The vapour pressure of water at $20^{\circ} C$ is $17.5 mmHg$ . If $18 g$ of glucose $(C_{6} H_{12} O_{6})$ is added to $178.2 g$ of water at $20^{\circ} C$ he vapour pressure of the resulting solution will be

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$15.750 mmHg$

$16.500 mmHg$

$17.675 mmHg$

$17.325 mmHg$

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{P^{\circ} - P_{s}}{P^{\circ}} = \frac{n}{n + N} \\ n = \frac{18}{180} = 0.1, N = \frac{178.2}{18} = 9.9 \end{array}$

$\begin{array}{l} \frac{17.5 - p_{s}}{17.5} = \frac{0.1}{0.1 + 9.9}; \frac{17.5 - p_{s}}{17.5} = \frac{1}{100} \\ 1750 - 100 p_{s} = 17.5; 1732.5 = 100 p_{s} \\ p_{s} = 17.325 mmHg . \end{array}$

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