JEE Advanced 2026 — Paper 2 — CHEM-IV solution

JEE Advanced2026Paper 2CHEM-IV

Question Stem (Q1-Q2). 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: Gas constant $R = 0.08$ L atm K $^{- 1}$ mol $^{- 1}$ ; molar mass of $A$ is $50$ g mol $^{- 1}$ ; molar mass of $B$ is $57$ g mol $^{- 1}$ ; density of liquid $B$ at $300$ K is $0.5$ g/mL; $1$ atm $= 760$ mm Hg.

Q2. The mole fraction of $B$ in vapour phase which is in equilibrium with this solution is _____.

Your answer

Solution

Partial pressure of B in vapour. $P_{B} = x_{B} P_{B}^{\circ} = 0.2 \times 80 = 16$ mm Hg.

Mole fraction in vapour. $y_{B} = \frac{P _{B}}{P _{T}} = \frac{16}{100} = 0.16$ .

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