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The 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 mercury

A.

0.8

B.

0.6

C.

0.4

D.

0.2

Answers (1)

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.

Posted by

Satyajeet Kumar

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