Liquid A and B form an ideal solution. The vapour pressure of pure liquids A and B are 350 and 750 mm Hg respectively at the same temperature. If $\mathrm{x}_{\mathrm{A}}$ and $x_{B}$ are the mole fraction of $A$ and $B$ in solution while $y_{A}$ and $y_{B}$ are the mole fraction of $A$ and $B$ in vapour phase then :

(C) $\frac{x_{A}}{x_{B}}>\frac{y_{A}}{y_{B}}$

(A) $\frac{x_{A}}{x_{B}}<\frac{y_{A}}{y_{B}}$

(B) $\frac{x_{A}}{x_{B}}=\frac{y_{A}}{y_{B}}$

(D) $(x_{A}-y_{A})<(x_{B}-y_{B})$

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Solutions

Liquid A and B form an ideal solution. The vapour pressure of pure liquids A and B are 350 and 750 mm Hg respectively at the same temperature. If

x_{A}

and

x_{B}

are the mole fraction of

A

and

B

in solution while

y_{A}

and

y_{B}

are the mole fraction of

A

and

B

in vapour phase then :

Answer: (C)

For an ideal solution the vapour is richer in the more volatile component. Raoult's law gives the vapour ratio as

\frac{y _{A}}{y _{B}} = \frac{P _{A}^{\circ}}{P _{B}^{\circ}} \cdot \frac{x _{A}}{x _{B}}

. Here

P_{A}^{\circ} = 350

and

P_{B}^{\circ} = 750

mm Hg, so the factor

\frac{P _{A}^{\circ}}{P _{B}^{\circ}} = \frac{350}{750}

is less than one. Multiplying the liquid ratio by a number below one makes the vapour ratio smaller, so

\frac{x _{A}}{x _{B}} > \frac{y _{A}}{y _{B}}

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Single correct option