr pressure
Answer:
d = \(\frac{PM}{RT}\)
For the same gas at two different pressures and temperatures,
States of Matter Class 11 Important Extra Questions Chemistry 14
States of Matter Class 11 Important Extra Questions Chemistry 15

Question 16.
If the density of a gas at sea level and 0°C is 1.29 kg m-3, what will be its molar mass (Assume that pressure is equal to 1 bar).
Answer:
States of Matter Class 11 Important Extra Questions Chemistry 16

Question 17.
A gaseous mixture contains 56 g N2, 44 g CO2 and 16 g CH4 The total pressure of the mixture is 720 mm Hg. What is the partial pressure of CH4?
Answer:
States of Matter Class 11 Important Extra Questions Chemistry 17

Question 18.
A 2 L flask contains 1.6 g of methane and 0.5 g of hydrogen at 27°C. Calculate the partial pressure of each gas in the mixture and hence calculate the total pressure.
Answer:
1.6 g CH4 = \(\frac{1.6}{16}\) = 0.1 Mole
16
Partial pressure of CH4 (pCH4) = CH4 × \(\frac{RT}{P}\)
States of Matter Class 11 Important Extra Questions Chemistry 18

Question 19.
One mole of S02 gas occupied a volume of 350 mL at 27°C and 50 atm pressure. Calculate the compressibility factor of the gas. Comment on the type of deviation shown by the gas from ideal behaviour.
Answer:
Compressibility factor, Z =
Here n = 1, P = 50 atm, V = 350 × 10-3 L = 0.35 L
R = 0.0821 L atm K-1 mol-1; T = 27 + 273 = 300 K
∴ Z = \(\frac{50 \times 0.35}{1 \times 0.0821 \times 300}\) = 0.711
For an ideal gas Z = 1
As for the given gas Z < 1, it shows a negative deviation, i.e., it is more compressible than expected from ideal behaviour.

Question 20.
A mixture of CO and CO2 is found to have a density of 1.5 g L-1 at 20°C and 740 mm pressure. Calculate the composition of the mixture.
Answer:
1. Calculation of average molecular mass of the mixture
M = \(\frac{d R T}{P}=\frac{1.50 \times 0.0821 \times 293}{\frac{740}{760}}\) = 37.06

2. Calculation of percentage composition
Let mol% of CO in the mixture = x
Then mol% of CO? in the mixture = 100 – x

Average molecular mass
States of Matter Class 11 Important Extra Questions Chemistry 19
States of Matter Class 11 Important Extra Questions Chemistry 20

Question 21.
A 5-L vessel contains 14 g of nitrogen. When heated to 1800 K, 30% of molecules are dissociated into atoms. Calculate the pressure of the gas at 1800 K.
Answer:
N2 ⇌ 2N

Initial moles = \(\frac{1.4}{28}\) = 0.05
Moles left = 0.05 – \(\frac{30}{100}\) × 0.05 = 2 × 0.015 = 0.03

∴ Total no. of moles = 0.035 + 0.030 = 0.065
i.e. n = 0.065 mol, V = 5 L, T = 1800 K; P =?
P = \(\frac{n \mathrm{RT}}{\mathrm{V}}=\frac{0.065 \times 0.0821 \times 1800}{5}\)
= 1.92 atm.

Question 22.
Calculate the temperature of 2 moles of S02 gas contained in a 5 L vessel at 10 bar pressure. Given that for SO2 gas van der Waal’s constant is a – 6.7 bar L2 mol-2 and b = 0.0564 L mol-1.
Answer:
According to van der Waal’s equation
States of Matter Class 11 Important Extra Questions Chemistry 21
States of Matter Class 11 Important Extra Questions Chemistry 22

Question 23.
A given mass of a gas occupies 919.0 mL in a dry state at STP. The same mass when collected over water at 15°C and 750 mm pressure occupies on litre volume. Calculate the vapour pressure of water at 15°C.
Answer:
If p is the vapour pressure of water at 15°C, then P2 = 750 – p
From the gas equation \(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\), we get
\(\frac{760 \times 919}{273}=\frac{(750-p) \times 1000}{288}\)
or
p = 13.3 mm

∴ Vapour pr. of water = 13.3 mm.

Question 24.
A steel tank containing air at 15 atm pressure at 15°C ¡s provided with a safety valve that will yield at a pressure of 30 atm. To what minimum temperature must the air be heated to below the safety valve?
Answer:
\(\frac{P_{1}}{P_{2}}=\frac{T_{1}}{T_{2}}\)
i.e., \(\frac{15}{30}=\frac{288}{\mathrm{~T}_{2}}\)
or
T2 = 576 K
or
t2°C = 576 – 273 = 303°C

Question 25.
Calculate the pressure exerted by 110 g of CO2 in a vessel of 2 L capacity at 37°C. Given that the van der Waals constants are a = 3.59 L2 atm mol-2 and b = 0.0427 L mol-1. Compare the value with the calculated value if the gas were considered ideal.
Answer:
According to van der Waals equation
States of Matter Class 11 Important Extra Questions Chemistry 23
110
Here, n = \(\frac{110}{44}\) = 2.5 moles. Putting the given values, we get
P = \(\frac{2.5 \times 0.0821 \times 310}{(2-2.5 \times 0.0427)}-\frac{3.59 \times 2.5}{2}\)
= 33.61 atm – 5.61 atm = 28.0 atm

If the gas were considered an ideal gas. applying ideal gas equation
PV = nRT
or
P = \(\frac{n \mathrm{RT}}{\mathrm{V}}\)
∴ P = \(\frac{2.5 \times 0.0821 \times 310}{2}\) = 31.8 atm

The s-Block Elements Class 11 Important Extra Questions Chemistry Chapter 10

Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 10 The s-Block Elements. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

Class 11 Chemistry Chapter 10 Important Extra Questions The s-Block Elements

The s-Block Elements Important Extra Questions Very Short Answer Type

Question 1
Which element is found in chlorophyll?
Answer:
Magnesium.

