Q 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 of 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

SOLUTION:

Total mass of reactants = mass of sodium
carbonate + mass of ethanoic acid
= 5.3 g + 6 g = 11.3 g
Total mass of products = mass of sodium ethanoate + mass of carbon dioxide + mass
of water
= 8.2 g + 2.2 g + 0.9 g = 11.3 g
Thus, the mass of reactants is equal to the mass of products, therefore the observations are in agreement with the law of conservation of mass.

Q 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?

SOLUTION:

1 g of hydrogen reacts with = 8 g of oxygen
∴ 3 g of hydrogen reacts with = 8 × 3 = 24 g of oxygen
Thus, 24 g of oxygen gas would be required to react completely with 3 g of hydrogen gas.

Q 3.

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

SOLUTION:

The postulate that “atoms can neither be created nor destroyed in a chemical reaction” is the result of the law of conservation of mass.

Q 4.

Which postulate of Dalton’s atomic theory can explain the law of definite proportions?

SOLUTION:

The postulate that “A chemical compound always consists of the same elements combined together in the same proportion by mass” is the law of definite proportions.

Q 5.

Define the atomic mass unit.

SOLUTION:

Atomic mass unit is defined as the mass unit equal to exactly one-twelfth (1/12^th) of the mass of one atom of carbon-12. It is denoted by u (unified mass).
i.e. 1 u = 1.66 × 10^–24 g

Q 6.

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

SOLUTION:

It is not possible to see an atom with naked eye because of its extremely small size (atomic radius is of the order of 10^–10 m).

Q 7.

Write down the formulae of
(I) sodium oxide,
(ii) aluminium chloride,
(iii) sodium sulphide,
(iv) magnesium hydroxide

SOLUTION:

(I) Sodium oxide

(ii) Aluminium chloride

(iii) Sodium sulphide

(iv) Magnesium hydroxide

Q 8.

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

SOLUTION:

(I) Aluminium sulphate
(ii) Calcium chloride
(iii) Potassium sulphate
(iv) Potassium nitrate
(v) Calcium carbonate

Q 9.

What is meant by the term chemical formulae?

SOLUTION:

The chemical formula of a compound is a symbolic representation of its composition. e.g., formula of calcium oxide.

Thus, chemical formula of calcium oxide is CaO.

Q 10.

How many atoms are present in a
(I) H₂S molecule and
(ii) PO₄^3– ion?

SOLUTION:

(I) 3 atoms because H₂S molecule has two atoms of hydrogen and one atom of sulphur.
(ii) 5 atoms because ion has one atom of phosphorus and four atoms of oxygen.

Q 11.

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

SOLUTION:

Molecular mass of H₂ = 2 × atomic mass of H
= 2 × 1 u = 2 u
Molecular mass of O₂ = 2 × atomic mass of O
= 2 × 16 u = 32 u
Molecular mass of Cl₂ = 2 × atomic mass of Cl
= 2 × 35.5 u = 71 u
Molecular mass of CO₂ = atomic mass of C
+ 2 × atomic mass of O
= 12 + (2 × 16) = (12 + 32) u = 44 u
Molecular mass of CH₄ = atomic mass of C
+ 4 × atomic mass of H
= 12 + (4 × 1) u = (12 + 4) u = 16 u
Molecular mass of C₂H₆ = 2 × atomic mass
of C + 6 × atomic mass of H
= (2 × 12 + 6 × 1) u = (24 + 6) u = 30 u
Molecular mass of C₂H₄ = 2 × atomic mass
of C + 4 × atomic mass of H
= (2 × 12 + 4 × 1) u = (24 + 4) u = 28 u
Molecular mass of NH₃ = atomic mass of N
+ 3 × atomic mass of H
= (14 + 3 × 1) u = (14 + 3) u = 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 + 16 + 1) u = (12 + 3 + 17) u = 32 u

Q 12.

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

SOLUTION:

Formula unit mass of ZnO = atomic mass of Zn
+ atomic mass of O
= (65 + 16) u = 81 u
Formula unit mass of Na₂O = 2 × atomic mass
of Na + atomic mass of O
= (2 × 23 + 16) u = 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) u = (78 + 12 + 48) u
= 138 u

Q 13.

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

SOLUTION:

Molecular mass of carbon = 12 g
∵ 6.022 × 10²³ atoms of carbon have mass
= 12 g
∴ 1 atom of carbon has mass

Q 14.

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)?

SOLUTION:

100 g of sodium :
Number of sodium atoms

Thus, 100 g of sodium has more atoms.

Q 15.

