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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 K_p for the reaction 3Fe (s) + 4H₂O (g) ⇌ Fe₃O₄ (s) + 4H₂ (g)
Answer:
K_p = \(\frac{p_{\mathrm{H}_{2}}^{4}}{p_{\mathrm{H}_{2} \mathrm{O}}^{4}}=\frac{p_{\mathrm{H}_{2}}}{p_{\mathrm{H}_{2} \mathrm{O}}}\)

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

Question 3.
What is the equilibrium constant expression for the reaction
Al (s) + 3H⁺ (aq) ⇌ Al³⁺ (aq) + \(\frac{3}{2}\)H₂ (g)
Answer:
K_c = [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/K_b.
K_b 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:
K_c = [CaO (s)][CO₂ (g)J/[CaCO₃ (s)]
K_c = [CO₂ (g)]
Similarly, K_p = P_CO2.

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) ⇌ 2NH^3-(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 K_p = K_c × (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 p^H of 0.1 M HCl?
Answer:
p^H = – 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) + CrO₄^2- (aq)
Answer:
K_sp = [Ag+]²[CrO₄^2-].

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

Question 29.
The K_sp 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 p^H 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 p^H of ammonium acetate solution when a few drops of acid are added to it?
Answer:
p^H 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 K_sp 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.
K_sp 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 p^H of a buffer solution?
Answer:
Dilution with water of a buffer solution has no effect on the p^H of a buffer solution.

Question 49.
What is the p^H of our blood? Why does it not change in spite of the variety of foods and Spices we eat?
Answer:
the p^H 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⁺] [SO₄^2- ] 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

[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 p^K_a and p^K_b values where K_a, K_b represent ionization constants of the acid and its conjugate base respectively?
Answer:
p^K_a + p^K_b = p^K_w =14.

Question 57.
What is the relationship between p^H and p^OH?
Answer:
p^H + p^OH= p^K_w = 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 p^OH? What is its value for pure water at 298 K?
Answer:
p^OH = – log [OH^–]
p^H + p^OH = 14 for pure water at 298 K
p^H = 7
or
p^OH of water at 298 = 7.

Question 3.
Calculate the p^H 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:

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
K_c = \(\frac{x^{2}}{4(1-x)^{2}}\)
where x = degree of dissociation

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

K_c for HI does not have a volume factor- and dissociation is independent of volume and hence pressure.
K_c 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:

Total moles at eqbm. = 1 – α + 2α = 1 + α
If P is total pressure

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 K_p to K_c for the
2SO₂ + O₂ (g) ⇌ 2SO₃ (g)
Answer:

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:

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:
K_c = [CH₃OH]/[CO][H₂]², K_p = PCH₃OH/PCO × PH₂
When the volume of the vessel is reduced to half, the concentration of each reactant or product becomes double. Thus
Q_c = 2[CH₃OH]/2[CO] × {2[H₂]}² = i K..
As Q_c < K_p, 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.
(K_i 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:
K_a for HCOOH > K_a 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 p^K_w? What is it? value at 25°C?
Answer:
p^K_w = – log ^K_w = – log 10^-14 = 14.

Question 19.
What are p^H and p^OH values of a neutral solution at a temperature at which = 10-13?
Answer:
p^K_w = p^H + p^OH, But p^K_w =13
Hence p^H = p^OH = 6.5.

Question 20.
The ionization constants of HF = 6.8 × 10^-4. Calculate the ionization constant of the corresponding conjugate base.
Answer:
K_b = \(\frac{\mathrm{K}_{w}}{\mathrm{~K}_{c}}=\frac{10^{-14}}{6.8 \times 10^{-4}}\) = 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⁺][S^2-] 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 p^H of an enzyme-catalyzed reaction has to be maintained between 7 and 8. What indicator should be used to monitor and control the p^H?
Answer:
Bromothymol blue or phenol red or cresol red.

Question 26.
The ionization constant of formic acid is 1.8 × 10^-4. Around what p^H will its mixture with sodium formate give buffer solution of highest capacity?
Answer:
Buffer solution of highest capacity is formed at which
p^H = P^K_a = – 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. PO₄^3- ion can accept a proton(s) but cannot donate any proton. Hence, PO₄^3- 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

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 p^H 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., p^H will decrease. Thus p^H 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.

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 p^H 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 p^H = – log [H⁺], the lesser the no. of H⁺ ions, the more the value of p^H. Therefore p^H 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:
CO₃^2-, H₂CO₃

(ii) H₂PO4-
Answer:
HPO₄^–, H₃PO₄

(iii) NH₃
Answer:
NH̅₂, NH₄⁺

(iv) HS̅.
Answer:
S^2-, 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

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:

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 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.

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 (p^H > 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 p^H 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 (K_w = 1.0 × 10^-14).
Answer:
Since NaOH is completely ionized
∴ [NaOH] = [OH^–] = 10^-3 N = 10^-3 M

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

Question 2.
Calculate the p^H of a 0.01 N solution of acetic acid. K_a for acetic acid is 1.8 × 10^-5 at 25°C.
Answer:

Applying the law of chemical equilibrium
K_a = [CH₃COO̅][H₃O⁺]/[CH₃COOH]

Question 3.
Calculate the p^H value of a solution of 0.1 M NH₃ (Kb = 1.8 × 10^-5). ,
Answer:
NH₃ H₂O ⇌ NH₄⁺ + OH^–

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

p^H = 10
∴ [H3O⁺] = 10^-10 M
or [OH^–] = 10^-4 M
Thus, they will exactly neutralize each other, and the p^H 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 p^H and p^OH.
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

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 K_p = 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

K_p = 4.28 × 10^-5
As Q > > K_p 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 > K_p 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


∴ Total initial pressure of PCl₃ and Cl₂ = 2.33 + 2.33 = 4.66 atm.

Question 8.
At 773 K, the equilibrium constant K_c for the reaction N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g) is 6.02 × 10^-2 L² mol^-2. Calculate the value of K_p at the same temperature.
Answer:

Question 9.
K_p for the reaction N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g) is 49 at a certain temperature. Calculate the value of K_p 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), K_p = 49
∴ for the reverse reaction 2NH₃ (g) ⇌ N₂ (g) + 3H₂ (g)

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

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:
K_c = \(\frac{\left[\mathrm{CO}_{2}\right]}{[\mathrm{CO}]}\)
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:

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, K_p.
Answer:

Total no. of moles = 1.5
If P is the total required pressure, then

Question 14.
The equilibrium constant for the reaction A₂ + B₂ ⇌ 2AB is K_p. 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 = K_p
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
p^H = – log [H₃O⁺] = – log (1.3 × 10^-5-5)
= 5 – log 1.3 = 5 – 0.1139 = 4.8861
As p^H 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
∴ p^H = – log 10^-1 = 1.

Question 17.
The value of K_w is 9.55 × 10^-14 at a certain temperature. Calculate the p^H of water at this temperature.
Answer:
K_w = 9.55 × 10^-14
Now for water [H₃O⁺] = [OH^–]
∴ K_w = [H₃O⁺] × [OH^–] = [H₃O⁺]²
or
[H₃O⁺] = \(\sqrt{\mathrm{K}_{w}}=\sqrt{9.55 \times 10^{-14}}\) = 3.09 × 10^-7M
p^H = – 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 p^H = 1.301.
p^H = – 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+ + SO₄^2- 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]

Question 20.
What would be the p^H 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]
p^H = – 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
p^H = – log (11 × 10^-8) = 8 – log 11
= 8 – 1.04 = 6.96.