Question 2.
Name the elements (alkali metals) which form superoxide when heated in excess of air.
Answer:
Potassium, rubidium and caesium.

Question 3.
Why is the oxidation state of Na and K always + 1?
Answer:
It is due to their high second ionisation enthalpy and stability of their ions [Na+, K+].

Question 4.
Name the metal which floats on the water without any apparent reaction with water.
Answer:
Lithium floats on the water without any apparent reaction to it.

Lithium Oxide Formula. Lithium oxide also lithia is an inorganic compound.

Question 5.
Why do group 1 elements have the lowest ionisation enthalpy?
Answer:
Because of the largest size in their respective periods, solitary electron present in the valence shell can be removed by supplying a small amount of energy.

Question 6.
Why does the following reaction
C – Cl + MF → C – F + MCI
proceed better with KF than with NaF?
Answer:
Because larger K+ cation stabilises larger anion.

Question 7.
Amongst Li, Na, K, Rb, Cs, Fr which one has the highest and which one has the lowest ionisation enthalpy?
Answer:
Li has the highest and Fr has the lowest ionisation enthalpy.

Question 8.
What is the general electronic configuration of alkali metals in their outermost shells?
Answer:
ns1 where n = 2 to 7.

Question 9.
What is meant by dead burnt plaster?
Answer:
It is anhydrous calcium sulphate (CaS04).

Question 10.
Name three forms of calcium carbonate.
Answer:
Limestone, chalk, marble.

Question 11.
Which member of the alkaline earth metals family has
(i) least reactivity
Answer:
Be

(ii) lowest density
Answer:
Ca

(iii) highest boiling point
Answer:
Be

(iv) maximum reduction potential?
Answer:
Be.

Question 12.
Why does lithium show anomalous behaviour?
Answer:
Due to its small size and high charge/size ratio.

Question 13.
Out of LiOH, NaOH, KOH which is the strongest base?
Answer:
KOH.

Question 14.
Write the balanced equations for the reaction between
(a) Na2O2 and water
Answer:
2Na2O2 + 2H2O → 4NaOH + O2

(b) KO2 and water
Answer:
2KO2 + 2H2O → 2KOH + H2O + O2

(c) Na2O and CO2.
Answer:
Na2O4 + CO2 → Na2CO3

Question 15.
Arrange the following in order of increasing covalent character MCI, MBr. MF, MI (where M = alkali metal]
Answer:
MF < MCI < MBr < Ml.
With the increasing size of anion, covalent character increases

Question 16.
Name an element that is invariably bivalent and whose oxide is soluble in excess of NaOH and its dipositive ion has a noble gas core.
Answer:
The element is beryllium (Be) which forms a divalent ion and has a noble gas core [He] 2s2
Be2+ = 1 s2
BeO + 2NaOH → Na2BeO2 + H2O

Question 17.
Arrange the following in the increasing order of solubility in water:
MgCl2, CaCl2, SrCl2, BaCl2.
Answer:
BaCl2 < SrCl2 < CaCl2 < MgCl2.

Question 18.
State the reason for the high solubility of beryllium chloride in organic solvents.
Answer:
Beryllium chloride is a covalent compound.

Question 19.
Which alkali carbonate decomposes on heating to liberate CO2?
Answer:
Lithium carbonate [Li2CO3]
The s-Block Elements Class 11 Important Extra Questions Chemistry 1
Question 20.
Why is the solution of an alkali metal in ammonia blue?
Answer:
Due to the presence of ammoniated electrons.

Question 21.
Beryllium oxide has a high melting point. Why?
Answer:
Due to its polymeric nature.

Question 22.
Mention the chief reasons for the resemblance between beryllium and aluminium.
Answer:
Both Be2+ and Al3+ ions have high polarising power.

Question 23.
State any one reason for alkaline earth metals, in general, having a greater tendency to form complexes than alkali metals.
Answer:
Because of their small size and high charge, alkaline earth metals have a tendency to form complexes.

Question 24.
Amongst alkali metals why is lithium regarded as the most powerful reducing agent in aqueous solution?
Answer:
Lithium is the best reducing agent because it has the lowest reduction potential:
Li+ + e → Li(s) ERed = – 3.07V.

Question 25.
Name the metal amongst alkaline earth metals whose salt does not impart any colour to a non-luminous flame.
Answer:
Beryllium does not impart colour to a non-luminous flame.

Question 26.
What is the difference between baking soda and baking powder?
Answer:
Baking soda is sodium bicarbonate (NaHCO3) while baking powder is a mixture of sodium bicarbonate fNaHC03) and Potassium hydrogen tartrate.

Question 27.
Why does table salt get wet in the rainy season?
Answer:
Table salt contains impurities of CaCl2 and MgCl which being deliquescent compounds absorb moisture from the air during the rainy season.

Question 28.
Sodium readily forms Na+ ion, but never Na2+ ion. Explain.
Answer:
After the removal of 1 electron from Na, it has been left with a noble gas core [Ne] which is closer to the nucleus and requires more energy.

Question 29.
Which compound of sodium is used
(i) as a component of baking powder
Answer:
NaHCO3

(ii) for softening hard water.
Answer:
Na2CO3.

Question 30.
Anhydrous calcium sulphate cannot be used as Plaster of Paris. Give reason.
Answer:
Because it does not have the ability to set like plaster of Paris.

Question 31.
Which of the following halides is insoluble in water?
CaF2, CaCl2, CaBr2 and Cal2.
Answer:
CaF2.

