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How many grams of sucrose (Molecuar weight = 342) should be dissolved in 100 gm water in order to have a solution with sum of lowering of freezing point and elevation of boiling point equal to $5^{0} C$ $(K_{b} = 0.51, K_{f} = 1.86)$

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72 gm

342 gm

34.2 gm

460 gm

answer is B.

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Detailed Solution

Given

$Δ T_{f} + Δ T_{b} = 5 (i = 1)$

$K_{f .} m + K_{b .} m = 5$

$m (k_{f} + k_{b}) = 5$

$m (2.37) = 5$

$m = \frac{5}{2.37}$

$\frac{ω}{G M M} \times \frac{1000}{W_{g}} = \frac{5}{2.37}$

$\frac{ω}{342} \times \frac{1000}{100} = \frac{5}{2.37}$

$ω = \frac{5 \times 342 \times 100}{2.37 \times 1000}$

= 72.1 g

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