Question

The azeotropic mixture of water and $HCl$ bolis at ${108.5}^{\circ }C$. When this mixture is distilled, it is possible to obtain:

• A. Pure $HCl$
• B. Pure water
• C. Neither pure $HCl$ nor water
• D. Both pure $HCl$ and water

Answer 1

The correct option is C Neither pure $HCl$ nor water

Boiling point of $HCl$ = $-{85.1}^{\circ }C$
Boiling point of $H_{2}O$ = ${100}^{\circ }C$
Since they have different boiling temperature, they can separated into pure $HCl$ and $H_{2}O$ by distillation unless it forms azeotropic mixture.

Azeotropic mixture is liquid mixture of two miscible liquids which distill over without changes in composition.
These mixture boils at constant temperature while the composition remains constant.

In $HCl$ and $H_{2}O$ solution, a strong intramolecular hydrogen bonding exist between them. So that the interaction between $HCl$ and $H_{2}O$ is stronger than interaction between $HCl-HCl$ and $H_{2}O-H_{2}O$.

Thus, the solution mixture have lower vapour pressure than the initial.
$P_T<X_A.p_A+X_B.p_B$
As a result the solution is showing a large negative deviation from Raoult's law and the mixture boils at higher temperature $\left({109.5}^{\circ }C\right)$ than its individual components.

Thus, the azeotropic mixture of $HCl$ and $H_{2}O$ boils at constant temperature in which the compositions of the mixtures remain same throught the boiling.
For such a solution mixture, the composition of vapour at its boiling point is same that of the liquid solution.
Hence, when the mixture is distilled neither pure $HCl$ nor $H_{2}O$ is obtained.