JEE MAIN - Chemistry (2025 - 23rd January Evening Shift - No. 2)
When a non-volatile solute is added to the solvent, the vapour pressure of the solvent decreases by 10 mm of Hg . The mole fraction of the solute in the solution is 0.2 . What would be the mole fraction of the solvent if decrease in vapour pressure is 20 mm of Hg ?
0.2
0.4
0.8
0.6
Explanation
$$\begin{aligned}
& \because \mathrm{P}^{\circ}-\mathrm{P} \propto \mathrm{X}_{\text {solute }} \\
& \text { and } \because 10 \propto 0.2 \\
& \therefore 20 \propto 0.4 \\
& \therefore \mathrm{X}_{\text {solvent }}=1-\mathrm{X}_{\text {solute }} \\
& \quad=1-0.4 \\
& \quad=0.6
\end{aligned}$$
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