Okay, let's use the provided vapor pressure values for pure benzene and pure methylbenzene. Given: Vapor pressure of pure benzene ($P_{\text{benzene}}^\circ$) $= 3.6 \times 10^4$ Vapor pressure of pure methylbenzene ($P_{\text{methylbenzene}}^\circ$) $= 1.12 \times 10^4$ Step 1: Calculate the vapor pressure contribution of benzene ($P_{\text{benzene}}$) using Raoult's Law: $$P_{\text{benzene}} = X_{\text{benzene}} P_{\text{benzene}}^\circ$$ Step 2: Calculate the vapor pressure contribution of methylbenzene ($P_{\text{methylbenzene}}$) using Raoult's Law: $$P_{\text{methylbenzene}} = X_{\text{methylbenzene}} P_{\text{methylbenzene}}^\circ$$ Remember that $X_{\text{methylbenzene}} = 1 - X_{\text{benzene}}$. Here are the updated calculations: | $X_{\text{benzene}}$ | $X_{\text{methylbenzene}}$ | $P_{\text{benzene}} = X_{\text{benzene}} \times (3.6 \times 10^4)$ | $P_{\text{methylbenzene}} = X_{\text{methylbenzene}} \times (1.12 \times 10^4)$ | | :------------------- | :------------------------- | :----------------------------------------------------------------- | :----------------------------------------------------------------------- | | 0 | 1 | $0 \times 3.6 \times 10^4 = 0$ | $1 \times 1.12 \times 10^4 = 1.12 \times 10^4$ | | 0.25 | 0.75 | $0.25 \times 3.6 \times 10^4 = 0.9 \times 10^4$ | $0.75 \times 1.12 \times 10^4 = 0.84 \times 10^4$ | | 0.5 | 0.5 | $0.5 \times 3.6 \times 10^4 = 1.8 \times 10^4$ | $0.5 \times 1.12 \times 10^4 = 0.56 \times 10^4$ | | 0.75 | 0.25 | $0.75 \times 3.6 \times 10^4 = 2.7 \times 10^4$ | $0.25 \times 1.12 \times 10^4 = 0.28 \times 10^4$ | | 1 | 0 | $1 \times 3.6 \times 10^4 = 3.6 \times 10^4$ | $0 \times 1.12 \times 10^4 = 0$ | The respective vapor pressure contributions for benzene are: $\boxed{\text{0, } 0.9 \times 10^4\text{, } 1.8 \times 10^4\text{, } 2.7 \times 10^4\text{, } 3.6 \times 10^4}$ The respective vapor pressure contributions for methylbenzene are: $\boxed{1.12 \times 10^4\text{, } 0.84 \times 10^4\text{, } 0.56 \times 10^4\text{, } 0.28 \times 10^4\text{, } 0}$