To solve this problem, we will use Raoult's Law, which applies to ideal solutions.$\\$Raoult's Law states that the partial vapor pressure of each component in the mixture is equal to$\\$the vapor pressure of the pure component multiplied by its mole fraction in the liquid mixture.$\\$Given:$\\$• Vapor pressure of pure benzene $(P^0_{benzene}) = 12.8 \,$kPa$\\$• Vapor pressure of pure toluene $(P^0_{toluene}) = 3.85 \,$kPa$\\$Assuming a 1:1 molar mixture of benzene and toluene:$\\$• Mole fraction of benzene $(x_{benzene}) = 0.5 \\$• Mole fraction of toluene $(x_{toluene}) = 0.5 \\$According to Raoult's Law, the partial vapor pressures are:$\\ P_{benzene} = x_{benzene} \cdot P^0_{benzene} = 0.5 \cdot 12.8 = 6.4 \,$kPa$\\ P_{toluene} = x_{toluene} \cdot P^0_{toluene} = 0.5 \cdot 3.85 = 1.925 \,$kPa$\\$The total vapor pressure of the mixture is:$\\ P_{total} = P_{benzene} + P_{toluene} = 6.4 + 1.925 = 8.325 \,$kPa$\\$The mole fraction of benzene in the vapor phase $(y_{benzene})$ is given by:$\\ y_{benzene} = \frac{P_{benzene}}{P_{total}} = \frac{6.4}{8.325} \approx 0.768 \\$The mole fraction of toluene in the vapor phase $(y_{toluene})$ is given by:$\\ y_{toluene} = \frac{P_{toluene}}{P_{total}} = \frac{1.925}{8.325} \approx 0.231 \\$Since $y_{benzene} > y_{toluene},$ the vapor will contain a higher percentage of benzene.$\\$Therefore, the correct statement is:$\\$Option 1: The vapour will contain a higher percentage of benzene.