$A solution of sucrose (molar mass = 342 g {mol}^{- 1}) has been prepared by dissolving 68 .5 g of sucrose in 1000 g of water. The freezing point of the$
$solution obtained will be (K_{f} for water = 1 .86 {Kkgmol}^{- 1})$

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$- {0.372}^{\circ} C$

$- {0.520}^{\circ} C$

$+ {0.372}^{\circ} C$

$- {0.570}^{\circ} C$

answer is A.

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Detailed Solution

${ΔT}_{f} = K_{f} \frac{1000 W_{2}}{M_{2} W_{1}} = \frac{1 .86 \times 1000 \times 68 .5}{342 \times 1000} = 0 .372 T_{f} = T^{\circ} C_{f} - {ΔT}_{f} \Rightarrow T_{f} = - 0 {.372}^{\circ} C$

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