Online Education for Atoms and Molecules Class 9 Extra Questions and Answers Science Chapter 3

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Online Education for Class 9 Science Chapter 3 Extra Questions and Answers Atoms and Molecules

Extra Questions for Class 9 Science Chapter 3 Atoms and Molecules with Answers Solutions

Atoms and Molecules Class 9 Extra Questions Very Short Answer Type

Atoms And Molecules Class 9 Extra Questions Question 1.
Define atomic mass unit.
Answer:
Atomic mass unit of an element is one twelfth (1/12th) of the mass of one atom of carbon-12.

Class 9 Science Chapter 3 Extra Questions And Answers Question 2.
Write the valency of sulphur in H₂S, SO₂ and SO₃.
Answer:
The valency of sulphur in H₂S, SO₂ and SO₃ are 2, 4, and 6 respectively.

Atoms And Molecules Extra Questions Question 3.
What is the difference between 2H and H₂?
Answer:
2H means two atoms of hydrogen, H₂ means a molecule of hydrogen which contains two atoms.

Class 9 Atoms And Molecules Extra Questions Question 4.
Where do we use the words mole and mol?
Answer:
We use the word mole to define the number of atoms, molecules, ions or particles having a mass equal to its atomic or molecular mass in grams, while as a unit, we call it mol.

Extra Questions On Atoms And Molecules Class 9 Question 5.
The valency of an element A is 4. Write the formula of its oxide.
Answer:
The formula of its oxide is A₂O₂ or AO₂.

Extra Questions Of Atoms And Molecules Class 9 Question 6.
Why are Dalton’s symbols not used in chemistry?
Answer:
Dalton was the first scientist to use the symbol for the name of the elements in a specific sense but it was difficult to memorise and in use, so Dalton’s symbols are not used in chemistry.

Class 9 Science Ch 3 Extra Questions Question 7.
Which postulate of Dalton’s Atomic theory is the basis of law of conservation of mass?
Answer:
“Atoms can neither be created nor destroyed a physical or a chemical change”.

Class 9 Chapter 3 Science Extra Questions Question 8.
Calculate the formula unit mass of CaCl₂. [NCERT Exemplar]
Answer:
Atomic mass of Ca + (2 × atomic mass of Cl)
= 40 + 2 × 35.5 = 40 + 71 = 111 u.

Ch 3 Science Class 9 Extra Questions Question 9.
What does the symbol ‘u’ represent?
Answer:
The symbol ‘u’ represents unified mass.

Class 9 Chemistry Chapter 3 Extra Questions Question 10.
Avogadro’s number represents how many particles?
Answer:
Avogadro’s numbers (N₀) represents 6.022 × 10²³ particles.

Mole Concept Class 9 Extra Questions Question 11.
Formula of the carbonate of a metal M is M₂CO₃. Write the formula of its chloride.
Answer:
The valency of the metal (M) in M₂CO₃ is (1+) i.e., metal exists as M⁺ ion. Therefore, the formula of metal chloride is MCl.

Atom And Molecules Class 9 Extra Questions Question 12.
Sample A contains one gram molecule of oxygen molecules and Sample B contains one mole of oxygen molecules. What is the ratio of the number of molecules in both the samples?
Answer:
One gram molecules is the same as one gram mole of a substance. Therefore, both the samples A and B contain the same number of molecules (6.022 × 10)²³ and the ratio is 1: 1.

Class 9 Ch 3 Science Extra Questions Question 13.
Name the compound Al₂(SO₄)₃ and mention the ions present in it.
Answer:
The compound is called aluminium sulphate; cation: Al³⁺; anion: (SO₄)^2-

Class 9 Science Chapter 3 Extra Questions Question 14.
Chemical symbol of sodium is Na. What is its Latin name?
Answer:
Natrium.

Class 9 Science Atoms And Molecules Extra Questions Question 15.
Give one example of a polyatomic cation.
Answer:
NH₄⁺ (Ammonium ion).

Atoms And Molecules Class 9 Extra Numerical Questions Question 16.
MNO₃ is the formula of nitrate of metal M. Write the formula of its oxide.
Answer:
M₂O

Ncert Class 9 Science Chapter 3 Extra Questions Question 17.
A vessel contains W molecules of oxygen at a certain temperature and pressure. How many molecules of sulphur dioxide can the vessel accommodate at the same temperature and pressure?
Answer:
N molecules.

