JEE Chemistry 2024 Question with Solution

Given: Four solutions are compared for maximum depression in freezing point. Find: Which solution has the highest $\Delta T_f$ and therefore the lowest freezing point.

Freezing point depression is given by

where $i$ is the van 't Hoff factor, $K_f$ is the cryoscopic constant, and $m$ is the molality.

From the solution text:

• acetic acid in water partially ionizes, so the number of solute particles increases.
• acetic acid in benzene shows less effective particle count than in water.
• benzoic acid in benzene forms associated species, lowering effective particle count.
• glucose in water does not ionize, so $i = 1$.

Thus, acetic acid dissolved in water gives the greatest effective value of $i \cdot m$ and hence the largest value of

Therefore, the solution with the highest depression in freezing point is $180 \, \text{g}$ of acetic acid dissolved in water. The correct option is A.

Note: The solution incorrectly displays "The Correct Option is C", but its own working and conclusion clearly identify acetic acid dissolved in water, which corresponds to option A in the given options.

Given: The same mass, $180 \, \text{g}$, of different solutes is dissolved in water or benzene. Find: Which solution shows the lowest freezing point.

For colligative properties, the important comparison is the total number of solute particles present in solution.

1. Acetic acid in water: partial ionization increases particle count.
2. Acetic acid in benzene: association is favored, so effective particle count is lower.
3. Benzoic acid in benzene: association lowers the effective number of particles.
4. Glucose in water: no ionization, so it remains as single particles.

Hence the solution producing the maximum number of effective particles gives the maximum depression in freezing point.

Therefore, acetic acid dissolved in water has the lowest freezing point, so the correct option is A.