Air contains $ {\text{23% }} $ oxygen and $ {\text{77% }} $ nitrogen by weight. What is the percentage of $ {{\text{O}}_{\text{2}}} $ by volume?
(A) $ 28.1 $
(B) $ 20.7 $
(C) $ 21.8 $
(D) $ 23.0 $

Hint: In the above question, percentage of oxygen and nitrogen is given and we have to find out the percentage of oxygen by volume. As we know that molar mass of a substance occupy $ {\text{22}}{\text{.4L}} $ volume and we can find out the percentage of $ {{\text{O}}_{\text{2}}} $ after knowing the volume occupied by oxygen and nitrogen.

Complete step by step solution
In the question, as the percentage of oxygen and nitrogen are given and hence, we should take a fixed amount of air in order to perform further calculations.
Suppose, $ {\text{100g}} $ of air is taken. This implies that weight of $ {{\text{O}}_{\text{2}}} $ is $ {\text{23g}} $ whereas weight of $ {{\text{N}}_{\text{2}}} $ is $ {\text{77g}} $ .
From the atomic table, we know that atomic mass of oxygen is 16g and atomic mass of nitrogen is 14g.
Molar mass of oxygen ( $ {{\text{M}}_{\text{O}}} $ ) = $ 2 \times $ atomic mass of oxygen
So, $ {{\text{M}}_{\text{O}}} $ = $ 2 \times 16{\text{ = 32g}} $
So, $ {\text{32g}} $ of oxygen occupies $ {\text{22}}{\text{.4L}} $ space.
 $ {\text{23g}} $ of oxygen occupies $ \dfrac{{22.4}}{{32}} \times 23{\text{ = 16}}{\text{.1L}} $ space.
Molar mass of oxygen ( $ {{\text{M}}_{\text{N}}} $ ) = $ 2 \times $ atomic mass of nitrogen
So, $ {{\text{M}}_{\text{N}}} $ = $ 2 \times 14{\text{ = 28g}} $
So, $ {\text{28g}} $ of nitrogen occupies $ {\text{22}}{\text{.4L}} $ space.
 $ {\text{77g}} $ of nitrogen occupies $ \dfrac{{22.4}}{{28}} \times 77{\text{ = 61}}{\text{.6L}} $ space.
Total volume occupied ( $ {\text{V}} $ ) = Volume occupied by oxygen ( $ {{\text{V}}_{\text{O}}} $ ) + Volume occupied by nitrogen ( $ {{\text{V}}_{\text{N}}} $ )
 $ {\text{V}} = {\text{16}}{\text{.1 + 61}}{\text{.6 = 77}}{\text{.7L}} $
Percentage volume of oxygen is $ \dfrac{{{{\text{V}}_{\text{O}}}}}{{\text{V}}}{\text{ }} \times {\text{ 100 = }}\dfrac{{16.1}}{{77.7}}{\text{ }} \times {\text{ 100 = 20}}{\text{.72}} $
 $ \therefore $ Percentage of $ {{\text{O}}_{\text{2}}} $ by volume is $ 20.7 $ .

Hence, the correct option is option B.

Note
In these types of questions where percentages of individual substances are given, first we have to take $ {\text{100g}} $ of air in order to make calculation easier. Remember that the molar mass of a substance occupies $ {\text{22}}{\text{.4L}} $ volume.