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An aqueous solution of glucose is made by dissolving 10g of glucose $(C_{6} H_{12} O_{6})$ in 90g of water at 303K. If the vapour pressure of pure water at 303 K is 32.8mm Hg, what would be the vapour pressure (in mm Hg) of the solution?

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34.8

35.5

31.8

32.44

answer is D.

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Detailed Solution

$\frac{P^{0} - P_{S}}{P^{0}} = \frac{W_{A}}{M_{A}} . \frac{M_{B}}{W_{B}}$

$\begin{array}{l} \frac{32.8 - P_{S}}{32.8} = \frac{10}{180} . \frac{18}{90} \\ P_{S} = 32.435 m m o f H g \end{array}$

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