# NCERT Solutions Class 9 Science Chapter 3 Atoms and Molecules

### In-Text Questions Solutions

#### In-Text Questions Page 32

1. In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.
sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water

Ans:- Sodium carbonate + acetic acid → Sodium acetate + carbon dioxide +  water
5.3g  +  6g     →     8.2g  +  2.2g  +  0.9g
As per the law of conservation of mass, the total mass of reactants must be equal to the total mass of products.
As per the above reaction, 5.3g + 6g = 2.2g + 0.9 g + 8.2 g = 11.3 g
Hence, the observations are in agreement with the law of conservation of mass.
2. Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?
Ans:- We know hydrogen and water mix in the ratio 1: 8.
For every 1 g of hydrogen, it is 8 g of oxygen.
Therefore, for 3g of hydrogen, the quantity of oxygen required = 3 x 8 = 24 g.
3. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
Ans:- The postulate of Dalton’s Atomic theory which is the result of the law of conservation of mass is, “Atoms can neither be created nor destroyed”.
4. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Ans:- The postulate of Dalton’s atomic theory that can explain the law of definite proportions is, “the relative number and kinds of atoms are equal in given compounds”.

#### In-Text Questions Page 35

1. Define the atomic mass unit.
Ans:- Atomic mass unit is the mass unit equal to exactly one twelfth (1/12th) the mass of one atom of carbon-12.
2. Why is it not possible to see an atom with naked eyes?
Ans:- The size of atom is too small and measured in nanometers. Hence, an atom cannot be visible to the naked eyes.

#### In-Text Questions Page 39

1. Write down the formulae of:
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium suphide
(iv) magnesium hydroxide

Ans:- 2. Write down the names of compounds represented by the following formulae:
(i) Al2(SO4)3
(ii) CaCl2
(iii) K2SO4
(iv) KNO3
(v) CaCO3.

Ans:- (i) Al2(SO4)3 – Aluminium sulphate
(ii) CaCl– Calcium chloride
(iii) K2SO4 – Potassium sulphate
(iv) KNO– Potassium nitrate
(v) CaCO3 – Calcium carbonate
3. What is meant by the term chemical formula?
Ans:- The symbolic representation of the composition of a compound is known as its chemical formula. Ex:- H2O, CO2, etc.
4. How many atoms are present in a
(i) H2S molecule and
(ii) PO43- ion?

Ans:- (i) H2S molecule has total 3 atoms of which 2 atoms are of hydrogen and 1 atom is of sulphur.
(ii) PO43- ion has total 5 atoms of which 1 atom is of phosphorus and 4 atoms are of oxygen.

#### In-Text Questions Page 40

1. Calculate the molecular masses of H2, O2, Cl2, CO2, CH4, C2H6,
C2H4, NH3, CH3OH.

Ans:- The molecular mass of H2
= 2 x atoms atomic mass of H
= 2 x 1 u = 2 u
The molecular mass of O2
= 2 x atoms atomic mass of O
= 2 x 16 u = 32 u
The molecular mass of Cl2
= 2 x atoms atomic mass of Cl
= 2 x 35.5u = 71 u
The molecular mass of CO2
= atomic mass of C + 2 x atomic mass of O
= 12 + ( 2×16) = 44 u
The molecular mass of CH4
= atomic mass of C + 4 x atomic mass of H
= 12 + ( 4 x 1) = 16 u
The molecular mass of C2H6
= 2 x atomic mass of C + 6 x atomic mass of H
= (2 x 12) +(6 x 1) = 24 + 6 = 30 u
The molecular mass of C2H4
= 2 x atomic mass of C + 4 x atomic mass of H
= (2 x 12) + (4 x 1) = 24 + 4 = 28 u
The molecular mass of NH3
= atomic mass of N + 3 x atomic mass of H
= 14 + (3 x 1) = 17 u
The molecular mass of CH3OH
= atomic mass of C + 3 x atomic mass of H + atomic mass of O + atomic mass of H
= (12 + 3×1 + 16 + 1) u = (12+3+17) u = 32 u
2. Calculate the formula unit masses of ZnO, Na2O, K2CO3, given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.
Ans:- The formula unit mass of ZnO
= Atomic mass of Zn + Atomic mass of O
= 65 u + 16 u = 81 u
The formula unit mass of Na2O
= 2 x Atomic mass of Na + Atomic mass of O
= (2 x 23) u +16 u
= 46 u + 16 u = 62 u
The formula unit mass of K2CO3
= 2 x Atomic mass of K + Atomic mass of C + 3 x Atomic mass of O
= 2 x 39 u + 12 u + 3 x 16 u = 138 u