Question 22.
Calculate the hydrolysis constant and degree of hydrolysis and p^H 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
∴ K_h = Hydrolysis constant

The hydrolysis reaction of KCN is

Question 23.
Calculate the p^H of 0.01 M solution of NH₄CN. Given that the K_a for HCN = 6.2 × 10-10 and K_b for NH₃ = 1.6 /10“5.
Answer:
Applying the formula
P^H = 7+ [p^K_a – p^K_b] = 7 + [- log K_a + log K_b]
= 7 + \(\frac{1}{2}\)[- log (6.2 × 10^-10) + log (1.6 × 10^-5)]
= 7+ \(\frac{1}{2}\)[(10- 0.7924) + (5 – 0.2041)]
p^H = 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

Ionic product of BaSO4 = [Ba++][SO4^—]
= 10^-3 × 10^-4 = 10^-7

As Ionic product 10^-7 is > K_sp 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 K_sp × 1 × 10^-10
Hence, no precipitate will be formed in this case.

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 2 Solutions. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 2 Important Extra Questions Solutions

Solutions Important Extra Questions Very Short Answer Type

Question 1.
What is meant by reverse osmosis? (CBSE 2011)
Answer:
The direction of osmosis can be reversed if a pressure larger than the osmotic pressure is applied to the solution side through the semi-permeable membrane.

Question 2.
Define an ideal solution and write one of its characteristics. (CBSE Delhi 2014)
Answer:
An ideal solution may be defined as the solution which obeys Raoult’s law exactly over the entire range of temperature and pressure. For ideal solution

• Heat of mixing is zero
• Volume change of mixing is zero.

Therefore first find the number of moles of each compound present and then use the mole fraction equation.

Question 3.
(i) Write the colligative property which is used to find the molecular mass of macromolecules.
(ii) In non-ideal solution, what type of deviation shows the formation of minimum boiling azeotropes? (CBSE Delhi 2016)
Answer:
(i) Osmotic pressure
(ii) Minimum boiling azeotropes show positive deviation from Raoult’s law.

How to find molar concentration with density how to calculate molarity of liquid solutions tank mixing problem dilution of two solutes

Question 4.
What is the relation between normality and molarity of a given solution of sulphuric acid?
Answer:
Normality = 2 × Molarity.

Question 5.
Given below is the sketch of a plant for carrying out a process.

(i) Name the process occurring in the above plant.
(ii) To which container does the net flow of solvent take place?
(iii) Name one SPM which can be used in this plant.
(iv) Give one practical use of the plant. (CBSE Sample Paper 2007)
Answer:
(i) Reverse osmosis
(ii) To fresh water container
(iii) Film of cellulose acetate
(iv) This can be used as a desalination plant to meet potable water requirements.

Free Molarity Calculator Chemistry determines the molarity of the chemical solution in a short span of time.

Question 6.
A and B liquids on mixing produce a warm solution. Which type of deviation from Raoult’s law is there? (CBSE Sample Paper 2011)
Answer:
Negative deviation.

Question 7.
Define ebullioscopic constant. (CBSE AI 2011)
Answer:
Ebullioscopic constant is the elevation in boiling point of a solution containing 1 mole of solute dissolved in 1000 g of the solvent.

This colloid osmotic pressure calculator determines the pressure induced by proteins in blood plasma.

Question 8.
What are isotonic solutions?
Answer:
The solutions having same osmotic pressure at the same temperature are called isotonic solutions. These have equimolar concentrations at same temperature.

Question 9.
Some liquids on mixing form ‘azeotropes’. What are ‘azeotropes’?
Answer:
The solutions (liquid mixtures) which boil at constant temperature and can distil unchanged in composition are called azeotropes or azeotropic mixtures.

Question 10.
Define‘colligative properties’
Answer:
Colligative properties are those properties of solutions which depend only on the number of solute particles and not on the nature of the solute.

Question 11.
Define ‘Molality (m)’.
Answer:
Molality is the number of moles of solute dissolved per 1000 g (or 1 kg) of the solvent.

Question 12.
Define ‘Ideal solution’.
Answer:
The solutions which obey Raoult’s law over the entire range of concentration are called ideal solutions.

Question 13.
Define ‘Abnormal molar mass’.
Answer:
Molar mass of a substance, calculated based on its colligative properties, may not be correct if the solute particles undergo dissociation or association in the solution. Molar mass thus calculated is called abnormal molar mass.

Solutions Important Extra Questions Short Answer Type

Question 1.
State Raoult’s law for a solution containing volatile components. Write two characteristics of the solution which obeys Raoult’s law at all concentrations. (CBSE Delhi 2019)
Answer:
Raoult’s law states that at a given temperature, the partial vapour pressure of any volatile component of a solution is equal to the product of the vapour pressure of the pure component and its mole fraction in the solution.

• No heat is evolved or absorbed when these components are mixed to form a solution.
• Volume of the solution is exactly same as the sum of the volumes of the components.

Question 2.
Define the terms ‘osmosis’ and ‘osmotic pressure’. What is the advantage of using osmotic pressure as compared to other colligative properties for the determination of molar masses of solutes in solutions? (CBSE 2010)
Answer:
The flow of solvent from solution of low concentration to higher concentration through a semipermeable membrane is called osmosis.
The excess pressure which must be applied to a solution to prevent the passage of solvent through a semi-permeable membrane is called osmotic pressure. Osmotic pressure measurement is preferred over all other colligative properties because
1. even in dilute solutions, the osmotic pressure values are appreciably high and can be measured accurately.
2. osmotic pressure can be measured at room temperature. On the other hand, elevation in boiling point is measured at high temperature where the solute may decompose. The depression in freezing point is measured at low temperatures.

Question 3.
State the following:
(i) Raoult’s law in its general form in reference to solutions.
(ii) Henry’s law about partial pressure of a gas in a mixture. (CBSE 2011)
Answer:
(i) For a solution of volatile liquids, at a given temperature, the partial vapour pressure of each component in solution is equal to the product of vapour pressure of the pure component and its mole fraction.

(ii) Henry’s law states that the mass of a gas dissolved per unit volume of the solvent at a constant temperature is directly proportional to the pressure of the gas in equilibrium with the solution.

Question 4.
The experimentally determined molar mass for what type of substances is always lower than the true value when water is used as solvent? Explain. Give one example of such a substance and one example of a substance which does not show a large variation from the true value. (CBSE Sample Paper 2019)
Answer:
Ionic compounds when dissolved in water dissociate into cations and anions. When there is dissociation of solute into ions, in dilute solutions, the number of particles increases if we ignore the interionic interactions.

As the value of the colligative properties depends on the number of particles of the solute, the experimentally observed value of colligative property will be higher than the true value. Therefore the experimentally determined molar mass is always lower than the true value.

If we dissolve KCl in water, the experimentally determined molar mass is always lower than the true value. Glucose (non-electrolyte) does not show a large variation from the true value.

Question 5.
What mass of ethylene glycol (molar mass = 62.0 g mol^-1) must be added to 5.50 kg of water to lower the freezing point of water from 0 to – 10.0°0 (k_f for water = 1.86 K kg mol^-1). (CBSE 2010)
Answer:
ΔT_f = \(\frac{\mathrm{K}_{f} \times 1000 \times \mathrm{w}_{\mathrm{B}}}{\mathrm{w}_{\mathrm{A}} \times \mathrm{M}_{\mathrm{B}}}\)
K_f = 1.86 K kg mol^-1, w_A = 5.50 kg = 5500 g
W_B = ?
M_B = 62.0 g/mol, ΔT_f = 0 – (- 10) = 10°C
10 = \(\frac{1.86 \times 1000 \times w_{B}}{5500 \times 62}\)
W_B = \(\frac{10 \times 5500 \times 62}{1.86 \times 1000}\) = 1833.3 g = 1.833 kg

Question 6.
15.0 g of an unknown molecular material was dissolved in 450 g of water. The resulting solution was found to freeze at – 0.34°C. What is the molar mass of this material? (k_f for water = 1.86 K kg mol^-1) (CBSE Delhi 2012)
Answer:
Molecular mass, M_B is
M_B = \(\frac{\mathrm{K}_{f} \times \mathrm{w}_{\mathrm{B}} \times 1000}{\mathrm{w}_{\mathrm{A}} \times \Delta \mathrm{T}_{f}}\)
ΔT_f = 0 – (- 0.34) = 0.34°,
K_f = 1.86 K Kg mol^-1, w_B = 15.0 g, W_A = 450 g
M_B = \(\frac{1.86 \times 15.0 \times 1000}{450 \times 0.34}\) = 182.3 g/mol

