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Chapter 13: Problem 68

At \(20^{\circ} \mathrm{C}\), the vapor pressure of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) is \(10 \mathrm{kPa}\), and that of toluene \(\left(\mathrm{C}_{7} \mathrm{H}_{8}\right)\) is \(2.9 \mathrm{kPa}\). Assume that benzene and toluene form an ideal solution. (a) What is the composition in mole fraction of a solution that has a vapor pressure of $4.7 \mathrm{kPa}\( at \)20^{\circ} \mathrm{C} ?$ (b) What is the mole fraction of benzene in the vapor above the solution described in part (a)?

Short Answer

Expert verified
The composition in mole fraction of the solution is 0.2535 for benzene and 0.7465 for toluene. The mole fraction of benzene in the vapor above the solution is 0.5394.

Step by step solution

01

Raoult's Law for ideal solutions

Raoult's Law states that the partial pressure of a component in an ideal solution is equal to the mole fraction of that component multiplied by its vapor pressure: \(P_{A} = x_{A}P_{A}^{\circ}\). In our problem, component A will be benzene, and component B will be toluene. We are given the vapor pressures of both components, and the total vapor pressure of the solution.
02

Write the expression for the partial pressure of benzene and toluene

Using Raoult's Law, let's write the expressions for the partial pressures of benzene (A) and toluene (B): \(P_{A} = x_{A}P_{A}^{\circ}\) \(P_{B} = x_{B}P_{B}^{\circ}\)
03

Find the mole fractions

We know that the total vapor pressure of the solution is 4.7 kPa: \(P_{total} = P_A + P_B = 4.7\) Since \(x_B = 1 - x_A\), we can substitute the expression for \(P_B\) in terms of \(x_A\): \(4.7 = x_A \cdot 10 + (1 - x_A) \cdot 2.9\) Now, we can solve for the mole fraction of benzene, \(x_A\): \(4.7 = 10x_A + 2.9 - 2.9x_A\) \(1.8 = 7.1x_A\) \(x_A = 0.2535\) Now, we can find the mole fraction of toluene, \(x_B\): \(x_B = 1 - x_A = 1 - 0.2535 = 0.7465\) Thus, the composition in mole fraction of the solution is 0.2535 for benzene and 0.7465 for toluene.
04

Find the mole fraction of benzene in the vapor above the solution

To find the mole fraction of benzene in the vapor, we will use the definition of mole fraction: \(y_A = \frac{P_A}{P_{total}}\) We already know the partial pressure of benzene from the previous steps: \(P_A = x_A P_A^{\circ} = 0.2535 \cdot 10 = 2.535\) Now plug in the values and find the mole fraction of benzene in the vapor: \(y_A = \frac{2.535}{4.7} = 0.5394\) Hence, the mole fraction of benzene in the vapor above the solution is 0.5394.

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