JEE Main 2026 — Jan 24, Shift 2 — Chemistry solution

JEE Main2026Jan 24, Shift 2Chemistry

Two liquids A and B form an ideal solution at temperature T K. At T K, the vapour pressures of pure A and B are 55 and 15 $kNm^{- 2}$ respectively. What is the mole fraction of A in solution of A and B in equilibrium with a vapour in which the mole fraction of A is 0.8 ?

A
0.5217
B
0.480
C
0.663
D
0.340

Solution

For an ideal solution, the ratio of vapour mole fractions equals the ratio of partial pressures:

$\frac{Y _{A}}{Y _{B}} = \frac{P _{A}^{o} X _{A}}{P _{B}^{o} X _{B}}$

With $Y_{A} = 0.8$ , $Y_{B} = 0.2$ , $P_{A}^{o} = 55$ , $P_{B}^{o} = 15$ :

$\frac{0.8}{0.2} = \frac{55}{15} \cdot \frac{X _{A}}{X _{B}} \Rightarrow \frac{X _{A}}{X _{B}} = \frac{4 \times 15}{55} = \frac{12}{11}$

Since $X_{A} + X_{B} = 1$ , we get $X_{A} = \frac{12}{23} \approx 0.5217$ .

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