Question 7.
State Henry’s law and write its two applications. (CBSE Delhi 2019)
Answer:
Henry’s law states that the mole fraction of the gas in the solution is directly proportional to the partial pressure of the gas over the solution.
P = K_H.X
Applications of Henry’s law

• In the production of carbonated beverages.
• In deep sea diving (scuba diving)

Question 8.
A solution of glycerol (C₃H₈O₃) in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42°C while pure water boils at 100°C. What mass of glycerol was dissolved to make the solution?
(Kb for water = 0.512 K kg mol^-1)         (CBSE Delhi 2012, CBSE AI 2012)
Answer:
Elevation in boiling point
T_b = \(\frac{\mathrm{K}_{b} \times 1000 \times \mathrm{w}_{\mathrm{B}}}{\mathrm{w}_{\mathrm{A}} \times \mathrm{M}_{\mathrm{B}}}\)
ΔT_b = 100.42 – 100 = 0.42°, w_A = 500 g
w_A = ?, K_b = 0.512 k Kg mol^-1
M_B = 3 × 12 + 8 × 1 + 3 × 16 = 92 g/mol
0.42 = \(\frac{0.512 \times 1000 \times w_{B}}{92 \times 500}\)
or w_B = \(\frac{0.42 \times 500 \times 92}{0.512 \times 1000}\)
= 37.33 g
Mass of glycerol to be added = 37.33 g

Question 9.
A solution prepared by dissolving 1.25 g of oil of winter green (methyl salicylate) in 99.0 g of benzene has a boiling point of 80.31°C. Determine the molar mass of this compound. (B.P. of pure benzene = 80.10°C and K_b for benzene = 2.53 °C kg mol^-1) (CBSE Delhi 2010)
Answer:

k_b = 2.53°C kg mol^-1
w_A = 99.0 g, w_B = 1.25 g
M_B = \(\frac{2.53 \times 1000 \times 1.25}{0.21 \times 99.0}\)
= 152.1 g mol^-1

Question 10.
18 g of glucose, C₆H₁₂O₆ (molar mass = 180 g mol^-1), is dissolved in 1 kg of water in a sauce pan. At what temperature will this solution boil?
(K_b for water = 0.52 K kg mol^-1, boiling point of pure water = 373.15 K). (CBSE Delhi 2013)
Answer:
ΔT_b = \(\frac{\mathrm{K}_{b} \times 1000 \times \mathrm{w}_{\mathrm{B}}}{\mathrm{w}_{\mathrm{A}} \times \mathrm{M}_{\mathrm{B}}}\)
w_B = 18 g, w_A = 1000 g, M_B = 180 g/mol,
K_b = 0.52 K kg mol^-1
∴ ΔT_b = \(\frac{0.52 \times 1000 \times 18}{1000 \times 180}\) = 0.052 K
Boiling point of solution
= 373.15 + 0.052 = 373.202 K.

Question 11.
A solution containing 15 g urea (molar mass = 60 g mol^-1) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g mol^-1) in water. Calculate the mass of glucose present in one litre of its solution. (CBSE Al 2014)
Answer:
For isotonic solutions,
π (urea) = π (glucose)

Question 12.
A 1.00 molal solution of trichloroacetic acid (CCl₃COOH) is heated to its boiling point. The solution has the boiling point of 100.18°C. Determine the van’t Hoff factor for trichloroacetic acid (K_b for water = 0.512 K kg mol^-1). (CBSE Delhi 2012)
Answer:
Observed boiling point elevation,
ΔT_b = 100.18 – 100.0 = 0.18°C
Molality of solution = 1.00 m
Calculated boiling point elevation,
ΔT_b(calc.) = K_b × m
= 0.512 × 1 = 0.512
van’t Hoff factor,

Question 13.
Blood cells are isotonic with 0.9% sodium chloride solution. What happens if we place blood cells in a solution containing
(i) 1.2% sodium chloride solution?
(ii) 0.4% sodium chloride solution? (CBSE AI 2016)
Answer:
(i) 1.2% sodium chloride solution is hypertonic with respect to 0.9% sodium chloride solution or blood cells. When blood cells are placed in this solution, water flows out of the cells and they shrink due to loss of water by osmosis.

(ii) 0.4% sodium chloride solution is hypotonic with respect to 0.9% sodium chloride solution or blood cells. When blood cells are placed in this solution, water flows into the cells and they swell.

Question 14.
1.00 g of non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. Find the molar mass of the solute.
(K_f for benzene = 5.12 K kg mol^-1). (CBSE AI 2013)
Answer:

Question 15.
(a) Out of 0.1 molal aqueous solution of glucose and 0.1 molal aqueous solution of KCI, which one will have higher boiling point and why?
Answer:
(a) 0.1 m KCI solution will have higher boiling point. This is because KCl dissociates in water to give two ions (K⁺ and Cl^–) whereas glucose does not dissociate. Therefore, number of solute particles is greater in 0.1 m KCl as compared to 0.1 m glucose.

(b) Predict whether van’t Hoff factor
(i) is less than one or greater than one in the following:
(i) CH₃COOH dissolved in water
(ii) CH₃COOH dissolved in benzene (CBSE AI 2019)
Answer:
(i) i > 1 because CH₃COOH dissociates in water.
(ii) i < 1 because CH₃COOH associates in benzene.

Question 16.
Urea forms an ideal solution in water. Determine the vapour pressure of an aqueous solution containing 10% by mass of urea at 40°C. (Vapour pressure of water at 40°C = 55.3 mm of Hg) (CBSE AI 2006)
Answer:
w_A = 90 g. w_B = 10 g

55.3 – p_A = 1.84
∴ p_A = 53.46 mm Hg

Question 17.
A solution of chloroform and acetone is an example of maximum boiling azeotrope. Why? (CBSE Sample Paper 2012)
Answer:
The solution of chloroform and acetone has lower vapour pressure than ideal solution because of stronger interactions between chloroform and acetone molecules. As a result, total vapour pressure becomes less than the corresponding ideal solution of same composition (i.e. negative deviations). Therefore, the boiling points of solutions are increased and form maximum boiling azeotropes.

Question 18.
A solution of glucose (C₆H₁₂O₆) in water is labelled as 10% by weight. What would be the molality of the solution? (CBSEAI 2013)
Answer:
10% by weight means that 10 g of glucose is present in 100 g of solution.
Mass of solvent = 100 – 10 = 90 g 10
Moles of glucose = \(\frac{10}{180}\) = 0.0556 moles
Molality = \(\frac{0.0556}{90}\) × 1000
= 0.618 m.

Question 19.
What is meant by positive deviations from Raoult’s law? Give an example. What is the sign of Δ_mix H for positive deviation?
OR
Define azeotropes. What type of azeotrope is formed by positive deviation from Raoult’s law? Give an example. (Delhi 2015)
Answer:
When the observed vapour pressure of a liquid solution is higher than the value expected from Raoult’s law, it is called positive deviation from Raoult’s law.
P_total > P°_A X_A + P°_B X_B
for the two components A and B.
Example: Ethyl alcohol and cyclohexane
ΔH_mix is positive.
OR
Liquid mixtures which boil at constant temperature and can distil unchanged in composition are called azeotropes.
The mixture which shows positive deviation from Raoult’s law forms minimum boiling azeotrope.
Example: A mixture of ethanol and water containing 95.4% ethanol forms minimum boiling azeotrope.

Question 20.
Write two differences between an ideal solution and a non-ideal solution. (CBSE Delhi 2019)
Answer:

Question 21.
Define an ideal solution and write one of its characteristics. (CBSE 2014)
Answer:
An ideal solution may be defined as the solution which obeys Raoult’s law exactly over the entire range of temperature and pressure. For ideal solution

• Heat of mixing is zero
• Volume change of mixing is zero.

