Chapter 1 – Some Basic Concepts of Chemistry
Page No 22:
Question 1.1:
Calculate the molecular mass of the following:
(i) H 2 O (ii) CO 2 (iii) CH 4
Answer:
(i) H 2 O:
The molecular mass of water, H 2 O
= (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, CO 2
= (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, CH 4
= (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 (Na 2 SO 4 ).
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 Fe 2 O 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 (CH 3 COONa) 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
Page No 22:
Question 1.1:
Calculate the molecular mass of the following:
(i) H 2 O (ii) CO 2 (iii) CH 4
Answer:
(i) H 2 O:
The molecular mass of water, H 2 O
= (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, CO 2
= (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, CH 4
= (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 (Na 2 SO 4 ).
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 Fe 2 O 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 (CH 3 COONa) 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
