# QuestionThe vapour pressure of a solvent decreased by 10mm of mercury, when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent, if decrease in the vapour pressure is to be 20mm of mercuryA.B.C.D.

For very dilute solutions -

$\frac{\Delta P}{P^{0}}= \frac{n_{solute}}{n_{solvent}}$

- wherein

$since \: \: n_{solute}+n_{solvent}\simeq n_{solvent}$

$\therefore P^{0}-P_{s}=P^{0}\times mole\; fraction \; solute$

$10=P^{0}\times 0.2;\; 20=P^{0}\times n\Rightarrow n=0.4\therefore N=0.6.$

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