For an ideal solution made up of benzene and methylbenzene. given that the mole fraction of benzene at different concentration are 0, 0.25, 0.5, 0.75 and 1. find the respective mole fractions of methylbenzene at this concentration. Calculate the respective vapour pressure contribution of benzene at this concentration

For a binary solution, the sum of mole fractions is 1:

$$X_{\text{benzene}} + X_{\text{methylbenzene}} = 1$$

Therefore, $X_{\text{methylbenzene}} = 1 - X_{\text{benzene}}$.

According to Raoult's Law for an ideal solution:

$$P_{\text{benzene}} = X_{\text{benzene}} P_{\text{benzene}}^\circ$$

where $P_{\text{benzene}}^\circ$ is the vapor pressure of pure benzene.

Here are the results for the given concentrations:

$X_{\text{benzene}}$	$X_{\text{methylbenzene}}$	$P_{\text{benzene}}$ (Vapor Pressure Contribution)
0	$1 - 0 = 1$	$0 \cdot P_{\text{benzene}}^\circ = 0$
0.25	$1 - 0.25 = 0.75$	$0.25 \cdot P_{\text{benzene}}^\circ$
0.5	$1 - 0.5 = 0.5$	$0.5 \cdot P_{\text{benzene}}^\circ$
0.75	$1 - 0.75 = 0.25$	$0.75 \cdot P_{\text{benzene}}^\circ$
1	$1 - 1 = 0$	$1 \cdot P_{\text{benzene}}^\circ = P_{\text{benzene}}^\circ$

The respective mole fractions of methylbenzene are 1, 0.75, 0.5, 0.25, 0.
The respective vapor pressure contributions of benzene are 0, $0.25 P_{\text{benzene}}^\circ$, $0.5 P_{\text{benzene}}^\circ$, $0.75 P_{\text{benzene}}^\circ$, $P_{\text{benzene}}^\circ$.