#### In-Text Questions Page42

1. If one mole of carbon atoms weighs 12 gram, what is the mass (in gram) of 1 atom of carbon?
Ans:- 1 mole of carbon = 6.022 X 1023 atoms of carbon
∵ Mass of 1 mole of carbon = 12 g
Mass of 6.022 X 1023 atoms of carbon = 12 g
Mass of 1 atom of carbon = g = 1.99 X 10-23 g
2. Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na = 23 u, Fe = 56 u)?
Ans:-  Atomic mass of Na = 23 u
Atomic mass of Fe = 56 u
To calculate the number of atoms in 100 g of sodium:
∵ 23 g of Na contains = 6.022 X 1023 atoms
1 g of Na contains =  atoms
100 g of Na contains =  atoms = 2.6182 X 1024 atoms
To calculate the number of atoms in 100g of sodium:
∵ 56 g of Fe contains = 6.022 X 1023 atoms
1 g of Fe contains =  atoms
100 g of Fe contains =  x 100 atoms = 1.075 X 1024 atoms
Clearly, 100 g of Na has more atoms.

### Chapter End Exercises

1. A 0.24 g sample of 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.
Ans:- Mass of the sample compound = 0.24 g,
Mass of boron = 0.096 g,
Mass of oxygen = 0.144 g
Percentage of boron
=
Mass of boron
Mass of compound
x 100
= x 100  = 40%
Percentage of oxygen = 100 – percentage of boron = 100 – 40 = 60%
2. When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?
Ans:- 11.00g of carbon dioxide is formed when 3.00g carbon is burnt in 8.00g of oxygen.
Carbon and oxygen are combined in the ratio 3:8 to give carbon dioxide using up all the carbon and oxygen.
Hence, for 3g of carbon and 50g of oxygen, 8g of oxygen is used and 11g of carbon dioxide is formed, the left oxygen is unused i.e., 50-8 = 42 g of oxygen is unused.
This depicts the law of definite proportions – The elements in a compound are present in definite proportions by mass.
3. What are polyatomic ions? Give examples.
Ans:- A group of atoms having some charge is called polyatomic ion. Ex:- SO42-, NH4+ etc.
4. Write the chemical formulae of the following:
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.

Ans:- (a) Magnesium chloride – MgCl2
(b) Calcium oxide – CaO
(c) Copper nitrate – Cu(NO3)2
(d) Aluminium chloride – AlCl3
(e) Calcium carbonate – CaCO3
5. Give the names of the elements present in the following compounds:
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate.

Ans:- (a) Quick lime (CaO)– Calcium and oxygen
(b) Hydrogen bromide (HBr)– Hydrogen and bromine
(c) Baking powder (NaHCO3)– Sodium, Carbon, Hydrogen, Oxygen
(d) Potassium sulphate (K2SO4)– Sulphur, Oxygen, Potassium
6. Calculate the molar mass of the following substances:
(a) Ethyne, C2H2
(b) Sulphur molecule, S8
(c) Phosphorus molecule, P4 (Atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO3

