The vapour pressure of a solvent decreases by | vedclass

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

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(C) According to Raoult's law for non-volatile solutes,the relative lowering of vapour pressure is equal to the mole fraction of the solute: $\frac{P^0 - P_s}{P^0} = X_{solute}$.Given,$P^0 - P_s = 10 \ mm \ Hg$ and $X_{solute} = 0.2$.So,$\frac{10}{P^0} = 0.2 \Rightarrow P^0 = \frac{10}{0.2} = 50 \ mm \ Hg$.Now,for a new decrease in vapour pressure of $20 \ mm \ Hg$:$\frac{20}{P^0} = X'_{solute} \Rightarrow X'_{solute} = \frac{20}{50} = 0.4$.The mole fraction of the solute should be $0.4$.

The vapour pressure of a solvent decreases by $10 \ mm$ of mercury when a non-volatile solute is added to the solvent. The mole fraction of the solute in the solution is $0.2$. What should be the mole fraction of the solute if the decrease in the vapour pressure is to be $20 \ mm$ of mercury?

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