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JEE Advanced 2026 · Paper 2, Chemistry question 16

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Question Stem for Question Nos. 15 and 16

Two volatile liquids A and B form an ideal solution. Consider a 5 molal solution of B in A inside a closed container having a total vapour pressure of 100 mm Hg at 300 K. The vapour pressure of pure A at 300 K is 105 mm Hg. Assume that A and B behave as ideal gases in the vapour phase.

Given:
The gas constant R=0.08R=0.08 L atm K−1^{-1} mol−1^{-1}
Molar mass of A is 50 g mol−1^{-1}
Molar mass of B is 57 g mol−1^{-1}
Density of liquid B at 300 K is 0.5 g/mL
1 atm = 760 mm Hg

The mole fraction of B in vapour phase which is in equilibrium with this solution is ____.

Answer and worked solution

Official answer: 0.16

From the previous part, xB=0.2x_B=0.2 and pB∘=80p_B^\circ=80 mm Hg. So pB=0.2×80=16p_B=0.2\times80=16 mm Hg, and
yB=pBP=16100=0.16.y_B=\frac{p_B}{P}=\frac{16}{100}=0.16.

Topics:Raoult's law and vapour pressureIdeal and non-ideal solutions
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