JEE Main 2026 — Solutions Question with Solution
JEE Main 2026 (23 January Shift 2)
Question
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)
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Show full solutionCorrect answer: 200
Correct answer
200
Step-by-step explanation
For an ideal solution, Raoult's law states:
.
Initial condition: 3 mol of A and 1 mol of B give a total vapor pressure of 500 mm Hg.
Mole fractions are and .
This gives: ... (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 and .
This gives: ... (2).
From equation (1) multiplied by 4: .
From equation (2) multiplied by 5: .
Subtracting: .
Substituting back into equation (1): , so mm Hg.
.
Initial condition: 3 mol of A and 1 mol of B give a total vapor pressure of 500 mm Hg.
Mole fractions are and .
This gives: ... (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 and .
This gives: ... (2).
From equation (1) multiplied by 4: .
From equation (2) multiplied by 5: .
Subtracting: .
Substituting back into equation (1): , so mm Hg.
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This is a previous-year question from JEE Main 2026, covering the Solutions chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.