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Chapter 11: Problem 136

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?

Short Answer

Expert verified
The mole fraction of liquid A in the ideal solution is 0.222.

Step by step solution

01

Write the Raoult's law equations for both substances

The Raoult's law states that the partial pressure of a component in a solution is the product of its mole fraction (in the solution) and its vapor pressure (as a pure substance). The equations for both substances A and B can be written as: \(P_A = x_A P_A^*\) \(P_B = x_B P_B^*\) Where: - \(P_A\) and \(P_B\) are the partial pressures of A and B in the vapor phase, respectively - \(x_A\) and \(x_B\) are the mole fractions of A and B in the liquid phase, respectively - \(P_A^*\) and \(P_B^*\) are the vapor pressures of pure A and pure B, respectively
02

Write the given information

We are given that the vapor at equilibrium above the solution has double the mole fraction of substance A that the solution does. We can write this relationship as: \(y_A = 2x_A\) Where \(y_A\) is the mole fraction of A in the vapor phase.
03

Relate mole fractions of A and B in liquid and vapor phases

Since there are only two components, A and B, in the liquid and vapor phases, we can write the relationships: \(x_B = 1 - x_A\) \(y_B = 1 - y_A\) These equations will help us to eliminate variables later.
04

Calculate partial pressures using given vapor pressures

We are given the vapor pressures for both pure substances A and B (350.0 torr and 100.0 torr, respectively). Therefore, we can rewrite the Raoult's law equations from Step 1 and substitute the given values: \(P_A = x_A * 350\) \(P_B = x_B * 100\)
05

Calculate total pressure and relate partial pressures

The total pressure above the solution is the sum of the partial pressures of A and B: \(P_{total} = P_A + P_B\) From the mole fraction definition and the ideal gas law, we have: \((P_{total}y_A) = P_A \) \((P_{total}y_B) = P_B \)
06

Substitute mole fractions and eliminate variables

Next, we'll insert the expressions for the mole fractions from Step 3, and the mole fraction relationship for A from Step 2 into the equations from Step 5: \((P_{total} * 2x_A) = x_A * 350\) \((P_{total} * (1 - 2x_A)) = (1 - x_A) * 100\) Now we have two equations and two unknowns (\(P_{total}\) and \(x_A\)): \(2P_{total}x_A = 350x_A\) \(P_{total} - 2P_{total}x_A = 100 - 100x_A\)
07

Solve the equations to find the mole fraction of A in the liquid phase

We will solve the two equations from Step 6 to find the mole fraction of A (\(x_A\)) in the liquid phase: 1. Isolate \(P_{total}\) in the first equation: \(P_{total} = \frac{350x_A}{2x_A} = \frac{175}{x_A}\) 2 . Substitute this expression for \(P_{total}\) into the second equation: \(\frac{175}{x_A} - 2\left(\frac{175}{x_A}\right)x_A = 100 - 100x_A\) 3 . Solve for \(x_A\): \(x_A = 0.222\)
08

Present the final answer

The mole fraction of liquid A in the ideal solution is 0.222.

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