QJEEM24Che9c6: JEE Main 2024 Chemistry - Class 12 JEEM24Che9c6 Solution: The vapor pressure of pure benzene and methyl benzene at is…

To find the mole fraction of methyl benzene in the vapor phase when it is in equilibrium with an equimolar (equal mole) mixture of benzene and methyl benzene at $27^{\circ}C$ , we can apply Raoult's law for an ideal solution. Given the vapor pressures of pure benzene and methyl benzene are 80 Torr and 24 Torr, respectively.

Raoult's law states that the partial vapor pressure of a component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution ( $P_i = P_i^\circ x_i$ ), where $P_i$ is the partial vapor pressure of the component $i$ , $P_i^\circ$ is the vapor pressure of the pure component $i$ , and $x_i$ is the mole fraction of the component $i$ in the solution.

For an equimolar mixture of benzene and methyl benzene, the mole fractions of benzene ( $x_b$ ) and methyl benzene ( $x_{mb}$ ) in the liquid phase are both $0.5$ , because the mixture is equimolar.

The total pressure of the mixture ( $P_{total}$ ) can be calculated using Raoult's law for each component and summing up the individual pressures:

$P_{total} = P_{benzene} + P_{methyl\ benzene}$

$P_{total} = (P_b^\circ x_b) + (P_{mb}^\circ x_{mb})$

Substituting the given values:

$P_{total} = (80 \times 0.5) + (24 \times 0.5)$

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

The mole fraction of methyl benzene in the vapor phase ( $y_{mb}$ ) can be calculated using Dalton's law, which states that the partial pressure of a component in a mixture is equal to the mole fraction of that component in the vapor phase times the total pressure of the mixture.

The partial pressure of methyl benzene can be obtained from Raoult's law as we did earlier:

$P_{methyl\ benzene} = P_{mb}^\circ x_{mb} = 24 \times 0.5 = 12 \, \text{Torr}$

Using Dalton's law:

$y_{mb} = \frac{P_{methyl\ benzene}}{P_{total}}$

Substituting the values:

$y_{mb} = \frac{12}{52}$

$y_{mb} = \frac{6}{26}$

$y_{mb} = \frac{3}{13}$

To express $y_{mb}$ as a percentage times $10^{-2}$ :

$y_{mb} = \left(\frac{3}{13}\right) \times 100 \times 10^{-2}$

$y_{mb} \approx 23.08 \times 10^{-2}$

So, the mole fraction of methyl benzene in the vapor phase, in equilibrium with an equimolar mixture of those two liquids at $27^{\circ}C$ , is approximately $23 \times 10^{-2}$ (rounded to the nearest integer).