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JEE Main levelSingle correct optionDifficulty 2 of 3so189

$12 g$ of urea (molar mass $60 g mo l^{- 1}$ ) is dissolved in $400 g$ of water. Taking $K_{f} (water) = 1.86 K kg mo l^{- 1}$ and the freezing point of water as $0^{\circ} C$ , the freezing point of the solution is

A-0.93 $^{\circ} C$
B-0.37 $^{\circ} C$
C-1.86 $^{\circ} C$
D-0.465 $^{\circ} C$

Answer: A

m = (12/60) /0.400 = 0.50 mol k g^{- 1}

, so

Δ T_{f} = K_{f} m = 1.86 \times 0.50 = 0.93 K

. The freezing point is

0 - 0.93 = - 0.9 3^{\circ} C

.

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