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JEE Advanced levelSingle correct optionDifficulty 3 of 3so103

A solution is made by dissolving $0.10 mol$ of $NaCl$ and $0.050 mol$ of $CaCl X_{2}$ in $500 g$ of water. Assuming complete dissociation of both salts and $K_{f} = 1.86 K kg mo l^{- 1}$ , the freezing point of the solution is

A $- 1.12^{\circ} C$
B $- 1.30^{\circ} C$
C $- 0.56^{\circ} C$
D $- 1.86^{\circ} C$

Answer: B

Particles:

NaCl

gives

2 \times 0.10 = 0.20 mol

and

CaCl X_{2}

gives

3 \times 0.050 = 0.15 mol

, total

0.35 mol

in

0.500 kg

, so

m_{e ff} = 0.70

.

Δ T_{f} = 1.86 \times 0.70 = 1.302 K

, so

T_{f} = - 1.30^{\circ} C

.

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