Question 22.
State Henry’s law. What is the effect of temperature on the solubility of a gas in a liquid? (CBSE 2014)
Answer:
Henry’s law states that the mass of a gas dissolved per unit volume of the solvent at a constant temperature is directly proportional to the pressure of the gas in equilibrium with the solution.

The solubility of a gas decreases with increase in temperature. However, it may be noted that there are certain gases like hydrogen and inert gases whose solubility increases slightly with increase in temperature especially in non-aqueous solvents such as alcohols, acetone, etc.

Question 23.
State Raoult’s law for the solution containing volatile components. What is the similarity between Raoult’s law and Henry’s law? (CBSE 2014)
Answer:
Raoult’s law states that at a given temperature for a solution of volatile liquids, the partial vapour pressure of each component in solution is equal to the product of the vapour pressure of pure component and its mole fraction.
Similarity between Raoult’s law and Henry’s law

• Both Raoult’s law and Henry’s law apply to volatile component in solution. .
• Both laws state that the vapour pressure of one component is proportional to the mole fraction of that component.

Solutions Important Extra Questions Long Answer Type

Question 1.
Calculate the freezing point of a solution containing 0.5 g KCI (Molar mass = 74.5 g/ mol) dissolved in 100 g water, assuming KCI to be 92% ionised. K_f of water = 1.86 K kg / mol. (CBSE Sample Paper 2019)
Answer:
Let us consider one mole of KCI whose degree of dissociation is α. The dissociation of KCI can be represented as:
KCI → K⁺ + Cl^–
1 – α       α      α
Total number of moles after dissociation
= 1 – α + α + α
= 1 + α
Hence i = \(\frac{1+\alpha}{1}\)
If the solute dissolves in the solvent giving n ions, then here n = 2
i = 1 + (n -1) α
= 1 + (2 – 1) α = 1 + α
Now, ∆T_f = i K_fm
= (1 + 0.92) × 1.86 × \(\frac{0.5 \times 1000}{74.5 \times 100}\)
∆T_f = 0.24
∆T_f = T°_f – T’_f = 0 – 0.24
T’_f = -0.24°C

Question 2.
State Henry’s law. Why do gases always tend to be less soluble in liquids as the temperature is raised?
OR
State Raoult’s taw for the solution containing volatile components. Write two differences between an ideal solution and a non-ideal solution. (CBSE Delhi 2015)
Answer:
Henry’s law states that the mass of a gas dissolved per unit volume of the solvent is directly proportional to the pressure of the gas in equilibrium with the solution.

If m is the mass of the gas dissolved in a unit volume of the solvent and p is the pressure of the gas in equilibrium with the solution, then
m ∝ p
or m = K.p
(where K is the proportionality constant) or “partial pressure of the gas in its vapour phase (p) is directly proportional to the mole fraction of the gas (x) in the solution”.
P = K_H.X
The dissolution of a gas in a liquid is exothermic process. Therefore, in accordance with Le Chatelier’s principle, with increase in temperature, the equilibrium shifts in the backward direction. As a result, solubility decreases with increase in temperature.
OR
Raoult’s law states that for a solution of volatile liquids, the partial vapour pressure of each component in solution is equal to the product of the vapour pressure of the pure component and its mole fraction. For a binary solution of two components A and B,
P_A = P_A° × X_A
P_B = P_B° × X_B
Differences between ideal and non-ideal solutions

Ideal solution Non-ideal solution

Question 3.
Calculate the amount of KCI which must be added to 1 kg of water so that the freezing point is depressed by 2 K. (K_f for water = 1.86 K kg mol^-1) (CBSE Delhi 2012)
Answer:
ΔT_f = 2 K
KCl ⇌ K⁺ + Cl^–
i = 2
K_f = 1.86 K kg mol^-1
w_A = 1 kg = 1000 g
M_B = (39 + 35.5) = 74.5 g
w_B = ?

Question 4.
What mass of NaCl must be dissolved in 65.0 g of water to lower the freezing point of water by 7.50°C? The freezing point depression constant (K_f) for water is 1.86°C/m. Assume van’t Hoff factor for NaCl is 1.87 (Molar mass of NaCI = 58.5 g mol^-1). (CBSE 2011)
Answer:
Lowering in freezing point,
ΔT_f = 7.50°C
K_f = 1.86 °C/m,
Mass of water, W_A = 65.0 g
Molar mass of NaCl, M_B = 58.5 g
Mass of NaCl, W_B = ?
van’t Hoff factor, i = 1.87

Question 5.
Determine the osmotic pressure of a solution prepared by dissolving 2.5 × 10^-2 g of K₂SO₄ in 2 L of water at 25°C, assuming that it is completely dissociated. (CBSE Delhi 2013)
(R = 0.0821 L atm K^-1 mol^-1, Molar mass of K₂SO₄ = 174 g mol^-1)
Answer:
Since K₂SO₄ dissociates completely,
K₂SO₄ → 2K⁺ + SO₄^2-
One mole of K₂SO₄ will give 3 mole particles and therefore, the value of ‘i’ is 3.
Osmotic pressure, π = iCRT
= i\(\frac{\mathrm{W}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{B}} \times \mathrm{V}}\)RT
W_B = 2.5 × 10^-2 g, V = 2.0 L, M_B = 174 g/mol
R = 0.821 L atm mol^-1 K^-1
∴ π = \(\frac{3 \times 2.5 \times 10^{-2} \times 0.0821 \times 298}{174 \times 2.0}\)
= 5.27 × 10^-3 atm

Question 6.
Calculate the mass of compound (molar mass = 256 g mol^-1) to be dissolved in 75 g of benzene to lower its freezing point by 0.48 K (K_f = 5.12 K kg mol^-1). (CBSE Delhi 2014)
Answer:
ΔT_f = 0.48 K, K_f = 5.12 K kg mol^-1
M = 256 g mol^-1, w_A = 75g, w_B = ?

Question 7.
3.9 g of benzoic acid dissolved in 49 g of benzene shows a depression in freezing point of 1.62 K. Calculate the van’t Hoff factor and predict the nature of solute (associated or dissociated).
(Given: Molar mass of benzoic acid = 122 g mol^-1, K_f for benzene = 4.9 K kg mol^-1) (CBSE Delhi 2015)
Answer:
ΔT_f = i K_f × m
= \(\frac{i \times K_{f} \times W_{2} \times 1000}{M_{2} \times W_{1}}\)
W₂ = 3.9 g, W₁ = 49 g, ΔT_f = 1.62 K,
M₂ = 122 g mol^-1
K_f = 4.9 K kg mol^-1
1.62 = \(\frac{i \times 4.9 \times 3.9 \times 1000}{122 \times 49}\)
or i = \(\frac{1.62 \times 122 \times 49}{4.9 \times 3.9 \times 1000}\) = 0.506
Since i is less than 1, the solute is associated.

Question 8.
A solution of glucose (molar mass = 180 g mol^-1) in water is labelled as 10% by mass. What would be the molality and molarity of the solution? (Density of solution = 1.2 g mL^-1) (Delhi Al 2014)
Answer:
10% (by mass) solution of glucose means that 10 g of glucose is present in 100 g of solution or in 90 g of water.