Atoms And Molecules Class 9 Extra Questions With Answers Question 18.
What do you understand by a polyatomic ion? Give two examples.
Answer:
A group of atoms carrying positive or negative charge is called a polyatomic ion, e.g., NO₃^–, SO₄^2-, NH₄⁺

Chapter 3 Science Class 9 Extra Questions Question 19.
Give two examples each of bivalent cations and bivalent anions.
Answer:
Bivalent cations = Zn²⁺, Ca²⁺. Bivalent anions = SO₄^2-, C₃^2-.

Science Class 9 Chapter 3 Extra Questions Question 20.
What is ‘molar volume’? What is its value?
Answer:
The volume occupied by one mole of a gas under standard conditions of temperature and pressure, i.e., STP conditions (0°C and 1 atmosphere i.e., pressure) is called molar volume of the gas. Its value is 22.4 L at STP.

Question 21.
What is the valency of calcium in CaCO₃?
Answer:
The valency of Ca in CaCO₃ is 2⁺ [i.e. Ca²⁺].

Question 22.
Can we regard sodium as a monoatomic element?
Answer:
No, sodium is a metal and does not exist as a single atom.

Question 23.
What happens to an element ‘A’ if its atom gains two electrons?
Answer:
It changes to a divalent anion [A^2-].

Question 24.
Do atoms in some of the elements have actually fractional mass?
Answer:
No, the atoms do not have fractional mass. Only their average comes out to be in fraction in elements which exists as isotopes.

Atoms and Molecules Class 9 Extra Questions Short Answer Type 1

Question 1.
Cinnabar (HgS) is a prominent ore of mercury. How many grams of mercury are present in 225 g of pure HgS? Molar mass of Hg and S are 200.6 g mol^-1 and 32 g mol^-1 respectively. [NCERT Exemplar]
Answer:
Molar mass of HgS = 200.6 + 32 = 232.6 g mol^-1
Mass of Hg in 232.6 g of HgS = 200.6 g
Mass of Hg in 225 g of HgS = \(\frac { 200.6 }{ 232.6 }\) × 225 = 194.05 g.

Question 2.
A gold sample contains 90% of gold and the rest copper. How many atoms of gold are present in one gram of this sample of gold? [NCERT Exemplar]
Answer:
One gram of gold sample will contain = \(\frac { 90 }{ 100 }\) = 0.9 g of gold
Atomic mass of gold =
= \(\frac { 0.9 }{ 197 }\)
One mole of gold contains N₀ atoms = 6.022 × 10²³
0.0046 mole of gold will contain = 0.0046 × 6.022 × 10²³
= 2.77 × 10²¹

Question 3.
A sample of ethane (C₂H₆) gas has the same mass as 1.5 × 10²⁰ molecules of methane (CH₄). How many C₂H₆ molecules does the sample of the gas contain? [NCERT Exemplar]
Answer:

Question 4.
Write the cations and anions present (if any) in the following compounds
(a) CH₃COONa
(b) NaCl
(c) H₂
(d) NH₄NO₃
Answer:

Question 5.
How many molecules of water are present in a drop of water which has a mass of 50 mg?
Answer:
We know that:
1 mole of a compound = 6.023 × 10²³ atoms
= Gram molecular mass
Gram molecular mass of H₂O = 18 g
18 g = 6.023 × 10²³ atoms

Question 6.
Find the number of protons and neutrons in the nucleus of an atom of an element X which is represented as ²⁰⁷ X₈₂.
Answer:
The element is ²⁰⁷ X₈₂.
Now, 82 = Atomic Number
207 = Mass Number.

(a) Atomic number = Number of protons
82 = Number of protons.

(b) Mass number = Number of protons + Number of neutrons
207 = 82 + Number of neutrons
207 – 82 = Number of neutrons
125 = Number of neutrons.

Question 7.
Calculate the molar mass of Na₂SO₄ and CaCO₃.
Answer:
Molar Mass of Na₂SO₄ =
2 × Mass of sodium + 1 × Mass of sulphur + 4 × Mass of oxygen
= 2 × 23 + 1 × 32 + 4 × 16
= 46 + 32 + 64
= 142 a.m.u.

Molar Mass of CaCO₃ =
1 × Mass of calcium + 1 × Mass of carbon + 3 × Mass of oxygen
= 40 + 12 + 3 × 16
= 40 + 12 + 48
= 100 a.m.u.

Question 8.
What is wrong with the statement ‘1 mol of hydrogen’?
Answer:
The statement is not correct. We must always write whether hydrogen is in atomic form or molecular form. The correct statement is 1 mole of hydrogen atoms or one mole of hydrogen molecules.

