Question

2 molal solution of a weak acid HA has a freezing point of 3.885°C. The degree of dissociation of this acid is __________ × 10$^{-3}$. (Round off to the Nearest Integer). [Given : Molal depression constant of water = 1.85 K kg mol$^{-1}$ Freezing point of pure water = 0°C] [Note: Usually, freezing point is depressed, so the value would be -3.885°C]

Hint:

Van't Hoff factor $i = \frac{\text{Observed } \Delta T_f}{\text{Calculated } \Delta T_f}$.

Answer 1

Correct Answer: 50

Step 1: $\Delta T_f = i \cdot K_f \cdot m$.
Step 2: $3.885 = i \times 1.85 \times 2 = i \times 3.7$.
Step 3: $i = \frac{3.885}{3.7} = 1.05$.
Step 4: For $HA \rightleftharpoons H^+ + A^-$, $i = 1 + \alpha$. $1 + \alpha = 1.05 \implies \alpha = 0.05$.
Step 5: $\alpha = 50 \times 10^{-3}$.