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.

According to Raoult's law, the partial vapor pressure of a component in an ideal mixture is equal to the product of its mole fraction in the solution and its vapor pressure as a pure substance. Let's denote the mole fraction of benzene in the original solution as \(x_{B}\) and the mole fraction of toluene as \(x_{T}\), then we can write the partial vapor pressures of benzene and toluene as: \(P_{B} = x_{B}P_{B}^{o}\) and \(P_{T} = x_{T}P_{T}^{o}\) Where \(P_{B}^{o}\) and \(P_{T}^{o}\) are the vapor pressures of pure benzene and toluene, respectively, given as 750.0 torr and 300.0 torr.

The total vapor pressure in the system is the sum of the partial vapor pressures of benzene and toluene: \(P_{total} = P_{B} + P_{T}\) We know that the mole fraction of benzene in the condensed vapor is 0.714, which means the mole fraction of toluene is 0.286. From this information, we can calculate the total vapor pressure of the system. To do this, let's assume that the mole fraction of benzene in the original solution remains constant during the condensation process, so the ratio of partial pressures remains the same: \(\frac{P_{B}}{P_{total}} = 0.714\) Then \(P_{B} = 0.714P_{total}\).

Since the original solution is composed of only benzene and toluene, their mole fractions in the solution sum to 1: \(x_{B} + x_{T} = 1\) Thus, we can write the mole fraction of toluene in terms of the mole fraction of benzene: \(x_{T} = 1 - x_{B}\)

We can now substitute the expressions from steps 2 and 3 back into the Raoult's law equations from step 1: \(P_{B} = x_{B}P_{B}^{o}\) becomes \(0.714P_{total} = x_{B}(750.0)\) \(P_{T} = x_{T}P_{T}^{o}\) becomes \(P_{total} - 0.714P_{total} = (1 - x_{B})(300.0)\) Now we can solve the equations simultaneously for the mole fraction of benzene in the original solution: Divide the first equation by 750 to isolate \(x_{B}\): \(x_{B} = \frac{0.714P_{total}}{750.0}\) Substitute the expression for \(x_{B}\) into the second equation: \(P_{total} - 0.714P_{total} = (1 - \frac{0.714P_{total}}{750.0})(300.0)\) Solve for \(P_{total}\): \(P_{total} = 428.57\, torr\) Now substitute the value of \(P_{total}\) back into the expression for \(x_{B}\): \(x_{B} = \frac{0.714(428.57)}{750.0}\) Which gives: \(x_{B} \approx 0.406\) Thus, the mole fraction of benzene in the original solution is approximately 0.406.