(i) Calculation of molality
Mass of glucose = 10 g
Moles of glucose = \(\frac{10}{180}\) = 0.0556
(Molar mass of glucose = 180 g/mol)
Mass of water = 90 g

= \(\frac{0.0556}{90}\) × 1000
= 0.618 m

(ii) Calculation of molarity
Moles of glucose = 0.0556 Mass

Question 9.
A 4% solution (w/w) of sucrose (M = 342 g mol^-1) in water has a freezing point of 271.15 K. Calculate the freezing point of 5% glucose (M = 180 g mol^-1) in water. (Given: Freezing point of pure water = 273.15 K)
Answer:
For Sucrose solution,
W_B = 4 g, W_A = 100 – 4 = 96 g,
ΔT_f = 273.15 – 271.5 = 2°C

Freezing point of solution = 273.15 – 4.8
= 268.35 K

Question 10.
What would be the molar mass of a compound if 6.21 g of it dissolved in 24.0 g of chloroform form a solution that has a boiling point of 68.04°C. The boiling point of pure chloroform is 61.7°C and the boiling point elevation constant, K_b, for chloroform is 3.63 °C/m. (CBSE Delhi 2011)
Answer:
Elevation in boiling point,
ΔT_b = 68.04 – 61.7 = 6.34 °C
Mass of substance,
w_B = 6.21 g,
Mass of chloroform,
w_A = 24.0g
K_b = 3.63 °C/m

Question 11.
Two elements A and B form compounds having molecular formula AB₂ and AB₄. When dissolved in 20 g of benzene, 1 g of AB₂ lowers the freezing point by 2.3 K, whereas 1.0 g of AB₄ lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol^-1. Calculate the atomic mass of A and B. (Delhi Al 2004)
Answer:
Let us first calculate molar masses of AB₂ and AB₄.
For AB₂ compound
M_B = \(\frac{\mathrm{K}_{f} \times \mathrm{W}_{\mathrm{B}} \times 1000}{\mathrm{w}_{\mathrm{A}} \times \Delta \mathrm{T}_{f}}\)
ΔT_f = 2.3 K, w_B = 1.0 g,
w_A = 20.0 g
K_f = 5.1 K kg mol^-1
M_AB2 = \(\frac{5.1 \times 1.0 \times 1000}{20.0 \times 2.3}\)
∴ M_AB2 = 110.87
For AB₄ compound
ΔT_f= 1.3 K, w_B = 1.0 g,
w_A = 20.0 g
M_AB4 = \(\frac{5.1 \times 1.0 \times 1000}{20.0 \times 1.3}\)
∴ M_AB4 = 196.15
Let a be the atomic mass of A and b be the atomic mass of B, then
M_AB2 = a + 2b = 110.87 …… (i)
M_AB4 = a + 4b = 196.15 ……. (ii)
Subtracting eqn. (ii) from eqn. (i)
– 2b = – 85.28
∴ b = 42.64
Substituting the value of b in eqn (i)
a + 2 × 42.64 = 110.87
a = 110.87 – 85.28 = 25.59
Atomic mass of A = 25.59 g
Atomic mass of B = 42.64 g

Question 12.
A solution prepared by dissolving 8.95 mg of a gene fragment in 35.0 mL of water has an osmotic pressure of 0.335 torr at 25°C. Assuming the gene fragment is a non-electrolyte, determine its molar mass. (CBSE Delhi 2011, Delhi Al 2011)
Answer:
Mass of gene fragment = 8.95 mg
= 8.95 × 10^-3 g
Volume of water = 35.0 mL
= 35.0 × 10^-3 L
Osmotic pressure,
π = 0.335 torr
= 0.335/760 atm
Temperature = 25°C
= 273 + 25 = 298 K

= 14193.3 g mol^-1 or
1.42 × 1o⁴g mol^-1.

Question 13.
Calculate the boiling point of solution when 2 g of Na₂SO₄ (M = 142 g mol^-1) was dissolved in 50 g of water, assuming Na₂SO₄ undergoes complete ionization. (K_b for water = 0.52 K kg mol^-1) [Delhi Al 2016)
Answer:
ΔT_b = \(\frac{i \times K_{b} \times w_{B} \times 1000}{M_{B} \times w_{A}}\)
Weight of solute, w_B = 2 g
Molar mass = 142 g mol^-1
Weight of solvent = 50 g
K_b = 0.52 K kg mol^-1
Na₂SO₄ undergoes complete ionisation as:
Na₂SO₄ ⇌ 2Na⁺ + SO₄^2-
One mole of Na₂SO₄ gives 3 mole particles and therefore,
i = 3
∴ ΔT_b = \(\frac{3 \times 0.52 \times 2 \times 1000}{142 \times 50}\) = 0.439
Boiling point of solution = 373 + 0.439
= 373.439 K

Question 14.
Calculate the amount of CaCl₂ (molar mass = 111 g mol^-1) which must be added to 500 g of water to lower its freezing point by 2 K, assuming CaCl₂ is completely dissociated, (K_f for water = 1.86 K kg mol^-1) (Delhi Al 2015)
Answer:
CaCl₂ undergoes complete dissociation as:
CaCl₂ → Ca²⁺ + 2Cl^–
One mole of CaCl₂ will give 3 mole particles and therefore, the value of T will be equal to 3.
ΔT_f = i K_f × m
= \(\frac{i \times K_{f} \times w_{B} \times 100}{M_{B} \times W_{A}}\)
K_f = 1.86 K kg mol^-1, w_A = 500 g, w_B = ?, ΔT_f = 2 K, i = 3, M_B = 111 g mol^-1
2 = \(\frac{3 \times 1.86 \times w_{B} \times 1000}{111 \times 500}\)
∴ W_B = \(\frac{2 \times 111 \times 500}{3 \times 1.86 \times 1000}\) = 19.89 g

Question 15.
A solution 0.1 M of Na₂SO₄ is dissolved to the extent of 95%. What would be its osmotic pressure at 27°C? (R = 0.0821 L atm K^-1 mol^-1) (CBSE AI 2019)
Answer:
Na₂SO₄ ⇌ 2Na⁺ + SO₄^2-
n = 3
If α is the degree of dissociation, then
α = \(\frac{i-1}{n-1}\)
0.95 = \(\frac{i-1}{3-1}\)
0.95 × 2 = i – 1 or i = 1.90 + 1 = 2.90
π = iCRT
= 2.90 × 0.1 × 0.0821 × 300
= 7.143 atm

Question 16.
(i) Out of 1 M glucose and 2 M glucose, which one has a higher boiling point and why?
(ii) What happens when the external pressure applied becomes more than the osmotic pressure of solution? (Delhi Al 2016)
Answer:
(i) The elevation in boiling point is a colligative property and depends upon the number of moles of solute added. Higher the concentration of solute added, higher will be the elevation in boiling point. Thus, 2M glucose solution has higher boiling point than 1 M glucose solution.

(ii) When the external pressure applied becomes more than the osmotic pressure of the solution, then the solvent molecules from the solution pass through the semipermeable membrane to the solvent side. This process is called reverse osmosis.

Question 17.
Which of the following solutions has higher freezing point?
0.05 M Al₂(SO₄)₃, 0.1 M K₃[Fe(CN)₆] Justify. (CBSE Sample Paper 2017-18)
Answer:
0.05 M Al₂(SO₄)₃ has higher freezing point.
ΔT_f ∝ i × concentration
For 0.05 M Al₂(SO₄)₃ i = 5
ΔT_f ∝ 5 × 0.05 = 0.25 moles of ions
For 0.1 M K₃[Fe(CN)₆], i = 4
ΔT_f ∝ 4 × 0.1 = 0.40 moles of ions
∴ Depression in freezing point for 0.05 M Al₂(SO₄)₃ will be less and hence freezing point will be higher.

