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The vapour pressure of a solvent decrease by $2.5 \mathrm{~mm} \mathrm{Hg}$ by adding a solute. What is the mole fraction of solute? (Vapour pressure of pure solvent is $250 \mathrm{~mm} \mathrm{Hg}$ )
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$0.01$
$\begin{aligned} & \frac{\mathrm{P}^{\circ}-\mathrm{P}_{\mathrm{S}}}{\mathrm{P}^0}=\mathrm{X}_{\text {solute }} \\ & \mathrm{P}^{\circ}-\mathrm{P}_{\mathrm{S}}=2.5 \mathrm{~mm}, \mathrm{P}^{\circ}=250 \mathrm{~mm} \\ & \frac{2.5}{250}=\mathrm{X}_{\text {solute }} \\ & \mathrm{X}_{\text {solute }}=0.01\end{aligned}$
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