Question 9.
(a) Calculate the relative molecular mass of water (H₂O). [NCERT Exemplar]
(b) Calculate the molecular mass of HNO₃.
Answer:
(a) Atomic mass of hydrogen = 1 u,
oxygen = 16 u
So the molecular mass of water, which contains two atoms of hydrogen and one atom of oxygen is
= 2 × 1 + 1 × 16 = 18u

(b) The molecular mass of HNO₃ = the atomic mass of H + the atomic mass of N + 3 × the atomic mass of O
= 1 + 14 + 48 = 63 u

Question 10.
Identify the cations and anions in the following compounds:
(a) CH₃COONa
(b) NH₃
(c) NH₄
(d) SrCl₂
Answer:
(a) Na⁺, C₃H₃COO^–
(b) It is a molecular compound
(c) NH₄⁺, Cl^–
(d) Sr²⁺, 2Cl^–

Question 11.
Calculate the total number of electrons present in 1.6 g of methane.
Answer:
Molecular mass of CH₄ = 12+ 1 × 4 = 16 g
16 g = 1 mole
1.6 g = 0.1 mole
1 mole of CH₄ has 6.02 × 10²³ molecules.
0.1 mole has 6.02 × 10²² molecules.
1 molecule of methane contains (6 + 1 × 4) = 10 electrons
∴ 6.02 × 10²² molecule will contain = 10 × 6.02 × 10²² = 6.02 × 10²³ electrons.

Question 12.
A 0.24 g sample of a compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Answer:
% oi any element in a compound =
% of boron = \(\frac { 0.096 }{ 0.24 }\) × 100 = 40%
% of oxygen = \(\frac { 0.144 }{ 0.24 }\) × 100 = 60%.

Question 13.
The percentage composition of sodium phosphate as determined by analysis is 42.1% sodium, 18.9% phosphorus and 39% oxygen. Find the empirical formula of the compound.(Work upto two decimal places).
(Atomic masses: Na = 23, P = 31, O = 16)
Answer:

Empirical formula = Na₃PO₄.

Question 14.
A sample of vitamin C is known to contain 2.58 × 10²⁴ oxygen atoms. How many moles of oxygen atoms are present in the sample?
Answer:
1 mole of oxygen atom = 6.023 × 10²³ atoms
Number of moles of oxygen atoms = \(\frac{2.58 \times 10^{24}}{6.022 \times 10^{23}}\) = 4.28 mol
4.28 moles of oxygen atoms.

Question 15.
The empirical formula of a compound is C₂H₄O. Its relative molecular mass is 88. Find the molecular formula.
Answer:
Let the molecular formula be (C₂H₄O)n.
Molecular mass = n × empirical formula mass 88 = n × (2 × 12 + 4 × 1 + 16)
n = \(\frac { 88 }{ 44 }\) = 2
The molecular formula is (C₂H₄O)₂ = C₄H₈O₂.

Question 16.
What is the mass in grams of one molecule of caffeine (C₈H₁₀N₄O₂)?
Answer:
1 mole of caffeine molecules = Gram molecular mass of C₈H₁₀N₄O₂
= 8 × 12 + 10 × 1 + 4 × 14 + 2 × 16g
= 96 + 10 + 56 + 32 g = 194 g
Also, 1 mole of caffeine molecules = 6.022 × 10²³ molecules
Thus, 6.022 × 1023 molecules of caffeine have mass = 194 g
1 molecule of caffeine will have mass = \(\frac{194}{6.022 \times 10^{23}}\) g = 3.22 × 10^-22

Question 17.
What is the difference between cation and anion?
Answer:
Cation:

It is positively charged, exam: Na⁺
It is formed from metal atoms.
On passing electric current, it moves towards cathode.
It is smaller in size than its parent atom.

Anion:

It is negatively charged, example : Cl^–
It is formed from non-metal atoms.
On passing electric current, it moves towards anode.
It is larger in size than its parent atom.

Question 18.
Differentiate between:
(i) Atom and molecule.
(ii) Molecular mass and formula of unit mass.
Answer:
(i) Atoms cannot exist independently but molecules can.

(ii) Molecular mass in the sum of masses of the atoms in the molecule whereas, formula unit mass is sum of atomic masses of the atoms in its empirical formula.

Atoms and Molecules Class 9 Extra Questions Short Answer Type 2

Question 1.
Calcium chloride when dissolved in water dissociates into its ions according to the following equation.
CaCl₂(aq) → Ca²⁺ (aq) + 2Cl^–(aq)
Calculate the number of ions obtained from CaCl₂ when 222 g of it is dissolved in water. [NCERT Exemplar]
Answer:
1 mole of calcium chloride = 111 g
222 g of CaCl₂ is equivalent to 2 moles of CaCl₂ Since 1 formula unit of CaCl₂ gives 3 ions, therefore, 1 mol of CaCl₂ will give 3 moles of ions.
2 moles of CaCl₂ would give 3 × 2 = 6 moles of ions.
No. of ions = No. of moles of ions × Avogadro number
= 6 × 6.022 × 10²³
= 36.132 × 10²³
= 3.6 132 × 10²⁴ ions.

