Updated By SaraNextGen

On March 11, 2024, 11:35 AM

Question 1.1:

Calculate the molecular mass of the following:

(i) H₂O (ii) CO₂ (iii) CH₄

Answer:

(i) H₂O:

The molecular mass of water, H₂O

= (2 × Atomic mass of hydrogen) + (1 × Atomic mass of oxygen)

= [2(1.0084) + 1(16.00 u)]

= 2.016 u + 16.00 u

= 18.016

= 18.02 u

(ii) CO_2:

The molecular mass of carbon dioxide, CO₂

= (1 × Atomic mass of carbon) + (2 × Atomic mass of oxygen)

= [1(12.011 u) + 2 (16.00 u)]

= 12.011 u + 32.00 u

= 44.01 u

(iii) CH_4:

The molecular mass of methane, CH₄

= (1 × Atomic mass of carbon) + (4 × Atomic mass of hydrogen)

= [1(12.011 u) + 4 (1.008 u)]

= 12.011 u + 4.032 u

= 16.043 u

Question 1.2:

Calculate the mass percent of different elements present in sodium sulphate (Na₂SO₄).

Answer:

The molecular formula of sodium sulphate is .

Molar mass of   = [(2 × 23.0) + (32.066) + 4 (16.00)]

= 142.066 g

Mass percent of an element 

∴ Mass percent of sodium:

Mass percent of sulphur:

Mass percent of oxygen:

Question 1.3:

Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.

Answer:

% of iron by mass = 69.9 % [Given]

% of oxygen by mass = 30.1 % [Given]

Relative moles of iron in iron oxide:

Relative moles of oxygen in iron oxide:

Simplest molar ratio of iron to oxygen:

= 1.25: 1.88

= 1: 1.5

 2: 3

∴ The empirical formula of the iron oxide is Fe₂O_3.

Question 1.4:

Calculate the amount of carbon dioxide that could be produced when

(i) 1 mole of carbon is burnt in air.

(ii) 1 mole of carbon is burnt in 16 g of dioxygen.

(iii) 2 moles of carbon are burnt in 16 g of dioxygen.

Answer:

The balanced reaction of combustion of carbon can be written as:

(i) As per the balanced equation, 1 mole of carbon burns in1 mole of dioxygen (air) to produce1 mole of carbon dioxide.

(ii) According to the Question, only 16 g of dioxygen is available. Hence, it will react with 0.5 mole of carbon to give 22 g of carbon dioxide. Hence, it is a limiting reactant.

(iii) According to the Question, only 16 g of dioxygen is available. It is a limiting reactant. Thus, 16 g of dioxygen can combine with only 0.5 mole of carbon to give 22 g of carbon dioxide.

Question 1.5:

Calculate the mass of sodium acetate (CH₃COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol^–1

Answer:

0.375 M aqueous solution of sodium acetate

≡ 1000 mL of solution containing 0.375 moles of sodium acetate

∴Number of moles of sodium acetate in 500 mL

Molar mass of sodium acetate = 82.0245 g mole^–1 (Given)

∴ Required mass of sodium acetate = (82.0245 g mol^–1) (0.1875 mole)

= 15.38 g

Question 1.6:

Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL^–1 and the mass per cent of nitric acid in it being 69%.

Answer:

Mass percent of nitric acid in the sample = 69 % [Given]

Thus, 100 g of nitric acid contains 69 g of nitric acid by mass.

Molar mass of nitric acid (HNO₃)

= {1 + 14 + 3(16)} g mol^–1

= 1 + 14 + 48

= 63 g mol^–1

∴ Number of moles in 69 g of HNO₃

Volume of 100g of nitric acid solution

Concentration of nitric acid

∴Concentration of nitric acid = 15.44 mol/L

Question 1.7:

How much copper can be obtained from 100 g of copper sulphate (CuSO₄)?

Answer:

1 mole of CuSO₄ contains 1 mole of copper.

Molar mass of CuSO₄ = (63.5) + (32.00) + 4(16.00)

= 63.5 + 32.00 + 64.00

= 159.5 g

159.5 g of CuSO₄ contains 63.5 g of copper.

⇒ 100 g of CuSO₄ will contain   of copper.

Amount of copper that can be obtained from 100 g CuSO₄ 

= 39.81 g

Question 1.8:

Determine the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are 69.9 and 30.1 respectively. Given that the molar mass of the oxide is 159.69 g mol^–1.

Answer:

Mass percent of iron (Fe) = 69.9% (Given)

Mass percent of oxygen (O) = 30.1% (Given)

Number of moles of iron present in the oxide 

= 1.25

Number of moles of oxygen present in the oxide 

= 1.88

Ratio of iron to oxygen in the oxide,

= 1 : 1.5

= 2 : 3

The empirical formula of the oxide is Fe₂O_3.

Empirical formula mass of Fe₂O₃ = [2(55.85) + 3(16.00)] g

Molar mass of Fe₂O₃ = 159.69 g

Molecular formula of a compound is obtained by multiplying the empirical formula with n.

Thus, the empirical formula of the given oxide is Fe₂O₃ and n is 1.

Hence, the molecular formula of the oxide is Fe₂O_3.

Question 1.9:

Calculate the atomic mass (average) of chlorine using the following data:

The average atomic mass of chlorineAnswer:

= 26.4959 + 8.9568

= 35.4527 u

The average atomic mass of chlorine = 35.4527 u

Question 1.10:

In three moles of ethane (C₂H₆), calculate the following:

(i) Number of moles of carbon atoms.

(ii) Number of moles of hydrogen atoms.

(iii) Number of molecules of ethane.

Answer:

(i) 1 mole of C₂H₆ contains 2 moles of carbon atoms.

Number of moles of carbon atoms in 3 moles of C₂H₆

= 2 × 3 = 6

(ii) 1 mole of C₂H₆ contains 6 moles of hydrogen atoms.

Number of moles of carbon atoms in 3 moles of C₂H₆

= 3 × 6 = 18

(iii) 1 mole of C₂H₆ contains 6.023 × 10²³ molecules of ethane.

Number of molecules in 3 moles of C₂H₆

= 3 × 6.023 × 10²³ = 18.069 × 10²³