Question 32.
Predict giving reason the outcome of the reaction.
Lil + KF → …………
Answer:
LiI + KF → LiF + KI
Larger cation stabilises larger anion.

Question 33.
Name three metal ions that play important role in performing several biological functions in the animal body.
Answer:
Na+, K+, Ca2+, Mg2+, etc.

Question 34.
Calcium metal is used to remove traces of air from vacuum tubes. Why?
Answer:
Calcium has a great affinity for oxygen and nitrogen.

Question 35.
Why is sodium kept in kerosene oil?
Answer:
To prevent its contact with oxygen and moist air, because sodium reacts with them.

Question 36.
What is brine?
Answer:
An aqueous solution of NaCl in water.

Question 37.
Why caesium can be used in photoelectric cells while lithium cannot be?
Answer:
Cs has the lowest & Li highest ionisation enthalpy. Hence Cs can lose electron very easily while lithium cannot.

Question 38.
In an aqueous solution, the Li+ ion has the lowest mobility. Why?
Answer:
Li+ ions are highly hydrated in an aqueous solution.

Question 39.
Lithium has the highest ionisation enthalpy in group-1, yet it is the strongest reducing agent. Why?
Answer:
This is because Li has the highest oxidation potential.
Li → Li+ + e
ox = + 3.07

Question 40.
Name one reagent or one operation to distinguish between
(i) BeSO4 and BaSO4
Answer:
BeSO4 is soluble in water while BaSO4 is not.

(ii) Be(OH)2 and Ba(OH)2
Answer:
Be(OH)2 dissolves in alkali, while Ba(OH)2 does not.

Question 41.
Which alkaline earth metal hydroxide is amphoteric?
Answer:
Be(OH)2.

Question 42.
Which alkali metal is radioactive?
Answer:
Francium (Fr).

Question 43.
Which alkaline earth metal is radioactive?
Answer:
Radium (Ra).

Question 44.
Name the alkali metal which shows a diagonal relationship with Mg.
Answer:
Lithium (Li).

Question 45.
Give the chemical formula of carnallite.
Answer:
KCl.MgCl2.6H2O.

Question 46.
Arrange CaSO4, SrSO4 & BaSO4 in order of decreasing solubility.
Answer:
The order of decreasing solubility of these sulphates is CaSO4, SrSO4 & BaSO4.

Question 47.
Why is K more reactive than Na?
Answer:
Since the valence electron of K is relatively at a greater distance than Na & it requires lesser energy to remove it.

Question 48.
Why does Beryllium show similarities with Al?
Answer:
Because of their similarity in charge/radius ratio (Be2+ = 0.064 & Al3+ = 0.660)

Question 49.
What is magnesia?
Answer:
Magnesium oxide (MgO).

Question 50.
How is Potassium extracted?
Answer:
By electrolysis of a fused solution of KOH.

The s-Block Elements Important Extra Questions Short Answer Type

Question 1.
Why the solubility of alkaline metal hydroxides increases down the group?
Answer:
If the anion and the cation are of comparable size, the cationic radius ‘vill influence the lattice energy. Since lattice energy decreases much more than the hydration energy with increasing ionic size, solubility will increases as we go down the group. This is the case of alkaline earth metal hydroxides.

Question 2.
Why the solubility of alkaline earth metal carbonates and sulphates decreases down the group?
Answer:
If the anion is large compared to the cation, the lattice; energy will remain almost constant within a particular group. Since the hydration energies decrease down the group, solubility will decrease as found for alkaline earth metal carbonates and sulphates.

Question 3.
Why cannot potassium carbonate be prepared by the SOLVAY process?
Answer:
Potassium carbonate cannot be prepared by the SOLVAY process because potassium bicarbonate (KHCO3) is highly soluble in water, unlike NaHCO3 which was separated as crystals. Due to its high solubility KHCO3 cannot be precipitated by the addition of ammonium bicarbonate to a saturated solution of KCl.

Question 4.
What are the main uses of calcium and magnesium?
Answer:
Main uses of calcium:

  1. Calcium is used in the extraction of metals from oxides which are difficult to reduce with carbon.
  2. Calcium, due to its affinity for O2 and N2 at elevated temperatures, has often been used to remove air from vacuum tubes.

Main uses of Magnesium:

  1. Magnesium forms alloys with Al, Zn, Mn and Sn. Mg-Al alloys being light in mass are used in aircraft construction.
  2. Magnesium (powder and ribbon) is used in flashbulbs, powders incendiary bombs and signals.
  3. A suspension of Mg(OH)2 in water is used as an antacid in medicine.
  4. Magnesium carbonate is an ingredient of toothpaste.

Question 5.
What is meant by the diagonal relationship in the periodic table? What is it due to?
Answer:
It has been observed that some elements of the second period show similarities with the elements of the third period situated diagonally to each other, though belonging to different groups. This is called a diagonal relationship.
The s-Block Elements Class 11 Important Extra Questions Chemistry 2
The cause of the diagonal relationship is due to the similarities in properties such as electronegativity, ionisation energy, size etc. between the diagonal elements. For example on moving from left to right across a period, the electronegativity increases, which on moving down a group, electronegativity decreases. Therefore on moving diagonally, two opposing tendencies almost cancel out and the electronegativity values remain almost the same as we move diagonally.

Question 6.
Why is the density of potassium less than that of sodium?
Answer:
Generally, in a group density increases with the atomic number, but potassium is an exception. It is due to the reason that the atomic volume of K is nearly twice Na, but its mass (39) is not exactly double of Na (23). Thus the density of potassium is less than that of sodium.

