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Q.

Freezing point of an aqueous solution is — 0.186°C. If the values of K_b, and K_f, of water are respectively 0.52 K kg ${mol}^{- 1} and 1 .86 Kkg {mol}^{- 1}$ then the elevation of boiling point of the solution in K is

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a

0.52

b

1.04

c

0.052

d

1.34

answer is D.

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Detailed Solution

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$Δ T_{f} = i \cdot k_{f} \cdot m; Δ T_{b} = i \cdot k_{b} \cdot m \frac{Δ T_{f}}{Δ T_{b}} = \frac{k_{f}}{k_{b}} Δ T_{f} = 0 - (- 0 {.186}^{\circ} C) = 0 {.186}^{\circ} C \frac{0 .186}{Δ T_{b}} = \frac{1 .86}{0 .52} \Rightarrow Δ T_{b} = \frac{0 .52 \times 0 .186}{1 .86} = 0 .052$

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Freezing point of an aqueous solution is — 0.186°C. If the values of Kb, and Kf, of water are respectively 0.52 K kg mol−1 and 1.86Kkgmol−1then the elevation of boiling point of the solution in K is