JEE Chemistry 2024 Question with Solution

Given: A solution of two miscible liquids shows negative deviation from Raoult’s law.

Find: The correct combination of vapour pressure and boiling point.

For negative deviation from Raoult’s law, the interactions between unlike molecules are stronger than those in the pure liquids. Therefore, escaping tendency of molecules decreases, so vapour pressure decreases.

Using the vapour pressure relation:

so the total vapour pressure of the solution is lower than the ideal value.

Since boiling occurs when vapour pressure becomes equal to external pressure, a lower vapour pressure means more heating is required. Hence, the boiling point increases.

Therefore, the correct option is D: Decreased vapour pressure, increased boiling point.

Given: A non-ideal solution shows negative deviation from Raoult’s law.

Find: Its effect on vapour pressure and boiling point.

According to Raoult’s law:

for an ideal solution. In a negatively deviating solution, actual intermolecular attraction between different components is stronger than expected.

Because of these stronger attractions, molecules do not escape easily into the vapour phase. As a result, the vapour pressure becomes lower than the ideal value.

A liquid boils when its vapour pressure equals the external pressure. If the vapour pressure is lower, the liquid must be heated to a higher temperature to reach boiling. Thus, the boiling point increases.

The answer key points to option 3, but the solution explicitly concludes option D. Hence the solution is taken as authoritative.