Boiling point of water at $750 mm Hg$ is $99.63 ° C$ . Find the mass of water present in a solution containing $121.67 g$ sucrose, if this solution boils at $100 ° C (K_{b (H_{2} O)} = 0.52 K - Kg / mol)$

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$50 g$

$500 g$

$100 g$

$1000 g$

answer is B.

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Detailed Solution

$∆ T_{b} = K_{b} . m 100 - 99.63 = (0.52) (m) m = \frac{0.37}{0.52} m = \frac{n_{sucrose}}{W_{H_{2} O (g)}} \times 1000 = \frac{0.37}{0.52} W_{H_{2} O} = \frac{121.67}{342} \times \frac{1000}{0.37} \times 0.52 W_{H_{2} O} = 500 g$

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