Chemistry  (Chapter – 1) (Some Basic Concepts of Chemistry) (Class – XI) Questions & Answers

Chemistry Chapter 1 Question & Answers Class 11 NCERT 

Question 1.1: Calculate the molecular mass of the following: (i) H2O   (ii) CO2   (iii) CH4  

Answer 1.1: 

(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) CO2: 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) CH4: 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 1.2: 

The molecular formula of sodium sulphate is .  

Molar mass of = [(2 × 23.0) + (32.066) + 4 (16.00)] = 142.066 g

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 1.5: 

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 1.5: 

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