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(B) According to Raoult's law,the relative lowering of vapour pressure is equal to the mole fraction of the solute: $\frac{\Delta p}{p^o} = x_{solute}

Vedclass · Accepted Answer

$0.6$

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(B) According to Raoult's law,the relative lowering of vapour pressure is equal to the mole fraction of the solute: $\frac{\Delta p}{p^o} = x_{solute}$.Given $\Delta p = 10 \ mm \ Hg$ and $x_{solute} = 0.2$,we have $\frac{10}{p^o} = 0.2$,which gives $p^o = \frac{10}{0.2} = 50 \ mm \ Hg$.For the second case,where $\Delta p = 20 \ mm \ Hg$,the new mole fraction of the solute $(x'_{solute})$ is $\frac{20}{p^o} = \frac{20}{50} = 0.4$.Since the sum of mole fractions of solute and solvent is $1$,the mole fraction of the solvent is $x_{solvent} = 1 - x'_{solute} = 1 - 0.4 = 0.6$.

The vapour pressure of a solvent decreases by $10 \ mm$ of $Hg$ 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 solvent if the decrease in the vapour pressure is to be $20 \ mm$ of $Hg$?

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