Question

A 0.1 molal aqueous solution of a weak acid is 30% ionised. If Kf for water is \( 1.86^\circ C/\text{mol} \), the freezing point of the solution will be

• A. \( -0.18^\circ C \)
• B. \( -0.54^\circ C \)
• C. \( -0.36^\circ C \)
• D. \( -0.24^\circ C \)

Hint:

For solutions, the depression in freezing point is proportional to the molality of the solution and the van't Hoff factor.

Answer 1

The Correct Option is C

The depression in freezing point is given by \( \Delta T_f = K_f \times m \times i \), where \( i \) is the van't Hoff factor (number of particles produced per formula unit). For this acid, \( i = 1 \) because it only dissociates partially. The freezing point depression is calculated as \( \Delta T_f = 1.86 \times 0.1 \times 0.3 = -0.36^\circ C \).

Step 2: Conclusion.
The freezing point is \( -0.36^\circ C \), corresponding to option (c).