JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Solutions

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When 0.25 moles of a non-volatile, non-ionizable solute was dissolved in 1 mole of a solvent the vapor pressure of solution was

x %

of vapor pressure of pure solvent. What is

x %

(A) $50%$
(B) $60%$
(C) $70%$
(D) $80%$ ✓ Correct

Answer: (D)

Solution

For a non-volatile, non-ionising solute Raoult's law gives

P_{s} = P^{0} x_{solvent}

. The mole fraction of the solvent is

x_{solvent} = \frac{1}{1 + 0.25} = \frac{1}{1.25} = 0.80.

Hence

\frac{P _{s}}{P ^{0}} = 0.80

, so the vapour pressure of the solution is

80%

of that of the pure solvent.

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Single correct option

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