Question

The azeotropic mixture of water and ${\text{HCl}}$ boils at ${108.5^ \circ }{\text{C}}$. When this mixture is distilled, it is possible to obtain:
(A) Pure ${\text{HCl}}$
(B) Pure water
(C) Neither pure ${\text{HCl}}$ nor pure water
(D) Both pure ${\text{HCl}}$ and pure water

Answer 1

Hint: We know that a constant boiling mixture is known as an azeotropic mixture. The composition of an azeotropic mixture does not change when it is boiled. The composition of the vapour and the liquid solution is the same.

Complete step by step solution:
 We know that a constant boiling mixture is known as an azeotropic mixture. The composition of an azeotropic mixture does not change when it is boiled. The composition of the vapour and the liquid solution is the same.
Thus, it is not possible to obtain pure form of any component in the azeotropic mixture by boiling the azeotropic mixture.
We are given an azeotropic mixture of water and ${\text{HCl}}$. When this azeotropic mixture is boiled at ${108.5^ \circ }{\text{C}}$, the composition of the mixture remains the same throughout.
When an azeotropic mixture of water and ${\text{HCl}}$ is boiled at ${108.5^ \circ }{\text{C}}$, the liquid solution turns into vapours. The composition of the vapours is the same as that of the azeotropic mixture. When the vapours are condensed and collected in a receiving flask, the composition of the collected liquid is also the same.
Thus, when an azeotropic mixture of water and ${\text{HCl}}$ is distilled, it is possible to obtain neither pure ${\text{HCl}}$ nor pure water.

Thus, the correct option is (C) neither pure ${\text{HCl}}$ nor pure water.

Note: We must remember that distillation does not separate the individual components of an azeotropic mixture. This is because boiling does not change the composition of an azeotropic mixture. The composition of the liquid solution, vapour and the condensed vapour remains the same.