Atoms and Molecules

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NCERT Solutions for Chapter 3 Atoms and Molecules Class 9 Science

Book Solutions

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

Answer

In the given reaction, sodium carbonate reacts with ethanoic acid to produce sodium ethanoate, carbon dioxide, and water.
Sodium Carbonate + Ethanoic acid → sodium ethanoate + carbon dioxide + water
Mass of sodium carbonate = 5.3 g (Given)
Mass of ethanoic acid = 6 g (Given)
Mass of sodium ethanoate = 8.2 g (Given)
Mass of carbon dioxide = 2.2 g (Given)
Mass of water = 0.9 g (Given)
Now, total mass before the reaction = (5.3 + 6) g
= 11.3 g
And, total mass after the reaction = (8.2 + 2.2 + 0.9) g
= 11.3 g
∴ Total mass before the reaction = Total mass after the reaction

Hence, the given observations are in agreement with the law of conservation of mass.

Intext Ques Pg 32 Page Number 32

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 ?

Answer

It is given that the ratio of hydrogen and oxygen by mass to form water is 1:8.
Then, the mass of oxygen gas required to react completely with 1 g of hydrogen gas is 8 g.
Therefore, the mass of oxygen gas required to react completely with 3 g of hydrogen gas is 8 × 3 g = 24 g.
Intext Ques Pg 32 Page Number 33

3

Which postulate of Dalton's atomic theory is the result of the law of conservation of mass ?

Answer

The postulate of Dalton :"Atoms are indivisible particles, which can not be created or destroyed in a chemical reaction" is the result of the law of conservation of mass.
Intext Ques Pg 32 Page Number 33

4

Which postulate of Dalton's atomic theory can explain the law of definite proportion

Answer

The postulate of Dalton, “The relative number and kinds of atoms are constant in a given compound”, can explain the law of definite proportions.
Intext Ques Pg 32 Page Number 33

1

Define atomic mass unit.

Answer

Mass unit equal to exactly one-twelfth the mass of one atom of carbon-12 is called one atomic mass unit. It is written as 'u'.
Intext Ques Pg 35 Page Number 35

2

Why is it not possible to see an atom with naked eyes?

Answer

The size of an atom is so small that it is not possible to see it with naked eyes. Also, the atom of an element does not exist independently.
Intext Ques Pg 35 Page Number 35

1

Write down the formulae of

(i) sodium oxide
(ii) aluminium chloride
(iii) sodium sulphide
(iv) magnesium hydroxide

Answer

(i) Na2O

(ii) AlCl3

(iii) Na2S
(iv) Mg(OH)2

Intext Ques Pg 39 Page Number 39

2

Write down the names of compounds represented by the following formulae:
(i) Al2(SO4)3

(ii) CaCl2

(iii) K2SO4

(iv) KNO3

(v) CaCO3

Answer

(i) Aluminium sulphate
(ii) Calcium chloride
(iii) Potassium sulphate
(iv) Potassium nitrate
(v) Calcium carbonate
Intext Ques Pg 39 Page Number 39

3

What is meant by the term chemical formula ?

Answer

The chemical formula of a compound is a symbolic representation of its composition.
Intext Ques Pg 39 Page Number 39

4

How many atoms are present in a
(i) H2S molecule and
(ii) PO43-ion?

Answer

(i) In an H2S molecule, three atoms are present; two of hydrogen and one of sulphur.
(ii) In a PO43-ion, five atoms are present; one of phosphorus and four of oxygen.
Intext Ques Pg 39 Page Number 39

1

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

Answer

• Molecular mass of H2= 2 × Atomic mass of H
= 2 × 1
= 2 u

• Molecular mass of O2= 2 × Atomic mass of O
= 2 × 16
= 32 u

• Molecular mass of Cl2= 2 × Atomic mass of Cl
= 2 × 35.5
= 71 u

 Molecular mass of CO2= Atomic mass of C + 2 × Atomic mass of O
= 12 + 2 × 16
= 44 u

• Molecular mass of CH4= Atomic mass of C + 4 × Atomic mass of H
= 12 + 4 × 1
= 16 u

• Molecular mass of C2H6= 2× Atomic mass of C + 6× Atomic mass of H
= 2 × 12 + 6 × 1
= 30 u

• Molecular mass of C2H4= 2 x Atomic mass of C + 4 × Atomic mass of H
= 2 × 12 + 4 × 1
= 28 u

• Molecular mass of NH3 = Atomic mass of N + 3 × Atomic mass of H
= 14 + 3×1
= 17 u

• Molecular mass of CH3OH= Atomic mass of C + 3 × Atomic mass of H + Atomic mass of O + Atomic mass of H
= 12 + 3×1 + 8 + 1
= 24 u
Intext Ques Pg 40 Page Number 40

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.

Answer

• Formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O
= 65 + 16
= 81 u

• Formula unit mass of Na2O = 2 × Atomic mass of Na + Atomic mass of O
= 2 × 23 + 16
= 62 u

• Formula unit mass of K2CO3= 2 × Atomic mass of K + Atomic mass of C + 3 × Atomic mass of O
= 2 × 39 + 12 + 3 × 16
= 78 + 12 + 48
= 122 u
Intext Ques Pg 40 Page Number 40

1

If one mole of carbon atoms weighs 12 gram, what is the mass (in gram) of 1 atom of carbon?