Question 2.
Give the formulae of the compounds formed from the following sets of elements. [NCERT Exemplar]
(a) Calcium and fluorine
(b) Hydrogen and sulphur
(c) Nitrogen and hydrogen
(d) Carbon and chlorine
(e) Sodium and oxygen
(f) Carbon and oxygen
Answer:
(a) CaF₂
(b) H₂S
(c) NH₃
(d) CCl₄
(e) Na₂O
(f) CO, CO₂

Question 3.
A compound was found to have the following percentage composition by mass Zn = 22.65%, S = 11.15%, H = 4.88%, O = 61.32%. The relative molecular mass is 287 g/mol. Find the molecular formula of the compound, assuming that all the hydrogen in the compound is present in water of crystallisation.
Answer:
Zn : S : O : H = \(\frac { 22.65 }{ 65 }\) : \(\frac { 11.15 }{ 32 }\) : \(\frac { 61.32 }{ 16 }\) : \(\frac { 4.88 }{ 1 }\)
= 0.3485 : 0.3484 : 3.833 : 4.88
To obtained an integral ratio, we divide by smallest number
= \(\frac { 0.3485 }{ 0.3484 }\) : \(\frac { 0.3484 }{ 0.3484 }\) : \(\frac { 3.833 }{ 0.3484 }\) : \(\frac { 4.88 }{ 0.3484 }\)
= 1 : 1 : 11 : 14
∴ Empirical formula is Zn SO₁₁H₁₄.
Let molecules formula be (ZnSO₁₁H₁₄)n.
RMM for the molecules = n [65 + 32 + (11 × 16) + 14]
Formula = 287
287 n = 287
n = 1
∴ Molecular formula is ZnSO₁₁H₁₄.

Question 4.
What are the failures of Dalton Atomic theory?
Answer:
Failures of Dalton Atomic Theory are:

Atom is not the smallest particle as it is made up of protons, neutrons and electrons.
Atom’s mass can be transformed to energy (E = mc²) and hence can be created and destroyed.
Atoms of one element have been changed into atoms of another element through artificial transmutation of elements.
Atoms of same element need not resemble each other in all respects as isotopes (different forms of same element) exist.
Atoms of different elements need not differ in all respects as isobars (same forms of different elements) exist.

Question 5.
The following questions are about one mole of sulphuric acid [H₂SO₄]?
(a) Find the number of gram atoms of hydrogen in it.
(b) How many atoms of hydrogen does it have?
(c) How many atoms (in grams) of hydrogen are present for every gram atom of oxygen in it?
(d) Calculate the number of atoms in H₂SO₄?
1 mole of H₂SO₄ = Gram Molecular mass = 6.023 × 10²³ molecules
Answer:
(a) In H₂SO₄ → 2 gram atoms of hydrogen are present

(b) One mole of H₂SO₄ contains = 6.023 × 10²³ molecules
= 2 × 6.023 × 10²³ atoms of H + 6.023 × 10²³ atoms of S + 4 × 6.023 ×10²³ atoms of O
∴ 2H = 2 × 6.023 × 10²³
= 12.046 × 10²³

(c) In H₂SO₄;
For every 2 hydrogen there are oxygen atoms
So for 1 hydrogen = \(\frac { 4 }{ 2 }\) = oxygen are present
= 2 oxygen atoms are present o
For 1 oxygen = \(\frac { 2 }{ 4 }\) hydrogen atoms are present
= 0.5 hydrogen atoms are present.

(d) 1 mole of H₂SO₄ = 6.023 × 10²³ atoms

Question 6.
Write the formulae for the following and calculate the molecular mass for each one of them. [NCERT Exemplar]
(а) Caustic potash
(b) Baking powder
(c) Lime stone
(d) Caustic soda
(e) Ethanol
(f) Common salt.
Answer:
(a) KOH
[39 + 16 + 1] = 56 g mol^-1