Question 7.
The hydroxides and carbonates of sodium and potassium are easily soluble in water while the corresponding salts of magnesium and calcium are sparingly soluble in water. Explain.
Answer:
Sodium and potassium ions (Na+ and K+) are larger than the corresponding Mg2+ and Ca2+ ions. Due to this lattice energy of Mg(OH)2, Ca(OH)2, MgCO3 and CaCO3. (Lattice energy is defined as the energy required to convert one mole of the ionic lattice into gaseous ions.

Thus lattices with smaller ions have higher lattice energies). The hydration energies of Mg2+ and Ca2+ are higher than Na+ and K+ because of their smaller sizes. But the difference in lattice energies is much more. Therefore, the hydroxides and carbonates of Mg2+ and Ca2+ are insoluble in water because of their higher lattice energies.

Question 8.
Why is it that the s-block elements never occur in free state/nature? What are their usual modes of occurrence and how are they generally prepared?
Answer:
The elements belonging to the s-block in the periodic table (i.e. alkali and alkaline earth metals) are highly reactive because of their low ionisation energy. They are highly electropositive forming positive ions. So they are never found in a free state.

They are widely distributed in nature in the combined state. They occur in the earth’s crust in the form of oxides, chlorides, silicates and carbonates.

Generally, a group I metals are prepared by the electrolysis of fused solution.
For example:

1. The s-Block Elements Class 11 Important Extra Questions Chemistry 3
At cathode: Na+ + e → Na
At anode: Cl → Cl + e
Cl + Cl → Cl2

2. KOH ⇌ K+ + OH
At cathode: K+ + e→ K
At anode: 4OH → 4OH + 4e
4OH → 2H2O + O2
or
4OH → 2H2O + O2 + 4e
These metals are highly reactive and therefore cannot be extracted by the usual methods, because they are strong reducing agents.

Question 9.
Explain what happens when
(i) Sodium hydrogen carbonate is heated.
Answer:
The s-Block Elements Class 11 Important Extra Questions Chemistry 4

(ii) Sodium with mercury reacts with water.
Answer:
2Na-Hg + 2H2O → 2NaOH + H2 ↑ + 2Hg

(iii) Fused sodium metal reacts with ammonia.
Answer:
The s-Block Elements Class 11 Important Extra Questions Chemistry 5

Question 10.
What is the effect of heat on the following compounds?
(a) Calcium carbonate
Answer:
The s-Block Elements Class 11 Important Extra Questions Chemistry 6

(b) Magnesium chloride hexahydrate
Answer:
The s-Block Elements Class 11 Important Extra Questions Chemistry 7

(c) Gypsum
Answer:
The s-Block Elements Class 11 Important Extra Questions Chemistry 8

(d) Magnesium sulphate heptahydrate.
Answer:
The s-Block Elements Class 11 Important Extra Questions Chemistry 9

Question 11.
Why is it that the s-block elements never occur in a free state? What are their usual modes of occurrence?
Answer:
The elements belonging to s-block in the periodic table. These metals (Alkali & alkaline earth metals) are highly reactive because of their low ionization energy. They are highly electropositive forming positive ions. So they are never found in a free state. They are widely distributed in nature in a combined state. They occur in the earth’s crust in the form of oxides, chlorides, silicates & carbonates.

Question 12.
Explain what happens when:
(a) Sodium hydrogen carbonate is heated.
Answer:
Sodium hydrogen carbonate on heating decomposes to sodium carbonate.
The s-Block Elements Class 11 Important Extra Questions Chemistry 10

(b) Sodium with Mercury reacts with water.
Answer:
When sodium with mercury reacts with water. It produces sodium hydroxide.
The s-Block Elements Class 11 Important Extra Questions Chemistry 11

Question 13.
What is the effect of heat on the following compounds:
(a) Calcium carbonate
Answer:
The s-Block Elements Class 11 Important Extra Questions Chemistry 12

(b) Magnesium chloride hexahydrate
Answer:
The s-Block Elements Class 11 Important Extra Questions Chemistry 13

(c) Gypsum
Answer:
The s-Block Elements Class 11 Important Extra Questions Chemistry 14

Question 14.
State as to why
(a) An aqueous solution of sodium carbonate gives an alkaline test.
Answer:
Sodium carbonate gets hydrolise by water to form an alkaline solution.
CO32- + H2O → HCO3 + OH
Due to this, it gives an alkaline test.

(b) Sodium is prepared by electrolytic method & not by chemical method.
Answer:
Sodium is a very strong reducing agent. Therefore, it cannot be isolated by a general reduction of its oxides or other compounds. The metal formed by electrolysis will immediately react with water forming hydroxides. So, sodium is prepared by the electrolytic method only.

(c) Lithium on being heated in the air mainly forms mono-oxide & not the peroxides.
Answer:
Lithium is the least reactive but the strongest reducing agent of all the alkali metals. It combines with air, it forms mono-oxide only because it does not react with excess air.
The s-Block Elements Class 11 Important Extra Questions Chemistry 15
Question 15.
Like Lithium in group-I, beryllium shows anomalous behaviour in group II. Write three such properties of beryllium which makes it anomalous in the group.
Answer:
1. Beryllium has an exceptionally small atomic size, due to high ionization energy & small atomic size it forms compounds that are largely covalent & its salts are hydrolysed easily.

2. Beryllium does not exhibit coordination N. more than four as in its valence shell (n = 2). There are only four orbitals. The remaining member of the group has co-ordination no. six by making use of group have co-ordination no. six by making use of some d-orbitals.

3. The oxides & hydroxides of beryllium quite unlike the other elements, in this group are amphoteric in nature.

Question 16.
Explain which one Na or K has a larger atomic radius?
Answer:
Potassium has a larger atomic size than sodium because, in K, the outermost electron is in the fourth energy state (4s1) while in Na, the outer-most electron is in the third energy state (3s1). That is r (K) > r (Na) since the fourth energy state (n = 4) in K is farther away from the nucleus than the third energy state (n = 3) in Na.

