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.

Q1. At $300$ K, the ratio of the molar volume of pure $B$ in vapour phase to its molar volume in liquid phase is _____.

Your answer

Solution

Mole fractions in the liquid. $5$ molal means $5$ mol B per $1000$ g A, so moles of A $= 1000/50 = 20$ . $x_{A} = 20/25 = 0.8$ and $x_{B} = 5/25 = 0.2$ .

Find $P_{B}^{\circ}$ from Raoult. $P_{T} = x_{A} P_{A}^{\circ} + x_{B} P_{B}^{\circ}$ :

$100 = 0.8 (105) + 0.2 P_{B}^{\circ} = 84 + 0.2 P_{B}^{\circ} \Rightarrow P_{B}^{\circ} = 80$ mm Hg.

Molar volume of B vapour. Using $P V = n R T$ for one mole at $P = P_{B}^{\circ}$ :

$V_{m}^{vap} = \frac{R T}{P} = \frac{( 0.08 ) ( 300 )}{80/760} = \frac{24}{0.1053} \approx 228$ L/mol.

Molar volume of B liquid. $V_{m}^{liq} = \frac{M _{B}}{ρ _{B}} = \frac{57}{0.5} = 114$ mL/mol $= 0.114$ L/mol.

Ratio. $\frac{V _{m}^{vap}}{V _{m}^{liq}} = \frac{228}{0.114} = 2000$ .

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