(b) NaHCO₃
23 + 1 + 12 + (3 × 16) = 84 mol^-1

(d) NaOH
23 + 16 + 1 = 40 g mol^-1

(e) C₂H₆OH = C₂H₆O
2 × 12 + (6 × 1) + 16 = 46 g mol^-1

(f) Nacl
23 + 35.5 = 58.5 g mol^-1

Question 7.
Calculate the mass of the following: [NCERT Exemplar]
(i) 0.5 mole of N₂ gas (mass from mole of molecule)
(ii) 0.5 mole of N atoms (mass from mole of atom).
(iii) 3.011 × 10²³ number of N atoms (mass from number)
(iv) 6.022 × 10²³ number of N₂ molecules mass from number.
Answer:
(i) mass = molar mass x number of moles
⇒ m = M × n = 28 x 0.5 = 14 g

(ii) mass = molar mass x number of moles
⇒ m = Mxn = 14x 0.5 = 7 g

(iii) The number of moles, n

(iv) n = \(\frac{N}{N_{0}}\)
⇒ m = M x \(\frac{N}{N_{0}}\) = 28 x \(\frac{6.022 \times 10^{2}}{6.022 \times 10^{2}}\)
= 28 x 1 = 28 g.

Question 8.
Calculate the number of particles in each of the following:
(i) 46 g of Na atoms (number from mass)
(ii) 8 g O₂ molecules (number of molecules from mass)

(iii) 0.1 mole of carbon atoms (number from given moles)
Answer:
(i) The numbers of atoms

⇒ N = \(\frac { m }{ M }\) x N₀
⇒ N = \(\frac { 46 }{ 23 }\) x 6.022 x 10²³
⇒ N = 12.044 x 10²³

(ii) The number of molecules

Atomic mass of oxygen = 16 u
molar mass of O₂ molecules
= 16 x 2 = 32 g
⇒ N = \(\frac { 8 }{ 32 }\) x 6.022 x 10²³
⇒ N = 1.5055 x 10²³
= 1.51 x 10²³

(iii) The number of particles (atom) = numbers of moles of particles x Avogadro number
N = n x N₀ = 0.1 x 6.022 x 10²³.
= 6.022 x 10²².

Question 9.
A flask contains 4.4 g of CO₂ gas. Calculate
(a) How many moles of CO₂ gas does it contain?
(b) How many molecules of CO₂ gas are present in the sample?
(c) How many atoms of oxygen are present in the given sample? (Atomic mass of C = 12 u, O = 16 u)
Answer:
(a)

(b) 1 mole of CO₂ has molecules
= 6.022 x 10²³
0. 1 mole of CO₂ has molecules
= 6.022 x 10²³ x (0.1) = 6.022 x 10²²

(c) 1 mole of CO₂ has oxygen atoms = 2 x N₀
0. 1 mole of CO₂ has oxygen atoms = 2 x N₀ x 0.1
= 2 x 6.022 x 10² x 0.1 = 1.204 x 10²⁴ atoms.

Question 10.
Calculate the number of moles for the following: [NCERT Exemplar]
(ii) 12.044 x 10²³ number of He atoms (finding mole from number of particles).
Answer:
No of moles = n
Given mass = m
Molar mass = M
Given number of particles = N
Avogadro number of particles = N₀

(i) Atomic mass of He = 4 u
Molar mass of He = 4 g
Thus, the number of moles =
= n = \(\frac { m }{ M }\) = \(\frac { 52 }{ 4 }\) = 13

(ii) We know,
1 mole = 6.022 x 10²³
The number of moles

Question 11.
Write the molecular formulae of all the compounds that can be formed by the combination of following ions:
Cu²⁺, Na⁺, Fe³⁺, Cb, SO₄^2-, PO₄^3-
Answer:
(i) CuCl₂, CuSO₄, Cu₃(PO₄)₂
(ii) NaCl, Na₂SO₄, Na₃PO₄
(iii) FeCl₃, Fe₂(SO₄)₃, FePO₄

Question 12.
Give the names of any three elements whose names have been derived from Latin. Give their Latin names and symbols.
Answer:

Question 13.
What is the simplest formula of the compound which has the following percentage composition:
Carbon 80%, Hydrogen 20%
If the molecular mass is 30, calculate its molecular formula.
Answer:
Calculation of empirical formula:

∴ Empirical formula is CH₃.
Calculation of molecular formula:
Empirical formula mass = 12 x 1 + 1 x 3 = 15

\(\frac { 30 }{ 15 }\) = 2
Molecular formula = Empirical formula x 2 = CH₃ x 2 = C₂H₆.