Question 17.
The alkali metals (group-1 elements) form only unipositive cations (M+) & not bivalent cations (M+2), Give reason.

Answer:
Alkali metals have an ns1 valence shell configuration. They can give up this electron to form univalent cations & attain stable electronic configuration like their nearest inner gas. Now the removal of an electron from the closed-shell configuration requires a very large amount of energy. Which is not available in chemical reactions. Therefore, the alkali metals form a univalent cation (M+) & No divalent cations (M2+) are formed.

Question 18.
Complete the following equations:
(a) Ca + H2O →?
Answer:
The s-Block Elements Class 11 Important Extra Questions Chemistry 16

(b) Ca(OH)2 + Cl2 →?
Answer:
2Ca(OH)2 + l → Ca(OCl)2 + CaCl2 + H2O

(c) BeO + NaOH →?
Answer:
BeO + 2NaOH → Na2BeO2 + H2O

Question 19.
How calcium carbonate can be changed into
(a) Calcium sulphate
(b) Calcium oxide?
Write a balanced chemical equation in each case.
Answer:
(a) When calcium carbonate is heated with dil. sulphuric acid, calcium sulphate is obtained.
CaCO3(s) + H2SO4(aq) → CaSO4(s) + CO2(g) + H2O(l)

(b) Calcium carbonate is heated at a high temperature between 1070 – 1270 K. It decomposes to give calcium oxide & carbon dioxide.
The s-Block Elements Class 11 Important Extra Questions Chemistry 17
Question 20.
Explain the following phenomenon by means of balanced equations:
(a) When exhaling is made through a tube passing into a solution of lime water, the solution becomes turbid.
Answer:
Ca(OH)2 (l) + CO2 (g) → CaCO3 (s) + H2O (l)

(b) The turbidity of solution in (a) eventually disappears when continued exhaling is made through it.
Answer:
CaCO3(s) + H2O(l) + CO2(g) → Ca(HCO3)2(aq)

(c) When the solution obtained in (B) is heated turbidity re-appears.
Answer:
Ca(HCO3)2(aq) → CaCO3(s) + H2O(l) + CO2(g)

Question 21.
Describe the reactivity of alkaline earth metals with water.
Answer:
The reactivity of alkaline earth metals with water increases with increasing atomic numbers.

  1. Beryllium does not react readily even with boiling water.
  2. Magnesium reacts very slowly with cold water. While it reacts with hot water at an appreciable rate & liberates hydrogen gas.
    The s-Block Elements Class 11 Important Extra Questions Chemistry 18
  3. Calcium, strontium & Barium react vigorously even with cold water & form hydroxides & liberate hydrogen gas
    M(s) + H2O(l) → M(OH)2 + H2(g)
    [M = Ca, Sr, Ba]

Question 22.
Compare the reactivity of alkali metals with water.
Answer:
All the alkali metals displace hydrogen from water i.e.
2M + 2H2O → 2MOH + H2

Lithium reacts slowly with water, sodium reacts violently & Cs reacts so vigorously that the reaction if not controlled can lead to an explosion.
2Li(s) + 2H2O(aq) → 2LiOH (aq) + H2(g) Slow
2Na(s) + 2H2O(aq) → 2NaOH (aq) + H2(g) Fast
2Cs(s) + 2H2O(aq) → 2CsOH (aq) + H2(g) Violent

Question 23.
Alkaline earth metals are harder than alkali metals. Give reason.
Answer:
Alkaline earth metals are harder than alkali metals because:

  1. The atomic radius of alkaline metals is small, atomic mass is high & density is larger than those of alkali metals. Alkaline earth metals have closed packed crystal structure.
  2. In alkaline earth metals, the metallic bonding in its crystal is very strong as compared to the crystal of an alkali metal. Therefore, the atoms in the crystal of alkaline earth metals are strongly bonded.

Question 24.
Differentiate amongst Quick lime, Lime water & Slaked lime.
Answer:
The s-Block Elements Class 11 Important Extra Questions Chemistry 19

Question 25.
The mobilities of alkali metal ions in aqueous solution follow the order
Li+ < Na+ < K+ < Rb+ < Cs+ Give reasons.
Answer:
Smaller ions are expected to have more mobility. But smaller ions such as Li+, Na+ due to their higher charge density tend to undergo hydration. Hydration increases the mass & effective size of the smaller ion & as a result, the mobility of ions show the trend given.

Question 26.
How is sodium peroxide manufactured? What happens when it reacts with chromium (III) hydroxide? Give it’s two uses.
Answer:
Sodium peroxide is manufactured by heating sodium metal on aluminium trays in the air (free-from CO2).
2Na + O2 (air) → Na2O2

Sodium Peroxide oxidises chromium (III) hydroxide to sodium chromate.
The s-Block Elements Class 11 Important Extra Questions Chemistry 20
Uses:

  • It is used as a bleaching agent because of its oxidising, power.
  • It is used in the manufacture of dyes & many other chemicals such as benzoyl peroxide, sodium perborate etc.

Question 27.
List three properties of Lithium in which it differs from the rest of the alkali metals.
Answer:
(a) Lithium is much harder, its melting & boiling points are higher than the other alkali metals.
(b) Lithium is the least reactive but the strongest reducing agent among all the alkali metals.
(c) LiCl is deliquescent & crystallise as a hydrate, LiCl.2H2O whereas other alkali metal chlorides do not form hydrates.

Question 28.
Why is LiF almost insoluble in water whereas LiCl is soluble not only in water but also in acetone?
Answer:
The low solubility of LiF is due to its high lattice enthalpy. LiCl has much higher solubility in water. This is due to the small size of Li+ ion & much higher hydration energy.

