The azeotropic mixture of water (boiling point = \[100^\circ C\]) and HCl ( boiling point =\[85^\circ C\]) boils at about \[110^\circ C\]. During distillation of this mixture, it is possible to obtain
(A) pure $HCl$
(B) pure \[{H_2}O\]
(C) pure $HCl$ as well as \[{H_2}O\]
(D) neither pure \[{H_2}O\] nor pure $HCl$

Hint: In order to find which will be obtained during distillation, we must first know about azeotropic mixture. Azeotropic mixtures are the mixture of two liquids which can exhibit the same proportions at the liquid and vapour phase. Azeotropic mixture will show deviation from Raoult's law.

Complete Solution :
Let us first understand what an azeotropic mixture is. Azeotropic mixtures are the mixture of two or more liquids whose concentration or proportions remain unchanged by simple distillation. Boiling of azeotropes will give the same proportions of vapour as well as the mixture which is not boiled. Azeotropes are referred to as constant boiling mixtures. Azeotropes do not obey Raoult’s law. This is because azeotropes are not an ideal solution.
Azeotropes can be classified into four types, they are
- Heterogeneous and Homogeneous
- Positive azeotropes
- Negative azeotropes
- binary azeotropes
As we know that azeotropic mixtures are constant boiling liquid, it is not possible to separate the components of the mixture by the process of distillation. Hence neither pure HCl nor pure water will be obtained.
So, the correct answer is “Option D”.

Note: Azeotropic mixtures cannot be separated by fractional distillation. These mixtures require an alternative method for separation. They are
- Entraîner distillation
- Pressure swing distillation
- Pervaporation.