You have a solution of two volatile liquids, \(A\) and \(B\) (assume ideal behavior). Pure liquid A has a vapor pressure of 350.0 torr and pure liquid B has a vapor pressure of 100.0 torr at the temperature of the solution. The vapor at equilibrium above the solution has double the mole fraction of substance A that the solution does. What is the mole fraction of liquid A in the solution?

Raoult's law for liquid A can be written as: \(P_A = x_A \times P_A^\star\) And for liquid B, it can be written as: \(P_B = x_B \times P_B^\star\) Here, \(P_A\) and \(P_B\) are the partial pressures of liquids A and B in the vapor phase, whereas \(P_A^\star\) and \(P_B^\star\) are their vapor pressures when they are pure liquids. In our case, \(P_A^\star = 350.0\,\text{torr}\) and \(P_B^\star = 100.0\,\text{torr}\).

It is given that the mole fraction of A in the vapor phase is double that in the solution. Thus, we have: \(y_A = 2x_A\)

Since the sum of mole fractions in the vapor phase must be equal to 1, we can write: \(y_B = 1 - y_A\)

Using the expressions for mole fractions in step 2 and step 3, we can write the expressions for partial pressures of A and B as: \(P_A = (2x_A) P_{total}\) \(P_B = (1 - 2x_A) P_{total}\)

Applying Raoult's law, we obtain the following equations: \(P_A = x_A \times 350.0\) \(P_B = x_B \times 100.0\) Since \(x_A\) + \(x_B\) = \(1\), we can write the Raoult's law equation for B as: \(P_B = (1 - x_A) \times 100.0\)

Substitute the equations from steps 4 and 5 into each other to find the mole fraction of A in the solution: \(x_A \times 350.0 = (2x_A) P_{total}\) \((1 - x_A) \times 100.0 = (1 - 2x_A) P_{total}\) Divide the first equation by 350 and the second equation by -100: \(x_A = \frac{2}{350} P_{total}\) \(x_A - 1 = \frac{1}{100} P_{total}\) Now we can combine the two equations: \(\frac{2}{350} P_{total} = 1 + \frac{1}{100} P_{total}\) Solve for \(P_{total}\): \(P_{total} = \frac{700}{3}\) Substitute the value of \(P_{total}\) back into the first equation: \(x_A = \frac{2}{350} \cdot \frac{700}{3}\) \(x_A = \frac{1}{3}\) So the mole fraction of liquid A in the solution is \(\frac{1}{3}\), or approximately 0.333.