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(C) According to Raoult's Law for non-volatile solutes,the relative lowering of vapor pressure is equal to the mole fraction of the solute: $\frac{\De

Vedclass · Accepted Answer

$0.6$

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(C) According to Raoult's Law for non-volatile solutes,the relative lowering of vapor pressure is equal to the mole fraction of the solute: $\frac{\Delta P}{P^0} = X_{solute}$.Given $\Delta P_1 = 10 \ mm \ Hg$ and $X_{solute, 1} = 0.2$.So,$\frac{10}{P^0} = 0.2 \implies P^0 = \frac{10}{0.2} = 50 \ mm \ Hg$.Now,for a new decrease $\Delta P_2 = 20 \ mm \ Hg$,the new mole fraction of solute is $X_{solute, 2} = \frac{\Delta P_2}{P^0} = \frac{20}{50} = 0.4$.The mole fraction of the solvent is $X_{solvent} = 1 - X_{solute, 2} = 1 - 0.4 = 0.6$.

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

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