The vapor pressure of pure benzene is 750.0 torr and the vapor pressure of toluene is 300.0 torr at a certain temperature. You make a solution by pouring “some” benzene with “some” toluene. You then place this solution in a closed container and wait for the vapor to come into equilibrium with the solution. Next, you condense the vapor. You put this liquid (the condensed vapor) in a closed container and wait for the vapor to come into equilibrium with the solution. You then condense this vapor and find the mole fraction of benzene in this vapor to be 0.714. Determine the mole fraction of benzene in the original solution assuming the solution behaves ideally.

Raoult's Law can be expressed as: \(P_i = x_i P_i^*\) Where: - \(P_i\) is the partial pressure of component i in the vapor phase - \(x_i\) is the mole fraction of component i in the liquid phase - \(P_i^*\) is the vapor pressure of pure component i In this exercise, we denote benzene as component A and toluene as component B. Raoult's Law can then be applied to both components: For Benzene (component A): \(P_A = x_A P_A^*\) For Toluene (component B): \(P_B = x_B P_B^*\)

The total pressure of the vapor phase can be calculated by adding the partial pressures of its components, which are benzene and toluene: \(P_{total} = P_A + P_B\)

We are given the mole fraction of benzene in the vapor phase after condensing and waiting for equilibrium. We can calculate the mole fraction of toluene using the relationship: \(y_A + y_B = 1\) Where: - \(y_A\) is the mole fraction of benzene in the vapor phase - \(y_B\) is the mole fraction of toluene in the vapor phase We are given \(y_A = 0.714\). To calculate the mole fraction of toluene in the vapor phase: \(y_B = 1 - y_A = 1 - 0.714 = 0.286\)

We can use the mole fractions and Raoult's Law to calculate the total vapor pressure. For benzene: \(P_{A} = y_A P_{total}\) For toluene: \(P_{B} = y_B P_{total}\)

We can use the values for the vapor pressures of pure benzene and toluene and the given mole fractions of the vapor phase to solve for the mole fraction of benzene in the original solution: For benzene: \(0.714 P_{total} = x_A (750.0\, torr)\) For toluene: \(0.286 P_{total} = x_B (300.0\, torr)\) The total pressure can be expressed as the sum of \(0.714 P_{total}\) and \(0.286 P_{total}\): \(P_{total} = 0.714 P_{total} + 0.286 P_{total}\) Using the equations for benzene and toluene, we can solve the total pressure: \begin{align*} P_{total} &= x_A (750.0\, torr) + x_B (300.0\, torr) \\ P_{total} &= (x_A) (750.0\, torr) + (1-x_A) (300.0\, torr) \end{align*} Now we can substitute the known mole fraction of benzene in the vapor phase as: \begin{align*} 0.714 P_{total} &= x_A (750.0\, torr) \\ 0.714 (x_A (750.0\, torr) + (1-x_A) (300.0\, torr)) &= x_A (750.0\, torr) \end{align*} Now, we can solve for the mole fraction of benzene in the original solution: \(x_A = 0.464\) So the mole fraction of benzene in the original solution is 0.464.