Ans:- (a) Molar mass of Ethyne C2H2
= 2 x Gram atomic mass of C + 2 x Gram atomic mass of H
= (2 × 12) + (2 × 1) = 24 + 2 = 26 g
(b) Molar mass of Sulphur molecule S8
= 8 x Gram atomic mass of S
= 8  x 32 = 256 g
(c) Molar mass of  Phosphorus molecule, P4
= 4 x Gram atomic mass of P
= 4 x 31 = 124 g
(d) Molar mass of Hydrochloric acid, HCl
= Gram atomic mass of H + Gram atomic mass of Cl
= 1 + 35.5 = 36.5 g
(e) Molar mass of Nitric acid, HNO3
= Gram atomic mass of H + Gram atomic mass of Nitrogen + 3 x Gram atomic mass of O
= 1 + 14 + 3 × 16 = 63 g
7. What is the mass of—
(a) 1 mole of nitrogen atoms?
(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?
(c) 10 moles of sodium sulphite (Na2SO3)?

Ans:- (a) Atomic mass of nitrogen = 14 u
Mass of 1 mole of nitrogen atoms = Gram atomic mass of nitrogen
Therefore, mass of 1 mole of nitrogen atom is 14 g
(b) Atomic mass of aluminium = 27 u
Mass of 1 mole of aluminium atoms = 27 g
Mass of 4 moles of aluminium atoms = 4 x 27 = 108 g
(c) Mass of 1 mole of sodium sulphite Na2SO3
= Gram molecular mass of sodium sulphite
= 2 x Gram atomic mass of Na + Gram atomic mass of S + 3 x Gram atomic mass of O
= (2 x 23) + 32 +(3x 16) = 46+32+48 = 126g
Therefore, mass of 10 moles of Na2SO3  = 10 x 126 = 1260 g
8. Convert into mole:
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide

Ans:- Conversion of the above-mentioned molecules into moles is as follows:
(a) Given mass of oxygen gas, m = 12 g
Molar mass of oxygen gas, M = 2 x 16 = 32 g
Number of moles, n = m/M = 12/32 =  0.375 moles
(b) Given mass of water, m = 20 g
Molar mass of water (H2O), M = 2 x 1 + 16 = 18 g
Number of moles, n = m/M = 20/18 = 1.11 moles
(c) Given mass of carbon dioxide, m = 22 g
Molar mass of carbon dioxide (CO2), M = 12 + 2 x 16 = 12 + 32 = 44 g
Number of moles, n = m/M = 22/44 = 0.5 moles
9. What is the mass of:
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?

Ans:- (a) ∵ Mass of 1 mole of oxygen atoms = 16 g
Mass of 0.2 moles of oxygen atoms = 0.2 x 16 = 3.2 g
(b) ∵ Mass of 1 mole of water molecules = 18 g
Mass of 0.5 moles of water molecules = 0.5 x 18 = 9 g
10. Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.
Ans:- Given mass of sulphur, m = 16 g
Molar mass of Sulphur (S8), M = 8 × 32 = 256 g
Number of moles, n = m/M = 16/256 = 0.0625 moles
Number of molecules, N = Number of moles x Avogadro number
= 0.0625 x 6.022 x 10²³ = 3.763 x 1022 molecules
11. Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.
(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u)

Ans:- Given mass of aluminium oxide (Al2O3), m = 0.051 g
Molar mass of aluminium oxide (Al2O3), M
= (2 x 27) + (3 x 16) = 54 + 48 = 102 g
Number of moles, n = m/M = 0.051/102 = 0.0005
Number of aluminium oxide molecules
= 0.0005 x 6.022 x 1023 = 3.011 x 1020
∵ Number of aluminium ions present in one molecule of aluminium oxide = 2 ,
Number of aluminium ions present in 3.011 x 1020  molecule of aluminium oxide
= 2 x 3.011x 1020 = 6.022 x 1020
Hence number of aluminium ions present in 0.051g of aluminium oxide = 6.022 x 1020