Question

The azeotropic mixture of water (b.pt.\[{{100}^{0}}C\]) and HCl (b.pt. \[{{85}^{0}}C\]) boils at \[{{108.5}^{0}}C\]. When this mixture is distilled, it is possible to obtain:
A ) Pure \[HCl\]
B ) pure water
C ) pure water as well as \[HCl\]
D ) neither \[HCl\]nor water in their pure states.

Answer 1

Hint: An azeotropic mixture is constant boiling mixture. The composition of azeotropic mixture does not change on boiling. The composition of the vapour is the same as the composition of liquid solution.

Complete answer:
An azeotropic mixture is a constant boiling mixture. When an azeotropic mixture is boiled, its composition remains constant throughout boiling. When an azeotropic mixture is boiled, the composition of the vapour is the same as the composition of liquid solution. It is not possible to obtain any component of the azeotropic mixture in pure form by boiling the mixture.
Thus, suppose an azeotropic mixture of two components A and B has 66% A and 44% B . When this mixture is boiled, the composition of the vapour is also 66% A and 44% B. When the vapour is condensed through the condenser and collected in the receiving flask, the composition of the liquid mixture in the receiving flask is also the same, i.e., 66% A and 44% B.
Same is also valid for the azeotropic mixture of water and \[HCl\]. When this mixture is boiled, the composition remains the same throughout. The mixture collected in the receiving flask will also have the same composition.
The azeotropic mixture of water (b.pt. \[{{100}^{0}}C\]) and HCl (b.pt. \[{{85}^{0}}C\]) boils at \[{{108.5}^{0}}C\].When this mixture is distilled, it is possible to obtain neither HCl nor water in their pure states.

Thus, the option D ) is the correct answer.

Note: It is not possible to separate the individual components of azeotropic mixture by distillation. This is because the composition of azeotropic mixture does not change on boiling. The composition of the vapour is the same as the composition of liquid solution.