JEE MAIN - Chemistry (2025 - 7th April Evening Shift - No. 12)

Liquid A and B form an ideal solution. The vapour pressures 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,
$(x_A - y_A) < (x_B - y_B)$
$\frac{x_A}{x_B} = \frac{y_A}{y_B}$
$\frac{x_A}{x_B} < \frac{y_A}{y_B}$
$\frac{x_A}{x_B} > \frac{y_A}{y_B}$

Explanation

Liquid A and B form an ideal solution. The vapor pressures of pure liquids A and B are 350 mm Hg and 750 mm Hg, respectively, at the same temperature. Here, $ x_A $ and $ x_B $ represent the mole fractions of A and B in the solution, and $ y_A $ and $ y_B $ are their mole fractions in the vapor phase.

Let’s begin by comparing the vapor pressures:

$ \mathrm{P}_{\mathrm{A}}^{\mathrm{o}} < \mathrm{P}_{\mathrm{B}}^{\mathrm{o}} $

$ \frac{\mathrm{P}_{\mathrm{A}}^{\mathrm{o}}}{\mathrm{P}_{\mathrm{B}}^{\mathrm{o}}} < 1 $

The relationship between the mole fractions in the vapor phase and the solution can be expressed as:

$ \frac{y_A}{y_B} = \frac{\mathrm{P}_{\mathrm{A}}^{\mathrm{o}}}{\mathrm{P}_{\mathrm{B}}^{\mathrm{o}}} \cdot \frac{x_A}{x_B} $

Since $\frac{\mathrm{P}_{\mathrm{A}}^{\mathrm{o}}}{\mathrm{P}_{\mathrm{B}}^{\mathrm{o}}} < 1$, it follows that:

$ \frac{\frac{y_A}{y_B}}{\frac{x_A}{x_B}} < 1 $

Which implies:

$ \frac{y_A}{y_B} < \frac{x_A}{x_B} $

This indicates that the mole fraction ratio of A to B in the vapor phase is less than that in the solution.

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