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 $15 \mathrm{kNm}^{-2}$ 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 ?

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Solutions

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 ?

Answer: (A)

For an ideal binary solution the partial pressures obey Raoult's law,

p_{A} = x_{A} p_{A}^{0}

and

p_{B} = x_{B} p_{B}^{0}

, and the vapour composition is their ratio,

\frac{y _{A}}{y _{B}} = \frac{x _{A} p _{A}^{0}}{x _{B} p _{B}^{0}}

. With

y_{A} = 0.8

y_{B} = 0.2

p_{A}^{0} = 55

and

p_{B}^{0} = 15 kN m^{- 2}

\frac{0.8}{0.2} = \frac{55}{15} \cdot \frac{x _{A}}{x _{B}}

, so

\frac{x _{A}}{x _{B}} = \frac{4 \times 15}{55} = \frac{12}{11}

. Since

x_{A} + x_{B} = 1

x_{A} = \frac{12}{23} = 0.5217

. The mole fraction of A in the liquid is

0.5217

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Single correct option