Question

0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If Kf of water is 1.86 K kg mol⁻¹, the lowering in freezing point of solution is

• A. - 0.56 K
• B. - 1.12 K
• C. 1.12 K
• D. 0.56 K

Hint:

Freezing point depression depends on the molality, the cryoscopic constant, and the degree of ionization of the solute.

Answer 1

The Correct Option is B

Step 1: Understanding the formula for freezing point depression.
The freezing point depression is given by the formula: \[ \Delta T_f = K_f \times m \times i \] Where \( \Delta T_f \) is the freezing point depression, \( K_f \) is the cryoscopic constant, \( m \) is the molality, and \( i \) is the van't Hoff factor (degree of ionization). For a weak acid, \( i \) is calculated using the degree of ionization.

Step 2: Calculating the van't Hoff factor.
Given that the solution is 20% ionized, the van't Hoff factor \( i \) is 1 + degree of ionization. For 20% ionization, \( i = 1 + 0.2 = 1.2 \).

Step 3: Substituting the values.
\[ \Delta T_f = 1.86 \times 0.5 \times 1.2 = 1.12 \, \text{K} \]
Step 4: Conclusion.
The correct answer is (B) - 1.12 K, as the lowering in freezing point is 1.12 K.