Question 29.
What happens when
(1) Calcium nitrate is heated,
Answer:
The s-Block Elements Class 11 Important Extra Questions Chemistry 21
(2) Chlorine reacts with slaked lime
Answer:
The s-Block Elements Class 11 Important Extra Questions Chemistry 22
(3) Quick lime is heated with silica.
Answer:
The s-Block Elements Class 11 Important Extra Questions Chemistry 23
Question 30.
Arrange the following in order of property mentioned:
(1) Mg(OH)2, Sr(OH)2, Ba(OH)2, Ca(OH)2
(Increasing ionic solubility in water)
Answer:
Mg(OH)2, Ca(OH)2, Sr(OH)2, Ba(OH)2

(2) BeO, MgO, BaO, CaO
(Increasing basic character)
Answer:
BeO, MgO, CaO, BaO

(3) BaCl2, MgCl2, BeCl,, CaCl2
(Increasing ionic character)
Answer:
BeCl2, MgCl2, CaCl2, BaCl2

The s-Block Elements Important Extra Questions Long Answer Type

Question 1.
What raw materials are used for making cement? Describe the manufacture of Portland cement. What is its approximate composition?
Answer:
Raw materials: The raw materials required for the manufacture of cement are limestone, stone and clay, limestone in calcium carbonate, CaCO3 and it provides calcium oxide, CaO. Clay- is a hydrated aluminium silicate, Al2O3 2Si02.2H20 and it provides alumina as well as silica. A small amount of gypsum, CaS04.2H20 is also required. It is added in calculated quantity in order to adjust the rate of setting of cement.

Manufacture: Cement is made by strongly heating a mixture of limestone and clay in a rotatory kiln. Limestone and clay are finely powdered and a little water is added to get a thick paste called slurry. The slurry is led into a rotatory kiln from the top through the hopper.

The hot gases produce a temperature of about 1770-1870 K in the kiln. At this high temperature, the limestone and clay present in the slurry combine to form cement in the form of small pieces called clinker. This clinker is mixed with 2 – 3 % by weight of gypsum (CaSO4 .2H2O) to regulate the setting time and is then ground to an exceedingly fine powder.
The s-Block Elements Class 11 Important Extra Questions Chemistry 24
Manufacture of cement
The s-Block Elements Class 11 Important Extra Questions Chemistry 25
When mixed with water the cement reacts to form a gelatinous mass which sets to a hard mass when three-dimensional cross-links are formed between
…………..Si — O — Si……… and …….. Si — O — Al…… chains.

Composition of cement:
CaO = 50 – 60%
SiO2 = 20 – 25%
Al2O3 = 5 – 10%
MgO = 2 – 3%
Fe2O3 = 1 – 2%
SO3= 1 – 2%

For a good quality cement, the ratio of silica (SiO2) and alumina (Al2O3) should be between 2.5 to 4.0. Similarly, the ratio of lime (CaO) to the total oxide mixtures consisting of SiO2, Al2O3 and Fe2O3 should be roughly 2: 1. If lime is in excess, the cement cracks during setting. On the other hand, if lime is less than required, the cement is weak in strength. Therefore, the proper composition of cement must be maintained to get cement of good quality.

Statistics Class 11 Notes Maths Chapter 15

By going through these CBSE Class 11 Maths Notes Chapter 15 Statistics Class 11 Notes, students can recall all the concepts quickly.

Statistics Notes Class 11 Maths Chapter 15

Use this online frequency distribution calculator to calculate from discrete frequency distributions.

Before we discuss about the measure of deviations, we are expected to know how to calculate arithmetic mean \(\bar{x}\) and median.
To find mean of a distribution, we proceeded as :
For ungrouped data : x1, x1 …….. xn.
\(\bar{x}=\frac{x_{1}+x_{2}+\ldots+x_{n}}{n}=\frac{\Sigma x_{i}}{n}\)

For discrete frequency distribution
Statistics Class 11 Notes Maths Chapter 15 1

For Continuous Frequency distribution : Let a be the assumed mean.
∴ di = xi – a, h = class interval
and N = Total frequency = f1 + f2 + … fn
Then \(\bar{x}=a+\left(\frac{\Sigma f_{i} d_{i}}{N}\right) \times h\)
To calculate the median, the process is as follows :
For ungrouped data : Arrange the data in ascending or descending order.

If n is odd, \(\left(\frac{n+1}{2}\right)\) th value = Median
If n is even \(\frac{1}{2}\left[\frac{n}{2} \text { th value }+\left(\frac{n}{2}+1\right)^{\text {th }} \text { value }\right]\) = Median
Same method is adopted in case of discrete frequency distribution.

For continuous frequency distribution:
(i) Find the cumulative frequencies :
(ii) \(\frac{\mathrm{N}}{2}\)th or next of \(\frac{\mathrm{N}}{2}\)th cumulative frequency correspond to median class.
(iii) Cumulative frequency C just before \(\frac{\mathrm{N}}{2}\)th c.f. is denoted by C.
(iv) l = lower limit of median class
h = class interval of median class
f= frequency corresponding to median class
Then, median = M = l + \(\left(\frac{\frac{N}{2}-C}{f}\right)\)x h

Measures of dispersion are required to study the degree of scattemess of observations from the central value. The central value may be mean (or median).

Mean deviation from the mean : If a set of n observations x1 x2, … xn have their mean \(\bar{x}\), then mean deviation from the mean \(\bar{x}=\frac{\sum_{i=1}^{n}\left|x_{i}-\bar{x}\right|}{n}\) , where |xi – \(\bar{x}\)| is the abosolute value of the deviation of xi from\(\bar{x}\) .

