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

15.750 m m H g

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16.500 m m H g

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17.325 m m H g

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17.675 m m H g

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Solution

The correct option is A $17.325 m m H g$
$\frac{P}{P_{A}^{0}} = X_{s o l v e n t}$

$n_{g / u} = \frac{18}{180} = 0 \cdot 1$

$n_{w a t e r} = \frac{178 \cdot 2}{18} = 9 \cdot 9$

$X_{s o l v e n t} - \frac{9 \cdot 9}{0 \cdot 1 + 9 \cdot 9}$

$= 0.99$

$P = 17 \cdot 5 \times 0.99$

$= 17 \cdot 325 m m H g$

Hence, the correct option is $C$

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