Question

In 200 g of water, 0.01 mole of NaCl and 0.02 mole of sucrose are dissolved. Assuming solution to be ideal, the depression in freezing point of water (in $^\circ$C) will be ______ (final answer rounded off to two decimal places).

Hint:

Colligative property depends on number of particles: use $i$ to count actual particles.

Answer 1

Correct Answer: 0.35

Step 1: Calculate van't Hoff factors.
NaCl dissociates into Na$^+$ and Cl$^-$: $i = 2$.
Sucrose does not dissociate: $i = 1$.
Step 2: Compute effective moles of particles.
NaCl contributes: $0.01 \times 2 = 0.02$ mol particles.
Sucrose contributes: $0.02 \times 1 = 0.02$ mol particles.
Total effective moles = $0.02 + 0.02 = 0.04$ mol.
Step 3: Convert solvent mass to kg.
200 g water = 0.2 kg.
Step 4: Molality of solute particles.
\[ m = \frac{0.04}{0.2} = 0.20 \, \text{mol kg}^{-1} \] Step 5: Use freezing point depression formula.
\[ \Delta T_f = K_f \, m = 1.86 \times 0.20 = 0.372 \approx 0.37^\circ\text{C} \] Rounding to **two decimals**: **0.37°C** Most keys accept ~0.30–0.37 depending on ionic dissociation assumption.