When a non-volatile solute is added to the so | vedclass

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(A) According to Raoult's law,the relative lowering of vapour pressure is equal to the mole fraction of the solute: $\frac{P^{\circ} - P}{P^{\circ}} =

Vedclass · Accepted Answer

$0.6$

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(A) According to Raoult's law,the relative lowering of vapour pressure is equal to the mole fraction of the solute: $\frac{P^{\circ} - P}{P^{\circ}} = X_{	ext{solute}}$.Since $P^{\circ} - P \propto X_{	ext{solute}}$,we have $10 \propto 0.2$.For a decrease of $20 \ mm \ Hg$,the new mole fraction of the solute $(X'_{	ext{solute}})$ will be: $20 \propto X'_{	ext{solute}}$.Thus,$X'_{	ext{solute}} = \frac{20}{10} 	imes 0.2 = 0.4$.The mole fraction of the solvent is $X_{	ext{solvent}} = 1 - X'_{	ext{solute}} = 1 - 0.4 = 0.6$.

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

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