Calculate the respective vapour pressure contribution for methylbenzene at this concentration

According to Raoult's Law for an ideal solution:

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

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

Here are the results for the given concentrations:

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

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