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

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15.750 mmHg

17.675 mmHg

16.500 mmHg

17.325 mmHg

answer is A.

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Detailed Solution

Amount of glucose, $n_{2} = \frac{m_{2}}{M_{2}} = \frac{18 g}{180 g {mol}^{- 1}} = 0.1 mol$

Amount of water, $n_{1} = \frac{m_{1}}{M_{1}} = \frac{178.2}{18 g {mol}^{- 1}} = 9.9 mol$

Mole fraction of water, $x_{1} = \frac{n_{1}}{n_{1} + n_{2}} = \frac{9.9 g}{(9.9 + 0.1) mol} = 0.99$

Vapour pressure of solution, $p = x_{1} p_{1}^{*} = (0.99) (17.5 mmHg) = 17.325 mmHg .$

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