JEE Main 2026 — Jan 28, Shift 1 — Chemistry solution

JEE Main2026Jan 28, Shift 1Chemistry

At T(K), 2 moles of liquid A and 3 moles of liquid B are mixed. The vapour pressure of ideal solution formed is 320 mm Hg. At this stage, one mole of A and one mole of B are added to the solution. The vapour pressure is now measured as 328.6 mm Hg. The vapour pressure (in mm Hg) of A and B are respectively:

A
300, 200
B
600, 400
C
400, 300
D
500, 200

Solution

Let $P_{A}^{\circ}$ and $P_{B}^{\circ}$ be the pure component vapour pressures. Apply Raoult's law to each stage.

Stage 1 (2 mol A + 3 mol B): $x_{A} = 2/5$ , $x_{B} = 3/5$ .

$P_{S} = \frac{2}{5} P_{A}^{\circ} + \frac{3}{5} P_{B}^{\circ} = 320$

$\Rightarrow 2 P_{A}^{\circ} + 3 P_{B}^{\circ} = 1600$ ... (I)

Stage 2 (after adding 1 mol A and 1 mol B, total 3 mol A + 4 mol B): $x_{A}^{'} = 3/7$ , $x_{B}^{'} = 4/7$ .

$P_{S}^{'} = \frac{3}{7} P_{A}^{\circ} + \frac{4}{7} P_{B}^{\circ} = 328.6$

$\Rightarrow 3 P_{A}^{\circ} + 4 P_{B}^{\circ} = 2300.2$ ... (II)

Eliminate $P_{A}^{\circ}$ : multiply (I) by 3 and (II) by 2:

$6 P_{A}^{\circ} + 9 P_{B}^{\circ} = 4800$

$6 P_{A}^{\circ} + 8 P_{B}^{\circ} = 4600.4$

Subtracting: $P_{B}^{\circ} \approx 200$ mm Hg, and from (I): $P_{A}^{\circ} \approx 500$ mm Hg.

$∴$ Option (4) is correct.

← Back to full paper

Practice more Chemistry

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →