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The vapour pressure of a dilute aqueous solution of glucose is 750 mm of mercury at 373 K The mole fraction of solute is :

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$\frac{1}{76}$

$\frac{1}{35}$

$\frac{1}{7.6}$

$\frac{1}{10}$

answer is D.

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Detailed Solution

As stated by Raoult's Law,

$\frac{ΔP}{P} = X_{Solute}$

The pressure value at 373K=760mm

$\Rightarrow \frac{760 - 750}{760} = X_{Solute}$

$\Rightarrow mole fraction of solute = \frac{1}{76}$

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