JEE Main 2026 — Chemistry Physical Chemistry

2026integermedium

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 pure state is $\_\_\_\_$ . (Nearest integer)

Official previous-year question

Held on 23 Jan 2026 · Verified 6 Jul 2026.

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Solution

For an ideal solution, Raoult's law states:

$P_{total} = P_A^o x_A + P_B^o x_B$ .

Initial condition: 3 mol of A and 1 mol of B give a total vapor pressure of 500 mm Hg.

Mole fractions are $x_A = \frac{3}{4} = 0.75$ and $x_B = \frac{1}{4} = 0.25$ .

This gives: $500 = 0.75 P_A^o + 0.25 P_B^o$ ... (1).

After adding 1 mol of A: 4 mol of A and 1 mol of B give a total vapor pressure of 520 mm Hg.

Mole fractions are $x_A = \frac{4}{5} = 0.8$ and $x_B = \frac{1}{5} = 0.2$ .

This gives: $520 = 0.8 P_A^o + 0.2 P_B^o$ ... (2).

From equation (1) multiplied by 4: $2000 = 3 P_A^o + P_B^o$ .

From equation (2) multiplied by 5: $2600 = 4 P_A^o + P_B^o$ .

Subtracting: $600 = P_A^o$ .

Substituting back into equation (1): $2000 = 3(600) + P_B^o$ , so $P_B^o = 200$ mm Hg.

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JEE Main 2026 — Chemistry Physical Chemistry

2026integermedium

Official previous-year question

Held on 23 Jan 2026 · Verified 6 Jul 2026.

Did you get this right?

Solution

For an ideal solution, Raoult's law states:

$P_{total} = P_A^o x_A + P_B^o x_B$ .

Initial condition: 3 mol of A and 1 mol of B give a total vapor pressure of 500 mm Hg.

Mole fractions are $x_A = \frac{3}{4} = 0.75$ and $x_B = \frac{1}{4} = 0.25$ .

This gives: $500 = 0.75 P_A^o + 0.25 P_B^o$ ... (1).

After adding 1 mol of A: 4 mol of A and 1 mol of B give a total vapor pressure of 520 mm Hg.

Mole fractions are $x_A = \frac{4}{5} = 0.8$ and $x_B = \frac{1}{5} = 0.2$ .

This gives: $520 = 0.8 P_A^o + 0.2 P_B^o$ ... (2).

From equation (1) multiplied by 4: $2000 = 3 P_A^o + P_B^o$ .

From equation (2) multiplied by 5: $2600 = 4 P_A^o + P_B^o$ .

Subtracting: $600 = P_A^o$ .

Substituting back into equation (1): $2000 = 3(600) + P_B^o$ , so $P_B^o = 200$ mm Hg.

JEE Main 2026 — Chemistry Physical Chemistry

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Solution

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JEE Main 2026 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry