Question

The vapour pressure of pure benzene and methyl benzene at $27^\circ \text{C}$ is given as 80 Torr and 24 Torr, respectively. The mole fraction of methyl benzene in vapour phase, in equilibrium with an equimolar mixture of those two liquids (ideal solution) at the same temperature, is $\dots \times 10^{-2}$ (nearest integer).

Answer 1

The mole fraction $x_m$ of methyl benzene in the vapor phase can be calculated using Raoult's Law for ideal solutions:
$x_m = \frac{P_m}{P_1 + P_2}$
where: $P_m$ is the partial vapor pressure of methyl benzene in the vapor phase, given as 24 Torr.
$P_1$ and $P_2$ are the vapor pressures of pure benzene and methyl benzene, respectively.
The mole fraction $x_m$ is:
$x_m = \frac{24}{80 + 24} = \frac{24}{104} \approx 0.2308$
Thus, the mole fraction is $23 \times 10^{-2}$. The correct answer is (23).

Answer 2

The mole fraction $x_m$ of methyl benzene in the vapor phase can be calculated using Raoult's Law for ideal solutions:
$x_m = \frac{P_m}{P_1 + P_2}$
where: $P_m$ is the partial vapor pressure of methyl benzene in the vapor phase, given as 24 Torr.
$P_1$ and $P_2$ are the vapor pressures of pure benzene and methyl benzene, respectively.
The mole fraction $x_m$ is:
$x_m = \frac{24}{80 + 24} = \frac{24}{104} \approx 0.2308$
Thus, the mole fraction is $23 \times 10^{-2}$. The correct answer is (23).