JEE Chemistry 2025 Question with Solution

Given: Liquid A and B form an ideal solution with vapour pressures of pure components $P_A^\circ = 350 \, \text{mm Hg}$ and $P_B^\circ = 750 \, \text{mm Hg}$.

Find: The correct relation between $\frac{x_A}{x_B}$ and $\frac{y_A}{y_B}$.

For an ideal solution, Raoult's law gives

$$P_A = x_A P_A^\circ$$

and

$$P_B = x_B P_B^\circ$$

So,

$$P_A = x_A \cdot 350$$ $$P_B = x_B \cdot 750$$

The mole fractions in vapour phase are

$$y_A = \frac{P_A}{P_A + P_B}$$

and

$$y_B = \frac{P_B}{P_A + P_B}$$

Therefore,

$$\frac{y_A}{y_B} = \frac{P_A}{P_B} = \frac{x_A \cdot 350}{x_B \cdot 750}$$

Hence,

$$\frac{y_A}{y_B} = \frac{x_A}{x_B} \cdot \frac{350}{750} = \frac{x_A}{x_B} \cdot \frac{7}{15}$$

Since

$$\frac{7}{15} < 1$$

it follows that

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

Thus,

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

Therefore, the correct option is C.

Given: Component B has higher vapour pressure than A, so B is more volatile.

Find: How the liquid-phase ratio compares with the vapour-phase ratio.

In an ideal solution, the vapour phase is richer in the more volatile component. Since $P_B^\circ > P_A^\circ$, component B is present in relatively larger proportion in vapour than in liquid.

So the ratio of A to B decreases in the vapour phase:

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

Therefore,

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

Hence, the correct option is C.