JEE Main 2026 — Jan 23, Shift 2 — Chemistry solution

JEE Main2026Jan 23, Shift 2Chemistry

Two liquids A and B form an ideal solution. At 320 K, the vapour pressure of the solution, containing 3 mol of A and 1 mol of B is 500 mm Hg. At the same temperature, if 1 mol of A is further added to this solution, vapour pressure of the solution increases by 20 mm Hg. Vapour pressure (in mm Hg) of B in the pure state is _______. (Nearest integer)

Your answer

Solution

Let $P_{A}^{\circ}$ and $P_{B}^{\circ}$ be the vapour pressures of pure A and B.

Initial mixture: 3 mol A + 1 mol B. Mole fractions $X_{A} = 3/4$ , $X_{B} = 1/4$ .

$P_{S} = P_{A}^{\circ} X_{A} + P_{B}^{\circ} X_{B} = 500$

$\frac{3}{4} P_{A}^{\circ} + \frac{1}{4} P_{B}^{\circ} = 500 \Rightarrow 3 P_{A}^{\circ} + P_{B}^{\circ} = 2000$ ...(1)

After adding 1 mol A: 4 mol A + 1 mol B. $X_{A}^{'} = 4/5$ , $X_{B}^{'} = 1/5$ . New $P_{S}^{'} = 520$ .

$\frac{4}{5} P_{A}^{\circ} + \frac{1}{5} P_{B}^{\circ} = 520 \Rightarrow 4 P_{A}^{\circ} + P_{B}^{\circ} = 2600$ ...(2)

Subtracting (1) from (2): $P_{A}^{\circ} = 600$ mm Hg.

Substituting back: $P_{B}^{\circ} = 2000 - 1800 = 200$ mm Hg.

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