Question 14.
On heating, potassium chlorate decomposes to potassium chloride and oxygen. In one experiment 30.0 g of potassium chlorate generates 14.9 g of potassium chloride and 9.6 g of oxygen. What mass of potassium chlorate remains undecomposed?
Answer:
Potassium chlorate → Potassium chloride + Oxygen
Total mass of potassium chlorate before reaction = 30.0 g
After reaction, mass of potassium chloride = 14.9 g Mass of oxygen = 9.6 g
Let mass of undecomposed potassium chlorate = x g
Total mass after reaction = (x + 14.9 + 9.6) g
According to law of conservation of mass the total mass before and after the reaction remains constant.
∴ 30.0 = x + 14.9 + 9.6
x = 30 – 14.9 – 9.6 = 5.5
∴ Mass of undecomposed potassium chlorate
= 5.5 g

Question 15.
Naturally occurring Boron consists of two isotopes whose atomic mass are 10.01 and 11.01. The atomic mass of natural Boron is 10.81. Calculate the percentage of each isotope in natural Boron.
Answer:
Suppose percentage of isotope with atomic mass 10.01 = x
Then percentage of isotope with atomic mass 11.01 = 100 – x

∴ % of isotope with atomic mass 10.01 = 20%
∴ % of isotope with atomic mass 11.01 = 80%

Question 16.
Which amongst the following has more number of atoms, 11.5 g of sodium or 15 g of calcium?
How? (Given atomic mass of Na = 23, Ca = 40)
Answer:
23 g of Na = 1 mole
11.5 g of Na = 11.5/23 mole
Now, 1 mole of Na contains = 6.022 x 10²³ atoms
11.5/23 mole of Na contains = 6.022 x 10²³ x 11.5²³ mole
= 3.011 x 10²³ atoms
40 g of Ca = 1 mole
15 g of Ca = 15/40 moles
Now, 1 mole of Ca contains = 6.022 x 10²³ atoms
15/40 mole of Ca contains = 6.022 x 10²³ x 15/40 mole
= 2.258 x 10²³ atoms
Na has more number of atoms.

Question 17.
(i) Name the body which approves the nomenclature of elements and compounds.
(ii) The symbol of sodium is written as Na and not as S. Give reason.
(iii) Name one element which forms diatomic and one which forms tetraatomic molecules.
Answer:
(i) International Union of Pure and Applied Chemistry.

(ii) The symbol of Na (sodium) is derived from its Latin name Natrium.

(iii) (a) Element forming diatomic molecule: H₂, O₂, N₂ (only one).
(b) Element forming tetratomic molecule: P₄.

Atoms and Molecules Class 9 Extra Questions Long Answer Type Questions

Question 1.
(a) Define empirical formula and molecular formula of a compound. How are molecular formula and empirical formula related to each other?
(b) A hydrocarbon contains 10.5 g of carbon per gram of hydrogen. Calculate the empirical formula.
Answer:
(a) Empirical formula is the chemical formula which gives us the simplest whole number ratio between the atoms of various elements present in one molecule of a compound. Molecular formula gives us the actual number of atoms of various elements present in one molecule of a compound.
Molecular formula = n x Empirical formula
Where n is a simple whole number

(b) In the hydrocarbon,
1 g of hydrogen combines with 10.5 g carbon
∴ % of H in hydrocarbon = \(\frac { 1 }{ 11.5 }\) x 100 = 8.696%
% of C = 100 – 8.696 = 91.304%

Therefore, Empirical formula = CH

Question 2.
(a) What is meant by mole concept?
(b) Calculate the mass of:
(i) 10²² atoms of sulphur
(ii) 0.1 mole of carbon dioxide.
[Atomic mass S = 32, C = 12, O = 16 and Avogadro’s number = 6 x 10²³]
Answer:
(a) 1 mole of a compound has a mass equal to its relative molecular mass expressed in grams.
1 mole = 6.022 x 10²³ number
= Relative mass in grams.

(b) (i) v 6.022 x 10²³ atoms of sulphur weighs = 32 g
∴ 10²² atoms of sulphur weighs = \(\frac{32}{6.022 \times 10^{23}}\) x 10²² = 0.531 g

(ii) Gram molar mass of CO₂ = 12 + 2 x 16 = 44 g
∵ 1 mole of carbon dioxide weighs = 44 g
∴ 0.1 mole of carbon dioxide weighs = \(\frac { 44 g }{ 1 }\) x 0.1 = 4.4 g.

