When a non-volatile solute is added to a solv | vedclass

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(B) According to Raoult's Law,the relative lowering of vapor pressure is equal to the mole fraction of the solute: $\frac{\Delta P}{P^\circ} = X_{	ex

Vedclass · Accepted Answer

$0.6$

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(B) According to Raoult's Law,the relative lowering of vapor pressure is equal to the mole fraction of the solute: $\frac{\Delta P}{P^\circ} = X_{	ext{solute}}$.Given,$\Delta P_1 = 10 \ mm$ when $X_{	ext{solute}, 1} = 0.2$.So,$\frac{10}{P^\circ} = 0.2$,which gives $P^\circ = \frac{10}{0.2} = 50 \ mm$.Now,for $\Delta P_2 = 20 \ mm$,the new mole fraction of the solute is $X_{	ext{solute}, 2} = \frac{\Delta P_2}{P^\circ} = \frac{20}{50} = 0.4$.The mole fraction of the solvent is $X_{	ext{solvent}} = 1 - X_{	ext{solute}, 2} = 1 - 0.4 = 0.6$.

When a non-volatile solute is added to a solvent,the vapor pressure decreases by $10 \ mm$. The mole fraction of the solute is $0.2$. If the vapor pressure decreases by $20 \ mm$,what will be the mole fraction of the solvent?

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