In case, the distribution is a grouped one, the mean deviation from the mean is given by
M.D \((\bar{x})=\frac{\sum_{i=1}^{k} f_{i}\left|x_{i}-\bar{x}\right|}{\sum_{i=1}^{k} f_{i}}\)
where k is the number of classes.

Mean deviation from the median : If a set of n observations are x1, x2, x3, ………….xn, then mean deviation from the median = M.D. (Med.)
= \(\frac{\Sigma \mid x_{i}-\text { Median } \mid}{n}\)
In case of grouped data, M.D. (Med.) = \(\frac{\Sigma f_{i} \mid x_{i}-\text { Median } \mid}{\Sigma f_{i}}\)

Variance and Standard Deviation: In case of mean deviation from the mean, we take the absolute values of the deviations from the mean, so that their sum does not become zero. Another way is to find the square of the deviations and then add up. It may be possible that \(\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}\) in case of distributions having large number of observations but small deviations may be larger than that in case of a distribution having small number of observations but larger deviations which may not be true. As such, to overcome this difficulty, we take average of the squares of deviations.

Thus, variance = \(\frac{\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}}{n}\) and standard deviation = σ = \(\sqrt{\text { variance }}=\sqrt{\frac{\Sigma\left(x_{i}-\bar{x}\right)^{2}}{n}} .\)

Short-cut Method to Calculate Variance for discrete frequency distribution :
Statistics Class 11 Notes Maths Chapter 15 2

For continuous frequency distribution : Let a be the assumed mean and h be the class interval.
Now, we calculate the variable, yi as
yi = \(\frac{x_{i}-a}{h}\)
Statistics Class 11 Notes Maths Chapter 15 3

Comparison of variability of distribution : To compare the variability of two or more distributions, we calculate the coefficient of variation of each distribution, i.e.,
Coefficient of variance (C.V.)
= \(\frac{\text { Standard Deviation }}{\text { Mean }}\) x 100
= \(\frac{\sigma}{x}\) x 100
Greater is the value of coefficient of variation of a distribution, there is more variability in that distribution.

Ascending and descending order calculator is a key tool for users to generate step-by-step

Linear Inequalities Class 11 Notes Maths Chapter 6

By going through these CBSE Class 11 Maths Notes Chapter 6 Linear Inequalities Class 11 Notes, students can recall all the concepts quickly.

Complex Numbers and Quadratic Equations Notes Class 11 Maths Chapter 6

Important Definitions: (a) A statement involving variable(s) an a sign of inequality, viz. ,>, <, ≥ or ≤ is called an inequality.
(b) Linear inequality : An inequality involving one or two variables in linear power (in single power) is known as linear inequality.
(c) Quadratic inequality: An inequality involving one or two variables in quadratic powers (i.e., in second degree) is known as quadratic inequality.

Important Rules:

Rule 1: Equal numbers may be added to (or subtracted from) both sides of an inequality without affecting the sigh of inequality.
Rule 2: Both sides of an inequality can be multiplied (or divided) by the same positive number. But, when both sides are multiplied or divided by a negative number, then the sign of inequality is reversed.

Solution of System of Linear Inequalities in One Variable
Procedure adopted :

(a) Find all the values of the variables satisfying each of the given inequalities.
(b) Find the values of the variable, which are common to the given inequalities.
These common values of the variables are the solutions of the given inequalities.

Learn about solving inequality calculator using our free math solver with step-by-step solutions.

Note : Range of the values of the solutions of the given inequalities is known as the solution interval of the given inequality.

Graphical Solution of Linear Inequalities in Two Variables

Important definitions : The region containing all the solutions of an inequality is called the solution region.

Note: If an equation involves sign of equality, then the points on the line are also included in the solution. In other cases, these points are not included in the solution. In this case, the line is drawn dotted or broken.

Procedure to Draw Graph of an Inequality

1. First draw the graph of the equation corresponding to the given inequality. To draw the graph of an equation, find the point on the x-axis (by putting y = 0 in the equation) and on they-axis (by putting x = 0 in the equation). Join these two points.

2. This line will divide the coordinate plane in two half plane regions, viz.,
(a) Half plane region I below the line.
(b) Half plane region II above the line.

3. To identify the half plane represented by the inequality, we take a point (a, b) not lying on the line. If putting the point (a, 6) in the inequality, the inequality is satisfied, then the point is situated in the region represented by the inequality otherwise not.

Note : Generally take (0, 0) as the arbitrary point. Sometimes we take the point (1, 1) also.

Solutions of System of Linear Inequalities in Two Variables

Procedure adopted : (a) Draw the graph of the inequalities following the method learnt as stated above reference to the same coordinate axes.
(b) Shade the relevant solution region of each inequality.
(c) Multishaded region will be the solution region of the given inequalities.

Applications : Here in this section, knowledge of solving inequalities will be utilized in solving problems from different fields such as economics, science, mathematics, psychology, etc.

Sequences and Series Class 11 Notes Maths Chapter 9

By going through these CBSE Class 11 Maths Notes Chapter 9 Sequences and Series Class 11 Notes, students can recall all the concepts quickly.

Sequences and Series Notes Class 11 Maths Chapter 9

Sequence : A set of numbers arranged according to some definite rule is called a sequence.

The different numbers which form the sequence are called the terms of the sequence. Terms are denoted by the symbols T1 T2, T3, … etc. the subscript denotes the position of the term. The nth term is also called the general term of the sequence.

Sequence containing finite number of terms is called a finite sequence and a sequence is called infinite, if it is not a finite sequence.

In a sequence, we should not always expect that the terms of a sequence will be necessarily given by an algebraic formula. For example, in a sequence of prime numbers 2, 3, 5, 7, …, there is no known formula for the prime numbers. But in the sequence, we do have a rule or law for writing its numbers in order.

