JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Solutions

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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)

Answer: 200

Solution

For an ideal solution

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

. With

3

mol A and

1

mol B,

500 = \frac{3}{4} P_{A}^{\circ} + \frac{1}{4} P_{B}^{\circ}

, that is

3 P_{A}^{\circ} + P_{B}^{\circ} = 2000

. After adding

1

mol A the mole fractions become

4/5

and

1/5

with

P_{S} = 520

, giving

4 P_{A}^{\circ} + P_{B}^{\circ} = 2600

. Subtracting the first from the second,

P_{A}^{\circ} = 600

mm Hg, and then

P_{B}^{\circ} = 2000 - 3 (600) = 200

mm Hg.

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Integer answer

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JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Solutions

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