Question

1.00 g of a non-electrolyte solute (molar mass 250 g mol⁻¹) was dissolved in 51.2 g of benzene. If the freezing point depression constant Kf of benzene is 5.12 K kg mol^-1, the freezing point of benzene will be lowered by:

• A. 0.3 K
• B. 0.5 K
• C. 0.4 K
• D. 0.2 K

Hint:

For non-electrolytes, van’t Hoff factor i = 1.

Answer 1

The Correct Option is C

Step 1: Molality calculation. m = (1/250)/(0.0512) = 0.078
Step 2: Freezing point depression. Δ Tf = Kf m = 5.12 × 0.078 ≈ 0.4