JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Solutions

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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✓ Correct
(B)0.4
(C)0.2
(D)0.8

Answer: (A)

Solution

By Raoult's law, the relative lowering of vapour pressure equals the mole fraction of solute:

\frac{P ^{\circ} - P}{P ^{\circ}} = x_{so l u t e}

, so for a fixed solvent the drop in vapour pressure is directly proportional to

x_{so l u t e}

. A drop of 10 mm Hg corresponds to

x_{so l u t e} = 0.2

, so a drop of 20 mm Hg (twice as large) corresponds to

x_{so l u t e} = 0.4

. The mole fraction of the solvent is then

1 - 0.4 = 0.6

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Single correct option

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