Calculate the vapour pressure of an aqueous solution of 1.0 molal glucose solution at 100°C.

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7.463 mmHg

74.632 mmHg

0.7463 mm Hg

746.32 mm Hg

answer is B.

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Detailed Solution

$\frac{p_{0} - p_{s}}{p_{0}} = \frac{n_{1}}{n_{1} + n_{2}}$

$\frac{p_{0} - p_{s}}{p_{0}} = \frac{n_{1}}{n_{2}}$

$= \frac{W_{s o l u t e}}{M_{s o l u t e}} \times \frac{1000}{W_{s o l v e n t}} \times \frac{M_{s o l v e n t}}{1000}$

$= m \times \frac{M_{s o l v e n t}}{1000}$

$\frac{760 - p}{760} = 1 \times \frac{18}{1000}$

=746.32 mm of Hg

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