Q25 / 90 · JEE Main 2016 (03 Apr) · Chemistry

$18 g$ glucose $(C_{6} H_{12} O_{6})$ is added to $178.2 g$ water. The vapour pressure of water (in torr) for this aqueous solution is:

Correct answer
A
$752.4$
B
$759.0$
C
$7.6$
D
$76.0$

Solution

\frac{∆ P}{P^{o}} =

Mole fraction of glucose.

\frac{760 - P_{S o l n}}{760} = \frac{\frac{W_{1}}{M w t_{1}}}{\frac{W_{1}}{M . w t_{1}} + \frac{w_{2}}{M . w t_{2}}} = \frac{\frac{18}{180}}{\frac{18}{180} + \frac{178.2}{18}} = \frac{0.1}{0.1 + 9.9} = \frac{1}{100}

760 - P_{S o l n} = \frac{760}{100}

P_{S o l} = 752.4

Hence, the correct answer is option: 'A'.

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