A Solution of non-volatile solute in water freezes at $- 0. 30 C .$ The vapour-pressure of pure water at $298 K$ is $23.51 m m H g$ and $K_{1}$ for water is $1.86 K . k g {m o l}^{- 1}$ Calculate the vapour-pressure of this solution at $298 K .$

23.4 m m

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24.8 m m

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34.8 m m

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40 m m

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Solution

The correct option is A $23.4 m m$
We know that, $△ T_{f} = K_{f} m$
$∴ m = \frac{△ T_{f}}{K_{f}} = \frac{0.30}{1.86} = 0.161$
According to Raooh's law,
$P^{\circ} - P / P^{\circ} =$ No. of moles of Solute/No of moles of solvent
$∴ 23.51 - P / 23.51 = \frac{0.161}{1000 / 18} = 0.0020898$
$∴ P = 23.51 - 23.51 \times 0.0020898$
$∴ P = 23.44 m m H g$

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QUESTION 2.19

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