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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Aicharya · Accepted Answer

$\begin{aligned} & \frac{\mathrm{P}^{\circ}-\mathrm{P}_{\mathrm{S}}}{\mathrm{P}^0}=\mathrm{X}_{	ext {solute }} \ & \mathrm{P}^{\circ}-\mathrm{P}_{\mathrm{S}}=2.5 \mathrm{~mm}, \mathrm{P}^{\circ}=250 \mathrm{~mm} \ & \frac{2.5}{250}=\mathrm{X}_{	ext {solute }} \ & \mathrm{X}_{	ext {solute }}=0.01\end{aligned}$

The vapour pressure of a solvent decrease by $2.5 \mathrm{~mm} \mathrm{Hg}$ by adding a solute. What is the…

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