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(A) The relative lowering of vapour pressure is given by the mole fraction of the solute: $\frac{\Delta P}{P} = x_{solute} = \frac{n_{solute}}{n_{solu

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(A) The relative lowering of vapour pressure is given by the mole fraction of the solute: $\frac{\Delta P}{P} = x_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}}$.Since the number of moles of solute is very small compared to the solvent,$n_{solute} + n_{solvent} \approx n_{solvent}$.Thus,$\frac{\Delta P}{P} \approx \frac{n_{solute}}{n_{solvent}} = \frac{n_{solute} 	imes M_{solvent}}{w_{solvent}}$.For a $5 \ molal$ solution,$n_{solute} = 5 \ mol$ and $w_{solvent} = 1000 \ g$.Therefore,$\left( \frac{\Delta P}{P} ight) = \frac{5 	imes M_{solvent}}{1000}$.Given $\left( \frac{\Delta P}{P} ight)_X = m \left( \frac{\Delta P}{P} ight)_Y$,we have $\frac{5 	imes M_X}{1000} = m 	imes \frac{5 	imes M_Y}{1000}$.This simplifies to $M_X = m 	imes M_Y$.Given $M_X = \frac{3}{4} M_Y$,we get $m = \frac{3}{4}$.

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