JEE Chemistry 2024 Question with Solution

Given: Pure benzene vapor pressure = $80 \, \text{Torr}$, pure methyl benzene vapor pressure = $24 \, \text{Torr}$, and the liquid mixture is equimolar, so liquid-phase mole fractions are $X_{\text{benzene}} = 0.5$ and $X_{\text{methylbenzene}} = 0.5$.

Find: The mole fraction of methyl benzene in the vapor phase in the form $\_\_\_\_\_ \times 10^{-2}$.

For an ideal solution, apply Raoult's law:

$$P_{\text{benzene}} = X_{\text{benzene}} P^{\text{pure}}_{\text{benzene}}$$ $$P_{\text{methylbenzene}} = X_{\text{methylbenzene}} P^{\text{pure}}_{\text{methylbenzene}}$$

Substituting the given values:

$$P_{\text{benzene}} = 0.5 \times 80 = 40 \, \text{Torr}$$ $$P_{\text{methylbenzene}} = 0.5 \times 24 = 12 \, \text{Torr}$$

Using Dalton's law, the total vapor pressure is:

$$P_{\text{total}} = 40 + 12 = 52 \, \text{Torr}$$

Now the mole fraction of methyl benzene in the vapor phase is:

$$Y_{\text{methylbenzene}} = \frac{P_{\text{methylbenzene}}}{P_{\text{total}}} = \frac{12}{52} \approx 0.2308$$

Expressing this as $\times 10^{-2}$:

$$0.2308 = 23.08 \times 10^{-2}$$

Rounding to the nearest integer gives $23$.

Therefore, the required numerical answer is 23.

Given: The liquid mixture is equimolar.

Find: Vapor-phase mole fraction of methyl benzene.

Because both liquid mole fractions are equal, the partial pressures are directly proportional to their pure vapor pressures. So:

$$Y_{\text{methylbenzene}} = \frac{0.5 \times 24}{0.5 \times 80 + 0.5 \times 24} = \frac{24}{80 + 24}$$

Thus,

$$Y_{\text{methylbenzene}} = \frac{24}{104} \approx 0.2308 = 23.08 \times 10^{-2}$$

Hence, the nearest integer is 23.

This shortcut works because the common factor $0.5$ cancels from numerator and denominator.