Series : If the numbers forming the sequence are connected by the signs of addition (+), we get a series.
The numbers of the series are known as the terms of the series.

Progression : If the terms of a sequence constantly increase or decrease in a set pattern, it is called a progression.

Arithmetic Progression (A.P.) : A sequence in which the difference of any term from the previous term is constant, is called an arithmetic progression and the constant difference is called the common difference.

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Notation :
(i) The first term is denoted by a.
(ii) The common difference is denoted by d.
(iii) d = T2 – T1 = T3 – T2 = T4 — T3 = … = Tn — Tn-1

nth term or general term of an A.P.: If a be the first term and d, the common difference of an A.P., then Tn = a + (n – 1 )d.
i.e., nth term = first term + (n – 1) x common difference.

Properties of an A.P.: 1. If a constant is added to each term of an A.P., the resulting sequence is also an A.P.

2. If a constant is subtracted from each term of an A.P., then resulting sequence is also an A.P.

3. If each term of an A.P. is multiplied by a constant, then the resulting sequence is also an A.P.

4. If each term of an A.P. is divided by a non-zero constant, then the resulting sequence is also an A.P.

Sum to n terms of an A.P.
(i) Sn = \(\frac{n}{2}\) (a + 1), where T1 = a, Tn = 1.

(ii) Sn = \(\frac{n}{2}\) [2a + (n – 1)d], where T1, = a, c.d. = d.

Limit of sequence calculator is the value of the series is the limit of the particular sequence.

Notes:
1. Formula (i) is used, when the last term is known and formula (ii) is used when the common difference is known.
2. These formulae have four quantities each. If three are known, the fourth can be found out.

Arithmetic Means : When three quantities are in A.P., the middle quantity is said to be Arithmetic Mean (A.M.) between the other two.

Thus, if, a, A, 6 are in A.P., then A is the A.M. between a and b.
∴ A-a = b-A [∵Each-c..d.]
⇒ 2A – a + b
A = \(\frac{a+b}{2}\)
A.M. between two numbers = Half their sum

An important result: Sum of n A.M.’s between two quantities is n times the single A.M. between them.
n Arithmetic means : Let A1, A2, A3, …, An be the n A.M.’s between two numbers a and b.
The total number of A.M.’s and numbers a and b = (n + 2)
The last term = b.

b = a + (n + 2- 1 )d i.e., d = \(\frac{b-a}{n+1}\)
∴ We can find the A.M.’s as
A1 = a + d = a + \(\frac{b-a}{n+1}\) , A2 = a + 2d = a + \(2\frac{b-a}{n+1}\) ……
An = a + nd = a + \(n\frac{b-a}{n+1}\).

Geometric progression (G.P.): A sequence in which the ratio of any term to the proceeding term is the same throughout, is called a Geometric Progression and the constant ratio is called the common ratio of G.P.

Standard form of G.P. : If a is the first term and r is the common ratio, then a, ar, ar2, … is the standard form of G.P.

nth term or general term of G.P.:
Tn = arn-1>, where T1 = a and c.r. = r.
The above formula has four quantities a, r, n and Tn. Out of these four, if any three are given, fourth can be found out.

Sum of n terms of a G.P.: If G.P. is a, ar, ar2,…; arn-1, then
(i) For, r = 1,
Sn = a + a + a + … + a (n terms)
= na
(ii) For r ≠ 1,
Sn = \(\frac{a\left(1-r^{n}\right)}{1-r}\), if |r| < 1
and Sn =\(\frac{a\left(r^{n}-1\right)}{1-r}\), if |r| > 1

Geometric Means : When three numbers are in G.P., then, middle one is called the Geometric Mean (G.M.) between the other two.

Thus, if a, G, b are in G.P., then G is the G.M. between a and b. It can also be defined as under :
If any number of numbers are in G.P., then all the terms between the first and last terms are called Geometric Means (G.M.’s) between them.

If a, G, b are in G.P. then
\(\frac{\mathrm{G}}{a}=\frac{b}{\mathrm{G}}\)
[v each = common ratio]
G2 = ab ⇒ G = \(\sqrt{a b}\) .

An important result: The product of n G.M.’s between a and b is equal to the nth power of the G.M. between a and b.

n-Geometric Means : Let G1 G2, G3, …, Gn be the n G.M.’s between a and b. Total number of terms = n G.M’s and the numbers a and b = n + 2.
∴ (n + 2)th term = last term = b.
If r is common ratio, b = arn+2-1 + 2 – 1 = arn+1
r = \(\left(\frac{b}{a}\right)^{n \frac{1}{n+1}}\)
G1 = ar = a\(\left(\frac{b}{a}\right)^{\frac{1}{n+1}}\) ,
G2 = ar2 = a\(\left(\frac{b}{a}\right)^{\frac{2}{n+1}}\),……. Gn = arn = a\(\left(\frac{b}{a}\right)^{\frac{n}{n+1}}\)

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Sum of first n natural numbers :
Σn = 1 + 2 + 3 + … + n = \(\frac{n(n+1)}{2}\)

Sum of the squares of the first n natural numbers

Σn2 = 12 + 22 + 32 + … + n2 = \(\frac{n(n+1)(2 n+1)}{6}\)

Sum of the cubes of the first n natural numbers
Σn3 = 13 + 23 + 33 + … + n3 = \(\left[\frac{n(n+1)}{2}\right]^{2}\)

Clearly, Σn3 = (Σn)2

Sum of it terms of a series whose nth term is an3 + bn2 + cn + d :
If Tn = an3 + bn2 + cn + d, then Sn = aΣn3 + bΣn2 4 cΣz + dn.

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