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JEE Advanced levelSingle correct optionDifficulty 3 of 3so068

A solution contains $9.0 g$ of glucose (molar mass $180 g mo l^{- 1}$ ) and $6.0 g$ of urea (molar mass $60 g mo l^{- 1}$ ) in $500 g$ of water. Taking $K_{f} = 1.86 K kg mo l^{- 1}$ and both solutes as non-electrolytes, the freezing point of the solution is

A $- 0.372^{\circ} C$
B $- 0.558^{\circ} C$
C $- 0.930^{\circ} C$
D $- 0.186^{\circ} C$

Answer: B

Colligative properties add over all dissolved particles. Moles: glucose

0.050

, urea

0.100

, total

0.150

in

0.500 kg

, so

m = 0.30 mol k g^{- 1}

.

Δ T_{f} = 1.86 \times 0.30 = 0.558 K

, giving

T_{f} = - 0.558^{\circ} C

.

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