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

SOLUTION:

We know that, % of any element in a compound

Q 16.

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?

SOLUTION:

Total mass of reactants = mass of C + mass
of O₂ = 3 + 8 = 11 g
Total mass of reactants = total mass of
products
Hence, the law of conservation of mass is proved.
Further, it also shows carbon dioxide contains carbon and oxygen in a fixed ratio by mass, which is 3 : 8. Thus, it also proves the law of constant proportions. 3 g of carbon must also combine with 8 g of oxygen only. This means that (50 – 8) = 42 g of oxygen will remain unreacted.

Q 17.

What are polyatomic ions? Give examples

SOLUTION:

A polyatomic ion is a group of atoms carrying positive or negative charge.

Q 18.

Write the chemical formulae of the following :
(A) Magnesium chloride
(B) Calcium oxide
(C) Copper nitrate
(D) Aluminium chloride
(E) Calcium carbonate

SOLUTION:

(A) Magnesium chloride

(B) Calcium oxide

(C) Copper nitrate

(D) Aluminium chloride

(E) Calcium carbonate

Q 19.

Give the names of the elements present in the following compounds.
(A) Quick lime
(B) Hydrogen bromide
(C) Baking powder
(D) Potassium sulphate

SOLUTION:

(A) Quick lime is CaO. Elements present are calcium and oxygen.
(B) Hydrogen bromide is HBr. Elements present are hydrogen and bromine.
(C) Baking powder is NaHCO₃. Elements present are sodium, hydrogen, carbon and oxygen.
(D) Potassium sulphate is K₂SO₄. Elements present are potassium, sulphur and oxygen.

Q 20.

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.

SOLUTION:

(A) Molar mass of C₂H₂ = 2 × atomic mass of C + 2 × atomic mass of H = (2 × 12 + 2 × 1) u = (24 + 2) u = 26 u
(B) Molar mass of S₈ = 8 × atomic mass of S = (8 × 32) u = 256 u
(C) Molar mass of P₄ = 4 × atomic mass of P = (4 × 31) u = 124 u
(D) Molar mass of HCl = atomic mass of H + atomic mass of Cl = (1 + 35.5) u = 36.5 u
(E) Molar mass of HNO₃ = atomic mass of H+ atomic mass of N + 3 × atomic mass of O = (1 + 14 + 3 × 16) u = (1 + 14 + 48) u = 63 u

Q 21.

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)?

SOLUTION:

(A) Mass of nitrogen atom
= number of moles × atomic mass
= 1 × 14 = 14 g
(B) Mass of aluminium atom
= 4 × atomic mass = 4 × 27 = 108 g
(C) Mass of Na₂SO₃ = 2 × 23 + 32 + 16 × 3
= 46 + 32 + 48 = 126 g
1 mole of Na₂SO₃ = 126 g
∴ 10 moles of Na₂SO₃ = (10 × 126) g
= 1260 g

Q 22.

Convert into mole.
(A) 12 g of oxygen gas
(B) 20 g of water
(C) 22 g of carbon dioxide

SOLUTION:

(A) Molar mass of O₂ = 2 × 16 = 32 g
∴ Number of moles in 12 g of O₂

(B) Molar mass of H₂O = 2 × 1 + 16 = 18 g
∴ Number of moles in 20 g of H₂O

(C) Molar mass of CO₂ = 12 + 16 × 2
= 12 + 32 = 44 g
∴ Number of moles in 22 g of CO₂

Q 23.

What is the mass of
(A) 0.2 mole of oxygen atoms?
(B) 0.5 mole of water molecules?

SOLUTION:

(A) Mass of 1 mole of oxygen atom = 16 u
Mass of 0.2 mole of oxygen atoms
= (16 × 0.2) u = 3.2 u
(B) Mass of 1 mole of H₂O molecule
= (1 × 2 + 16) = 18 u
Mass of 0.5 mole of H₂O molecule
= (18 × 0.5) u = 9 u

Q 24.

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

SOLUTION:

Mass of 1 mole of S₈ = 8 × 32 = 256 g
∴ 256 g of S₈ contains = 6.022 × 10²³
molecules
∴ 16 g of S₈ contains

Q 25.

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)

SOLUTION:

Molar mass of Al₂O₃ = 2 × 27 + 3 × 16
= 54 + 48 = 102 g
∴ 102 g of Al₂O₃ contains
= 2 × 6.022 × 10²³ Al³⁺ ions
∴ 0.051 g of Al₂O₃ contains

= 6.022 × 10²⁰ Al³⁺ ions.