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(A) According to Raoult's law,the lowering of vapour pressure is given by $\Delta P = P^{\circ} - P_s = P^{\circ} 	imes X_{solute}$.Given $\Delta P =

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$\frac{3}{4}$

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(A) According to Raoult's law,the lowering of vapour pressure is given by $\Delta P = P^{\circ} - P_s = P^{\circ} 	imes X_{solute}$.Given $\Delta P = 20 \ mm \ Hg$ and $X_{solute} = 0.5$.So,$20 = P^{\circ} 	imes 0.5$,which gives $P^{\circ} = 40 \ mm \ Hg$.Now,we want the new lowering of vapour pressure $\Delta P' = 10 \ mm \ Hg$.Using $\Delta P' = P^{\circ} 	imes X'_{solute}$,we get $10 = 40 	imes X'_{solute}$.Therefore,$X'_{solute} = \frac{10}{40} = 0.25$.The mole fraction of the solvent is $X_{solvent} = 1 - X'_{solute} = 1 - 0.25 = 0.75 = \frac{3}{4}$.

The vapour pressure of a solvent decreases by $20 \ mm$ of $Hg$ when a non-volatile solute is added to the solvent. The mole fraction of the solute in the solution is $0.5$. What should be the mole fraction of the solvent for the decrease in the vapour pressure to be $10 \ mm$ of $Hg$?

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