Question

The boiling point of an azeotropic mixture of water and ethanol is less than that of pure water and ethanol. The mixture shows:

• A. Positive deviation from Raoult's law
• B. Negative deviation from Raoult's law
• C. No deviation from Raoult's law
• D. The solution is an ideal solution

Hint:

Minimum boiling azeotrope shows positive deviation; maximum boiling azeotrope shows negative deviation.

Answer 1

The Correct Option is A

Concept:
Azeotropes are constant boiling mixtures. If an azeotrope has boiling point lower than both pure components, it is called a minimum boiling azeotrope. Minimum boiling azeotropes show positive deviation from Raoult's law.

Step 1: Understand the given condition.
The question says that the boiling point of water-ethanol azeotropic mixture is less than that of pure water and pure ethanol. So it is a minimum boiling azeotrope. \[ T_b(\text{mixture}) \lt T_b(\text{pure components}) \]

Step 2: Relate boiling point with vapour pressure.
Lower boiling point means higher vapour pressure. If the observed vapour pressure is greater than expected by Raoult's law, the solution shows positive deviation. \[ P_{\text{observed}}\gt P_{\text{Raoult}} \]

Step 3: Final conclusion.
Since the azeotrope has lower boiling point, it shows positive deviation from Raoult's law. Hence: \[ \boxed{\text{(A) Positive deviation from Raoult's law}} \]