JEE Main 2025 Jan 23 Shift 2, Chemistry Q8: Vapour Pressure Of A Solution

JEE Main2025Jan 23, Shift 2Chemistry

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 ?

A
0.6
B
0.4
C
0.2
D
0.8

Solution

By Raoult's law for a non-volatile solute, the lowering of vapour pressure is proportional to the mole fraction of solute:

$P^{\circ} - P = P^{\circ} x_{so l u t e}$ , so $Δ P \propto x_{so l u t e}$ .

Given $Δ P = 10$ mm at $x_{so l u t e} = 0.2$ , doubling $Δ P$ to 20 mm doubles $x_{so l u t e}$ to $0.4$ .

$∴ x_{so l v e n t} = 1 - 0.4 = 0.6$ .

Concept behind this question

Vapour Pressure Of A SolutionNotes, formulas and examples →

Practice more Chemistry

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Jan 23 Shift 2, Chemistry Q8: Vapour Pressure Of A Solution

Concept behind this question

More previous year questions on Vapour Pressure Of A Solution

Practice more Chemistry