Question 18.
A 10% solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water, if freezing point of pure water is 273.15 K. Given: (Molar mass of sucrose = 342 g mol^-1) (Molar mass of glucose = 180 g mol^-1) (CBSE AI 2017, CBSE Delhi 2017)
Answer:
For sucrose solution,
K_f = \(\frac{\Delta T_{f} \times w_{A} \times M_{B}}{w_{B} \times 1000}\)
ΔT_f = 273.15 – 269.15 = 4.0 k, w_B = 10 g
W_A = 100 – 10 = 90 g
k_f = \(\frac{4.0 \times 90 \times 342}{10 \times 1000}\) = 12.31 km^-1
For glucose solution,
w_B = 10 g, w_A = 100 – 10 = 90 g
M_B = 180 g
ΔT_f = \(\frac{12.31 \times 10 \times 1000}{90 \times 180}\) = 7.6 k
Freezing point of glucose solution
= 273.15 – 7.6
= 265.55 K

Question 19.
30 g of urea (M = 60 g mol^-1) is dissolved in 846 g of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K is 23.8 mm Hg. (CBSE Al 2017)
Answer:

Question 20.
The vapour pressures of pure liquids A and B are 450 mm and 700 mm of Hg respectively at 350 K. Calculate the composition of liquid mixture if total vapour pressure is 600 mm of Hg. Also find the composition of the mixture in vapour phase.
(CBSE Sample Paper 2010)
Answer:
Let mole fraction of liquid A in solution = x_A
Mole fraction of liquid B in solution, x_B = 1 – x_A
P = P_A°X_A + P_B°X_B or = P_A°X_A + P_B°(1 – x_A)
p = 600 mm Hg
600 = 450 x_A + 700 (1 – x_A)
Solving, X_A = \(\frac{100}{250}\) = 0.4
Mole fraction of liquid A = 0.4
Mole fraction of liquid B = 1 – 0.4 = 0.6
Calculation of composition of vapour phase
p_A = P_A°X_A = 450 mm × 0.4 = 180 mm
p_B = P_B°X_B = 700 × 0.6 = 420 mm
P_total = p_A + p_B = 180 + 420 = 600 mm

Question 21.
Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to
250.0 g of water. (K_b for water = 0.512 K kg mol^-1, molar mass of NaCl = 58.44 g) (CBSE Delhi 2011)
Answer:
ΔT_b = \(\frac{\mathrm{iK}_{\mathrm{b}} \times 1000 \times \mathrm{W}_{2}}{\mathrm{~W}_{1} \times \mathrm{M}_{2}}\)
NaCl dissociates as:
NaCl → Na⁺ + Cl^–
∴ i = 2
W₂ = 15.0 g, W₁ = 250.0 g, M₂ = 58.44 g mol^-1
∴ K_b = 0.512 K kg mol^-1
∴ ΔT_b = \(\frac{2 \times 0.512 \times 1000 \times 15.0}{250.0 \times 58.44}\)
∴ Boiling point of solution = 100 + 1.05
= 101.5°C

Question 22.
(i) Define the following terms:
(a) Mole fraction
(b) van’t Hoff factor
(ii) 100 mg of a protein is dissolved in enough water to make 10.0 mL of a solution. If this solution has an osmotic pressure of 13.3 mm Hg at 25°C, what is the molar mass of protein? (R = 0.0821 L atm. mol^-1 K^-1 and 760 mm Hg = 1 atm.) (CBSE Delhi 2010)
Answer:
(i) (a) Mole fraction is the ratio of number of moles of one component to the total number of motes In a mixture. For example, in a binary mixture containing n₁ and n₂ moles of two components, Mole fraction of one component,
x₁ = \(\frac{n_{1}}{n_{1}+n_{2}}\)
Mote fraction of second component,
x₂ = \(\frac{n_{2}}{n_{1}+n_{2}}\)
(b) van’t Hoff factor is the ratio of the normal molar mass to the observed or abnormal molar mass of a solute in a solution due to association or dissociation.

(ii) Osmotic pressure,

Question 23.
(i) Differentiate between molarity and molality of a solution. How does a change in temperature influence their values?
(ii) Calculate the freezing point of an aqueous solution contaning 10.50 g of MgBr2 in 200 g of water (Molar mass of MgBr₂ = 184 g mol^-1, = for water = 1.86 K kg mol^-1). (CBSE Delhi 2011)
Answer:
(i) Molality is the number of moles of solute per 1000 g of solvent, whereas molarity is the number of moles of solute per 1000 ml of the solution. Molality is represented as m, whereas molarity is represented as M.

Molarity changes with change in temperature because of change in volume. On the other hand, there is no effect of temperature on the molality of the solution.

(if) Moles of MgBr₂
= \(\frac{10.50}{184}\)
= 0.0571 mol
Mass of water = 200 g
Molality = \(\frac{0.0571}{200}\) × 1000
= 0.2855 m
MgBr₂ ionises as:
MgBr₂ → Mg²⁺ + 2Br^–
Assuming complete dissociation of MgBr₂,
i = 3
Freezing point depression
ΔT_f = i × K_f × m
= 3 × 1.86 × 0.2855
= 1.59
Freezing point = 0 – 1.59°C
= – 1.59°C

Question 24.
(i) Define the terms osmosis and osmotic pressure. Is the osmotic pressure of a solution a colligative property? Explain.
(ii) Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250.0 g of water. (K_b for water = 0.512 K kg mol^-1, molar mass of NaCl = 58.44 g mol^-1) (CBSE Delhi 2011)
Answer:
(i) Osmosis is the flow of solvent from solution of lower concentration to higher concentration through a semi- permeable membrane.

Osmotic pressure is the excess pressure which must be applied to a solution to prevent the passage of solvent through a semipermeable memberane.

It has been found experimentally that for n moles of the solute dissolved in V litres of the solution, the osmotic pressure (π) at temperature T is
πV = nRT
where R is a gas constant.
or π = \(\frac{n}{V}\) RT
= C RT
where C is the molar concentration of the solution.
For a solution at given tempeature, both R and T are constant so that
π ∝ C
Thus, osmotic pressure depends upon the molar concentration of solution and therefore, is a colligative property.

(ii) ΔT_b = \(\frac{i \mathrm{~K}_{b} \times 1000 \times \mathrm{W}_{2}}{\mathrm{~W}_{1} \times \mathrm{M}_{2}}\)
NaCl dissociates as:
NaCl → Na⁺ + Cl^–
∴ i = 2
W₂ = 15.0 g, W₁ = 250.0 g,
M₂ = 58.44 g mol^-1
K_b = 0.512 K kg mol^-1 .
∴ ΔT_b = \(\frac{2 \times 0.512 \times 1000 \times 15.0}{250.0 \times 58.44}\)
= 1.05°C
∴ Boiling point of solution = 100 + 1.05
= 101.5°C

Question 25.
(a) A 5% solution (by mass) of cane sugar in water has a freezing point of 271 K. Calculate the freezing point of 5% solution (by mass) of glucose in water. The freezing point of pure water is 273.15 K.
(b) Why is osmotic pressure of 1 M KCl higher than 1 M urea solution?
(c) What type of liquids form ideal solutions?
OR
(a) 1.0 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol^-1. Find the molar mass of the solute.
(b) What is the significance of Henry’s law constant, K_H?
(c) What leads to anoxia? (CBSE 2019C)
Answer:
(a)

(b) On dissolving in water KCl dissociates into K⁺ and Cl^– ions but urea does not dissociate into ions.

(c) Ideal solutions: An ideal solution may be defined as the solution which obeys Raoult’s law exactly over the entire range of concentration.
Such solutions are formed by mixing the two components which are identical in molecular size, in structure and have almost identical intermolecular forces. In these solutions, the intermolecular interactions between the components (A – B attractions) are of same magnitude as the intermolecular interactions in pure components (A – A and B – B attractions).

The ideal solutions have also the following characteristics:
1. Heat change on mixing is zero. Since there is no change in magnitude of the attractive forces in the two components present, the heat change on mixing, i.e. Δ_mixingH in such solutions must be zero.

(ii) Volume change on mixing is zero. In ideal solutions, the volume of the solution is the sum of the volumes of the components before mixing, i.e. there is no change in volume on mixing or Δ_mixingV is zero.

For example, when we mix 100 cm³ of benzene with 100 cm³ of toluene, the volume of the solution is found to be exactly 200 cm³. Therefore, there is no change in volume on mixing, i.e. Δ_mixingV = 0. It has been noticed that the solutions generally tend to become ideal when they are dilute.

Examples of ideal solutions: In fact, ideal solutions are quite rare but some solutions are nearly ideal in behaviour at least when they are very dilute. A few examples of ideal solutions are:

• Benzene and toluene
• n-hexane and n-heptane
• Bromoethane and iodoethane
• Chlorobenzene and bromobenzene.