Question 3.
(a) The ratio of hydrogen and oxygen in water is 1: 8 by mass, find out their ratio by number of atoms, in one molecule of water. (Atomic mass: H = 1 u, O – 16 u)
(b) Write the formulae of the following compounds:
(i) Ammonium sulphate
(ii) Magnesium chloride
(Given , ammonium = NH₄²⁺ , sulphate = SO₄^2-, Magnesium = Mg²⁺, chloride = Cl^–)
Answer:
(a)

Ratio of number of atoms for water is H : O = 2 : 1

(b)
(i) Ammonium sulphate: (NH₄)₂SO₄
(ii) Magnesium chloride : MgCl₂

Question 4.
A compound of carbon and sulphur has a composition of 15.8% carbon and 84.2% sulphur.
(a) Find the empirical formula.
(b) The relative molecular mass of the compound is 76. Find the molecular formula.
Answer:
(a)

The empirical formula is CS₂.

(b) Let the molecular formula be (CS₂)_n.
Molecular mass = n x empirical formula mass
76 = n x (12 + 2 x 32)
n = \(\frac { 76 }{ 76 }\) = 1
The molecule formula is (CS₂)₁ = CS₂.

Question 5.
What do the following formulae stand for:
(a) (i) 20
(ii) O₂
(iii) O₃
(iv) H₂O

(b) Give the chemical formulae of the following compounds?
(i) Potassium carbonate
(ii) Calcium chloride

(c) Calculate the formula unit mass of Al₂(SO₄)₃.
(Given atomic mass of Al = 27u, S = 32 u, 0 = 16 u)
Answer:
(a)
(i) Two atoms of oxygen 20
(ii) Diatomic oxygen → O₂ molecule
(iii) Triatomic oxygen → O₃ molecule
(iv) Two atoms of hydrogen and 1 atom of oxygen forming one molecule of water.

(b) K₂CO₃ : Potassium carbonate
CaC₂ : Calcium chloride

Question 6.
(a) Write the chemical symbols of two elements:
(i) Which are formed from the first letter of the elements name.
(ii) Whose name has been taken from the names of the elements in Latin.

(b) Define Avogadro’s constant and molar mass. How are they related to one mole of an atom, molecule, or ion?
Answer:
(a)
(i) C, N, O, etc.
(ii) Na, Fe, K, etc.

(b) Avogadro’s constant:
6.22 x 10²³ is defined as the Avogadro’s constant. It is number of atoms in exactly 12 g of carbon-12.

Molar mass : Mass of 1 mole of substance.

Question 7.
(a) Calculate the mass of 0.5 mole of sulphuric acid. Atomic mass (H = 1 u, S = 32 u, O = 16 u).
(b) Find the number of atoms in 12 g of carbon.
(c) How many atoms are present in (i) H₂S molecule (ii) PO_3-⁴ ions.
(d) Write the names of elements present in (i) quicklime (ii) hydrogen bromide.
Answer:
(a) 1 mole of H₂SO₄ = 98
∴ 1/2 mole of H₂SO₄ = 49 g

(b) 6.023 x 10²³ atoms

(d) (i) Elements in quicklime (CaO)-calcium, oxygen.
(ii) Elements in hydrogen bromide (HBr)-hydrogen, bromine.

Question 8.
(a) Define polyatomic ions. Write an example.
(b) Calculate the formula unit mass of CaCO₃.
(Atomic mass of C = 12 u, Ca = 40 u, O = 16 u)
(c) Calculate the molecular mass of the following:
(i) HNO₃ (ii) CH₃COOH
(Atomic mass of H = 1 u, N = 14 u, O = 16 u, C = 12 u)
Answer:
(a) Cluster of atoms that acts as an ion are called polyatomic ions. example : NH₄⁺, PO_3-⁴, SO_2-⁴, etc.

(b) CaCO₃ = Mass of Ca + Mass of C + Mass of O
= 40 u + 12 u + 16 u x 3 = 100 u.

(ii) CH₃COOH = Mass of C + Mass of H + Mass of O
= 12 u x 2 + 1 u x 4 + 16 u x 2
= 24 u + 4 u + 32 u = 60 u

Atoms and Molecules Class 9 Extra Questions HOTS

Question 1.
In photosynthesis, 6 molecules of carbon dioxide combine with an equal number of water molecules through a complex series of reactions to give a molecule of glucose having a molecular formula C₆H₁₂O₆. How many grams of water would be required to produce 18 g of glucose? Compute the volume of water so consumed assuming the density of water to be 1 g cm^-3. [NCERT Exemplar]
Answer:

1 mole of glucose needs 6 moles of water
180 g of glucose needs (6 x 18) g of water
1 g of glucose will need \(\frac { 108 }{ 180 }\) g of water.
18 g of glucose would need \(\frac { 108 }{ 180 }\) x 18 g of water = 10.8 g
Volume of water used = \(\frac { Mass }{ Density }\) = \(\frac{10.8 \mathrm{g}}{1 \mathrm{g} \mathrm{cm}^{-3}}\) = 10.8 cm³

Question 2.
Fill in the missing data in the Table 3.2. [NCERT Exemplar]
Table 3.2

Answer:

Question 3.
Raunak took 5 moles of carbon atoms in a container and Krish also took 5 moles of sodium atoms in another container of same weight. [NCERT Exemplar]
(a) Whose container is heavier?
(b) Whose container has more number of atoms?
Answer:
(a) Mass of sodium atoms carried by Krish = (5x 23) g = 115 g while mass of carbon atom carried by Raunak = (5 x 12) g = 60 g. Thus, Krish’s container is heavy.

