JEE Main 2025 Apr 2 Shift 1, Chemistry Q7: Vapour Pressure Of A Solution

JEE Main2025Apr 2, Shift 1Chemistry

A solution is made by mixing one mole of volatile liquid A with 3 moles of volatile liquid B. The vapour pressure of pure A is 200 mm Hg and that of the solution is 500 mm Hg. The vapour pressure of pure B and the least volatile component of the solution, respectively, are :

A
1400 mm Hg, A
B
1400 mm Hg, B
C
600 mm Hg, B
D
600 mm Hg, A

Solution

Mole fractions: $x_{A} = \frac{1}{4}$ , $x_{B} = \frac{3}{4}$ .

Raoult's law for an ideal binary solution:

$P_{S} = P_{A}^{\circ} x_{A} + P_{B}^{\circ} x_{B}$

$500 = 200 \times \frac{1}{4} + P_{B}^{\circ} \times \frac{3}{4}$

$500 - 50 = \frac{3}{4} P_{B}^{\circ} \Rightarrow P_{B}^{\circ} = 600 mm Hg$

Since $P_{A}^{\circ} = 200 < P_{B}^{\circ} = 600$ , component A has the lower pure-liquid vapour pressure and is therefore the least volatile.

$∴$ Option (4): 600 mm Hg, A.

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JEE Main 2025 Apr 2 Shift 1, Chemistry Q7: Vapour Pressure Of A Solution

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