OR

(a)

(b) (i) Henry’s law constant, K_H depends upon the nature of the gas.
(ii) Higher the value of K_H at a particular pressure, the lower is the solubility of the gas in the liquid. (∵ x = \(\frac{1}{\mathrm{~K}_{\mathrm{H}}}\) . p)

(iii) The value of K_H increases with increase in temperature indicating that the solubility of gases decreases with increase of temperature. This is the reason that aquatic species are more comfortable in cold water rather than warm water.

(c) At high altitudes, the partial pressure of oxygen is less than that at the ground level. This leads to low concentration of oxygen in the blood and the tissues of the people living at high altitudes. As a result of low oxygen in the blood, the people become weak and unable to think clearly. These are the symptoms of a condition known as anoxia.

Question 26.
(i) State Raoutt’s law for a solution containing volatile components. How does Raoult’s law become a special case of Henry’s law?
(ii) 1.00 g of non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. Find the molar mass of the solute. (K_f for benzene = 512 K kg mol^-1)
OR
(i) Define the following terms:
(a) Ideal solution
(b) Azeotrope
(c) Osmotic pressure
(ii) A solution of glucose (C₆H₁₂O₆) in
water is labelled as 10% by weight. What would be the molality of the solution? (CBSE 2013)
(Molar mass of glucose = 180 g mol^-1)
Answer:
(i) Raoult’s law states that at a given temperature, for a solution of volatile liquids, the partial pressure of each component in solution is equal to the product of the vapour pressure of the pure component and its mole fraction. For example, for a binary solution of two volatile liquids A and B having mole fractions x_A and x_B,
P_A = P°_A X_A and P_B = P°_B X_B
where p_A and p_B are the vapour pressures of the components in solutions and P°_A and P°_B are vapour pressure of pure components.
According to Henry’s law for a gas dissolved in a liquid, the pressure of the gas is directly proportional to mole fraction, i.e.
p = K_Hx
where K_H is a proportionality constant known as Henry’s constant.
But Raoult’s law states that
p = p°x
∴ K_H = p°
This means that Raoult’s law is a special case of Henry’s law.

(ii) Molar mass of solute,
M_B = \(\frac{K_{f} \times W_{B} \times 1000}{W_{A} \times \Delta T_{f}}\)
W_B = 1.0g
W_A = 50.0 g,
ΔT_f = 0.40 K
K_f = 5.12 K kg mol^-1
M_B = \(\frac{5.12 \times 1.0 \times 1000}{50 \times 0.40}\)
= 256 g/mol

OR

(i) (a) Ideal solution. A solution which obeys Raoult’s law exactly over the entire range of concentration is called ideal solution.

(b) Azeotrope. The solutions or liquid mixtures which boil at constant temperature and can distil unchanged in composition are called azeotropes.

(c) Osmotic pressure. The excess pressure which must be applied to a solution to prevent the passage of solvent into it through a semipermeable membrane is called osmotic pressure.

(ii) 10% by weight means that 10 g of glucose is present in 100 g of solution.
Mass of solvent = 100 – 10 = 90 g
Moles of glucose = \(\frac{10}{180}\) = 0.0556
Molality = \(\frac{0.0556}{90}\) × 1000
= 0.618 m

Question 27.
(i) Define the following terms:
(a) Molarity
(b) Molal elevation constant (K_b)
(ii) A solution containing 15 g urea
(molar mass = 60 g mol^-1) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g mol^-1) in water. Calculate the mass of glucose present in one litre of its solution. (CBSE 2014)
OR
(i) What type of deviation is shown by a mixture of ethanol and acetone? Give reason.
(ii) A solution of glucose (molar mass = 180 g mol^-1) in water is labelled as 10% (by mass). What would be the molality and molarity of the solution?
(Density of solution = 1.2 g mL^-1)
Answer:
(i) (a) Molarity is defined as number of moles of solute dissolved per litre of solution.

Its unit is mol L^-1 or M.

(b) Molal elevation constant K_b is the elevation in boiling point for 1 molal solution.
(ii) For isotonic solutions,
π (urea) = π (glucose)

OR

(i) Mixture of ethanol and acetone shows positive deviations from Raoult’s law. This is because in ethanol, the molecules are held together due to hydrogen bonding as:

When acetone is added to ethanol, there are weaker interactions between acetone and ethanol than ethanol-ethanol interactions. Some molecules of acetone occupy spaces between ethanol molecules and consequently, some hydrogen bonds in alcohol molecules break and attractive forces between ethanol molecules are weakened.

Therefore, the escaping tendency of ethanol and acetone molecules from solution increases. Thus, the vapour pressure of the solution is greater than the vapour pressure as expected according to Raoult’s law.

(ii) W_B = 10 g, wt. of solvent = 90 g
M_B = 180 g mol^-1

Question 28.
(i) Calculate the freezing point of solution when 1.9 g of MgCl2 (M = 95 g mol-1) was dissolved in 50 g of water, assuming MgCl₂ undergoes complete ionisation.
(K_f for water = 1.86 K kg mol^-1)
(ii) (a) Out of 1 M glucose and 2 M glucose, which one has a higher boiling point and why?
(b) What happens when the external pressure applied becomes more than the osmotic pressure of solution? (CBSE Delhi 2016)
OR
(i) When 2.56 g of sulphur was dissolved in 100 g of CS₂, the freezing point lowered by 0.383 K. Calculate the formula of sulphur (S_x).
(K_f for CS₂ = 3.83 K kg mol^-1, Atomic mass of sulphur = 32 g mol^-1)
(ii) Blood cells are isotonic with 0.9% sodium chloride solution. What happens if we place blood cells in a solution containing
(a) 1.2% sodium chloride solution?
(b) 0.4% sodium chloride solution?
Answer:
(i) ΔT_f = \(\frac{i \mathrm{~K}_{f} \times \mathrm{w}_{\mathrm{B}} \times 1000}{M_{\mathrm{B}} \times \mathrm{w}_{\mathrm{A}}}\)
w_B = 1.9 g, w_A = 50 g, M_B = 95 g mol^-1
K_f = 1.86 K kg mol^-1
MgCl₂ ⇌ Mg²⁺ + 2Cl^–
i = 3
ΔT_f = \(\frac{3 \times 1.86 \times 1.9 \times 1000}{95 \times 50}\)
= 2.232 K
Freezing point of solution
= 273 – 2.232 = 270.768 K

(ii) (a) The elevation in boiling point is a colligative property and depends upon the number of moles of solute added. Higher the concentration of solute added, higher will be the elevation in boiling point. Thus, 2 M glucose solution has higher boiling point than 1 M glucose solution.

(b) When the external pressure applied becomes more than the osmotic pressure of the solution, then the solvent molecules from the solution pass through the semipermeable membrane to the solvent side. This process is called reverse osmosis.
OR
(i) M_B = \(\frac{\mathrm{K}_{f} \times 1000 \times \mathrm{W}_{\mathrm{B}}}{\mathrm{W}_{\mathrm{A}} \times \Delta \mathrm{T}_{f}}\)
W_B = 2.56 g, W_A = 100 g, ΔT_f = 0.383 K_f = 3.83 K kg mol^-1
Let the molecular formula of sulphur
= S_x
32 × x = 256
x = \(\frac{256}{32}\) = 8
∴ Molecular formula = S₈

(ii) (a) 1.2% sodium chloride solution is hypertonic with respect to 0.9% sodium chloride solution or blood cells. When red blood cells are placed in this solution, water flows out of the cell and they shrink due to loss of water by osmosis.

(b) 0.4% sodium chloride solution is hypertonic with respect to 0.9% sodium chloride solution or blood cells. When red blood cells are placed in this solution, water flows into the cells and they swell.