(b) Both the bags have same number of atoms as they have same number of moles of atoms.

Question 4.
How many molecules are present in 1 ml. of water?
Answer:
Molecular mass of H₂O = 18 g
Mass of 1 mole of water = 18 g
18 g of water contain = 6.022 x 10²³ molecules
Density of water = 1 g/ml, therefore 1 ml of water weighs 1 g.
1.0 g of water contain = (6.022 x 10²³)/18 molecules
= 3.34 x 10²² molecules.

Question 5.
A flask P contains 0.5 mole of oxygen gas. Another flask Q contains 0.4 mole of ozone gas. Which of the two flasks contains greater number of oxygen atoms?
Answer:
1 molecule of oxygen (O₂) = 2 atoms of oxygen
1 molecule of ozone (O₃) = 3 atoms of oxygen
In flask P:1 mole of oxygen gas = 6.022 x 10²³ molecules
0.4 mole of oxygen gas = 6.022 x 10²³ x 0.5 molecules
= 6.022 x 10²³ x 0.5 x 2 atoms = 6.022 x 10²³ atoms
In flask Q: 1 mole of ozone gas = 6.022 x 10²³ molecules
0.4 mole of ozone gas = 6.022 x 10²³ x 0.4 molecules
= 6.022 x 10²³ x 0.4 x 3 atoms = 7.23 x 10²³ atoms
∴ Flask Q has a greater number of oxygen atoms as compared to flask P.

Question 6.
Complete the following table:

Answer:

Question 7.
What is the valency of underlined element in the following compounds?
(i) MnO₂ (ii) Ca₃N₂ (iii) ZnSO₄ (iv) CCl₄
Answer:
(i) + 4
(ii) + 2
(iii) + 2
(iv) + 4

Question 8.
A student puts his signature with graphite pencil. If the mass of carbon in the signature is 10^-12 g. Calculate the number of carbon atoms in the signature.
Answer:
Mass of carbon = 10^-12 g
No. of moles of carbon = \(\frac{10^{-12}}{12}\) moles
1 mole of C has = 6.022 x 10²³
∵ 1 mole of C has atoms = 6.022 x 10²³
∴ \(\frac{10^{-12}}{12}\) mole of C will have = \(\frac{6.022 \times 10^{23}}{12}\) x 10¹⁰

Question 9.
Rudra took 5 moles of carbon atoms in a container and Ayush also took 5 moles of sodium atoms in another container of some weight.
(a) Whose container is heavier?
(b) Whose container has more number of atoms?
Answer:
(a) 1 mole of C atoms = 12 g
∴ 5 moles of C atoms = 5 x 12 g = 60 g
1 mole of Na atoms = 23 g
∴ 5 moles of Na atoms = 5 x 23 = 115 g
Thus, Ayush’s container is heavier.

(b) 1 mole of C atoms = 6.022 x 10²³ atoms
1 mole of Na atoms = 6.022 x 10²³ atoms
∴ 5 moles of each will contain the same number of atoms.

Question 10.
10²² atoms of an element ‘X’ are found to have a mass of 930 mg. Calculate the molar mass of the element ‘X’.
Answer:
Molar mass of an element is the mass of Avogadro’s number of atoms.
∴6.022 x 10²³ atoms will have mass = \(\frac{0.930}{10^{22}}\) x 6.022 x 10²³ g = 56.0 g
∴Molar mass of the element = 56 g mol^-1.

Question 11.
Which will contain large number of atoms, 1 g of gold or 1 g of silver? Explain with reason. (Atomic masses of Gold = 197 u, silver = 108 u).
Answer:
No. of moles (n) in 1 g of gold = \(\frac { m }{ M }\) = \(\frac { 1 }{ 197 }\)
No. of moles (n) in 1 g of silver = \(\frac { m }{ M }\) = \(\frac { 1 }{ 108 }\)
Greater the number of moles, greater is the number of atoms present. As \(\frac { 1 }{ 197 }\) > \(\frac { 1 }{ 108 }\), silver will contain greater number of atoms.