Answer

One mole of carbon atoms weighs 12 g (Given)
i.e., mass of 1 mole of carbon atoms = 12 g
Then, mass of 6.022 × 1023 number of carbon atoms = 12 g
Therefore, mass of 1 atom of carbon = 12 ÷ (6.022 × 1023)
= 1.9926 x 10-23 g
Intext Ques Pg 42 Page Number 42

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)?

Answer

Atomic mass of Na = 23 u (Given)
Then, gram atomic mass of Na = 23 g
Now, 23 g of Na contains = 6.022 × 1023 g number of atoms
Thus, 100 g of Na contains = 6.022 × 1023 / 23×100 number of atoms
= 2.6182 × 1024 number of atoms

Again, atomic mass of Fe = 56 u (Given)
Then, gram atomic mass of Fe = 56 g

Now, 56 g of Fe contains = 6.022 × 1023 g number of atoms

Thus, 100 g of Fe contains = 6.022 × 1023 / 56 × 100 number of atoms
= 1.0753 × 1024 number of atoms

Therefore, 100 grams of sodium contain more number of atoms than 100 grams of iron.

Intext Ques Pg 42 Page Number 42

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.

Answer

Total mass of Compound = 0.24 g (Given)
Mass of boron = 0.096 g (Given)
Mass of oxygen = 0.144 g (Given)
Thus, percentage of boron by weight in the compound = 0.096 / 0.24 × 100% = 40%
And, percentage of oxygen by weight in the compound = 0.144 / 0.24 × 100% = 60%
Exercises Page Number 43

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 combinations will govern your answer ?

Answer

3.0 g of carbon combines with 8.0 g of oxygen to give 11.0 of carbon dioxide.
If 3 g of carbon is burnt in 50 g of oxygen, then 3 g of carbon will react with 8 g of oxygen. The remaining 42 g of oxygen will be left un-reactive.
In this case also, only 11 g of carbon dioxide will be formed.
The above answer is governed by the law of constant proportions.
Exercises Page Number 43

3

What are polyatomic ions? Give examples?

Answer

A polyatomic ion is a group of atoms carrying a charge (positive or negative). For example, Nitrate (NO3-) , hydroxide ion (OH-).
Exercises Page Number 44

4

Write the chemical formulae of the following:

(a) Magnesium chloride

(b) Calcium oxide

(c) Copper nitrate

(d) Aluminium chloride

(e) Calcium carbonate

Answer

(a) MgCl2
(b) CaO
(c) Cu (NO3)2
(d) AlCl3
(e) CaCO3
Exercises Page Number 44

5

Give the names of the elements present in the following compounds:
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate

Answer

(a) Calcium and oxygen
(b) Hydrogen and bromine
(c) Sodium, hydrogen, carbon, and oxygen
(d) Potassium, sulphur, and oxygen
Exercises Page Number 44

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

Answer

(a) Molar mass of ethyne, C2H2 = 2×12 + 2 × 1 = 26 g
(b) Molar mass of sulphur molecule, S8 = 8×32 = 256 g
(c) Molar mass of phosphorus molecule, P4 = 4 × 31 = 124 g
(d) Molar mass of hydrochloric acid, HCl = 1 + 35.5 = 36.5 g
(e) Molar mass of nitric acid, HNO3 = 1 + 14 + 3×16 = 63 g
Exercises Page Number 44

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)?

Answer

(a) The mass of 1 mole of nitrogen atoms is 14 g.

(b) The mass of 4 moles of aluminium atoms is (4 × 27) g = 108 g.

(c) The mass of 10 moles of sodium sulphite (Na2SO3) is 10×[2×23 + 32 + 3×16] g = 10×126 g = 1260 g.
Exercises Page Number 44

8

Convert into mole.
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide

Answer

(a) 32 g of oxygen gas = 1 mole
Then, 12 g of oxygen gas = 12 / 32 mole = 0.375 mole

(b) 18 g of water = 1 mole
Then, 20 g of water = 20 / 18 mole = 1.111 mole

(c) 44 g of carbon dioxide = 1 mole
Then, 22 g of carbon dioxide = 22 / 44 mole = 0.5 mole
Exercises Page Number 44

9

What is the mass of:
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?

Answer

(a) Mass of one mole of oxygen atoms = 16 g
Then, mass of 0.2 mole of oxygen atoms = 0.2 × 16g = 3.2 g

(b) Mass of one mole of water molecule = 18 g

Then, mass of 0.5 mole of water molecules = 0.5 × 18 g = 9 g

Exercises Page Number 44

10

Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.

Answer

1 mole of solid sulphur (S8) = 8 × 32 g = 256 g
i.e., 256 g of solid sulphur contains = 6.022 × 1023 molecules
Then, 16 g of solid sulphur contains = 6.022 × 1023 / 256  = 16 molecules
= 3.76375 × 1022 molecules
Exercises Page Number 44

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)

Answer

Mole of aluminium oxide (Al2O3) = 2×27 + 3×16 = 102 g
i.e., 102 g of Al2O3= 6.022 × 1023molecules of Al2O3
Then, 0.051 g of Al2O3contains = 6.022 × 1023 / 102 × 0.051 molecules
= 3.011×1020 molecules of Al2O3

The number of aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.

Therefore, the number of aluminium ions (Al3+) present in 3.011 × 1020molecules (0.051 g ) of aluminium oxide (Al2O3) = 2 × 3.011 × 1020
= 6.022 × 1020
Exercises Page Number 44

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