Question 29.
A 4% solutioin (w/w) of sucrose (M = 342 g mol^-1) in water has a freezing point of 271.15 K. Calculate the freezing point of 5% glucose (M = 180 g mol^-1) in water.
(Given: Freezing point of pure water = 273.15 K)
Answer:
In case of sucrose:
ΔT_f = (273.15 – 271.15) K = 2.00 K
Molar mass of sucrose (C₁₂H₂₂O₁₁)
= (12 × 12) + (22 × 1) + (11 × 16) = 342 g mol^-1
4% solution (w/w) of sucrose in water means 4 g of cane sugar in (100 – 4) g = 96 g of water.
Number of moles in 4 g sucrose in water
= 4/342 mol or 0.01169 mol
Therefore, molality of the solution,
m = 0.011696 mol / 0.096 kg
Or
m = 0.1217 mol kg^-1
Now applying the relation,
ΔT_f = K_f × m
⇒ K_f = ΔT_f / m
⇒ 2.00 K/0.1217 mol kg^-1
= 16.4338 K kg mol^-1
Molar mass of glucose (C₆H₁₂O₆) = (6 × 12) + (12 × 1) + (6 × 16) = 180 g mol^-1
5% glucose in water means 5 g of glucose is present in (100 – 5) g = 95 g of water.
∴ Number of moles of glucose = 5/180 mol
= 0.0278 mol
Therefore, molality of the solution, m = 0.0278 mol / 0.095 kg
= 0.2926 mol kg^-1
Applying the relation, ΔT_f = K_f × m
ΔT_f = (16.4338 K kg mol^-1) × (0.2926 mol kg^-1 )
= 4.81 K (approx.)
Hence, the freezing point of 5% glucose solution is (273.15 – 4.81) K = 268.34 K.

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 = H_p – H_r and H_r < H_p.

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 + Δn_gRT.

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 q_p and q_v?
Answer:
q_p = q_v + Δng RT, where Δn = n_p– n_r (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 Δ_fH of SO₂?
Answer:
S(s) + O₂(g) → SO₂(g); Δ_fH = – 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:

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) N^2-(g) + e- → N^3-(g).
Answer:
Endothermic (higher electron affinities are required).

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

Δ_r G^θ = – 13.6 kJ mol^-1.
Answer:
-Δ_r G^θ = 2.303 RT log K

Question 17.
At 60°C, N₂O₄ is 50% dissociated. Calculate Δ_r G^θ 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

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

∴ 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 (C_v): 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 (C_v).

Heat capacity at constant pressure (C_p): 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 (C_p).

Relationship between C_p and C_v: C_p – C_v = 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 q_p = q_v + Δn_gRT?
Answer:
Enthalpy change ΔH = q_p; where q_p = heat change at constant pressure,
Internal energy change ΔU = q_v; where q_v = 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 + Δn_gRT
or
q_p = q_v + Δn_gRT

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 C_p – C_v = R.
Answer:

Question 28.
Derive the expression -ΔG = w non-expansion
Answer:
From I law of thermodynamics

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); Δ_f 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); ΔH^θ = – 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) Δ_cHe = 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.

(vi) Thermochemical equation.
Answer:
Thermochemical Equation: A balanced chemical equation together with the value of its A^H is called a thermochemical equation.

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, K_c = 1.8 × 10^-7. What is ΔG° for the reaction? (R = 8.314 JK^-1 mol^-1).
Answer:

Question 2.
Calculate the equilibrium constant, K, for the following reaction at 400 K?
2NOCl(g) ⇌ 2NO(g) + Cl₂(g)
Given that Δ_rH° = 80.0 kJ mol^-1 and Δ_rS° = 120 JK^-1 mol^-1.
Answer:

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 Δ_rG° = 38.484 kJ.
Answer:

= -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:

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
Δ_fH = – 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:

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.

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:

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:

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 Δ_r, 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); Δ_rH° =?

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:

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:

= [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 (Δ_fG°) 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 Δ_fG°(NH₃) = – 16.8 kJ mol^-1, Δ_fG°(NO) = + 86.7 kj mol^-1; Δ_fG°(H₂O) = – 237.2 kJ mol^-1

= – [4 × ΔfG°(NH3) + 5 × ΔfG°(O2)]
= [4 × 86.7 + 6(- 237.2)] – [4 × (- 16.8) + 5 × 0]
= – 1009.2 kJ
Since Δ_rG° 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:
Δ_rG° = – 2.303 RT log K
= – 2.303 × 8.314 × 298 × log (1.06 × 10⁵)
On solving, Δ_rG° = – 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:
Δ_rG° = – 2.303 RT log K
R = 8.314 JK^-1 mol^-1
∴ Δ_fG° = – 2.303 (8-314 JK^-1 mol^-1) × 298 × (log 2.47 × 10^-29)

On solving
Δ_rG° = 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, Δ_rG° at the given temperature is – 13.6 kJ mol^-1.
Answer:
-Δ_rG° = 2.303 RT log K

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? (Δ_rH° = -1648 × 103 J mol^-1).
Answer:
The spontaneity of a reaction is determined from ΔS_total
(which is = Δ_sys + Δ_sutu)

Now ΔS_surr = – \(\frac{\Delta_{r} \mathrm{H}^{0}}{\mathrm{~T}}\) 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
Δ_r S_total = 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:

No. of moles in 18 g H₂O(l) = \(\frac{18 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 1 mol.
Δ_vap U = Δ_vap H° – pΔV = Δ_vapH – Δn_g RT (assuming steam behaving as an ideal gas)
Δ_vap H° – Δn_g RT = 40.66 kJ mol^-1 – (1)(8.314 JK^-1 mol^-1)(373 K) (10^-3 kJ) ,
∴ Δ_vap 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 = C_v × 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 = – C_v × 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 Δng_g = 0
∴ ΔH = ΔU = – 2.48 × 10² kJ mol^-1.

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?

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:
V_av = \(\frac{v_{1}+v_{2}+v_{3}+v_{4}}{4}\)

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

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 C_rms = \(\sqrt{\frac{3 \mathrm{P}}{\rho}}=\sqrt{\frac{3 \mathrm{P}}{\mathrm{mn}}}\)

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 S₁  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 \(\frac{1}{2}\) k_BT.

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,

( ∵ 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.

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 C_rms ∝ \(\sqrt{T}\).

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 C_rms ∝ \(\frac{1}{\sqrt{\rho}}\)

Also ρ₀ = 16 ρ_H. So C_rms 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

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

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,

Also, we know that

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

Also, we know that r.m.s. velocity is directly proportional to the rate of diffusion (r) of the gas, i.e.

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:

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

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 = \(\frac{\text { Mass }}{\text { Density }}\)

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.

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. ρ_Hg = 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

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,

(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

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

Question 9.
At what temperature will the average velocity of O₂ molecules be sufficient so as to escape from earth? V_e = 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,
V_e = 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

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 = hρ_water 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,

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:
V_rms = \(\sqrt{\frac{3 R T}{M}}\)
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
C_v = \(\frac{5}{2}\) 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 = ?

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
\(\frac{C_{2}^{2}}{C_{1}^{2}}=\frac{T_{2}}{T_{1}}\)

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
C_rms = ?

Using the relation,

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, C_rms = ?
Using the relation,
K.E = E ∝ T, we get

Also, we know that

(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

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

∴ Using the relation,

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

∴ According to the statement,

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

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 = ?
T_H = temp. of hydrogen
= 150°C = 150 + 273 = 423 K
C₀ = C_H

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, C_p = 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
C_rms =?

Using the relation,


Now using the relation,

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

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,

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 l^st, 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:

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.

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 = Nm^2- =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:

= 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

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.
P_{dry gas}= P_{moist 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
P_A = n_A \(\frac{RT}{V}\)
P_B = n_B \(\frac{RT}{V}\)
P_C = n_C \(\frac{RT}{V}\)
where n_A, n_B n_C stand for their respective moles in the same vessel .
[V = constant, keeping T constant]
Then according to Dalton’s law ‘

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,

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]

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

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

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:

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.

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

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

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:

∴ 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,

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:

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:

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 × \(\frac{RT}{P}\)

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

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

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

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

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]

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) E_Red = – 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^–
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: