Benzene and toluene form an ideal solution. Consider a solution of benzene and toluene prepared at \(25^{\circ} \mathrm{C}\). Assuming the mole fractions of benzene and toluene in the vapor phase are equal, calculate the composition of the solution. At \(25^{\circ} \mathrm{C}\) the vapor pressures of benzene and toluene are 95 and 28 torr, respectively.

Short Answer

Expert verified
The composition of the solution is 50% benzene and 50% toluene (by mole fraction).

Step by step solution

01

Recall Raoult's law

Raoult's law states that the partial pressure of a component in a solution is equal to its mole fraction in the solution multiplied by its vapor pressure. Mathematically, it is given by: \(P_{A} = x_{A} \cdot P^{o}_{A}\) and \(P_{B} = x_{B} \cdot P^{o}_{B}\) where \(P_{A}\) and \(P_{B}\) are the partial pressures of component A (benzene) and B (toluene), respectively. \(P^{o}_{A}\) and \(P^{o}_{B}\) are their respective vapor pressures, and \(x_{A}\) and \(x_{B}\) are their mole fractions in the solution.
02

Write the given information in terms of Raoult's law

We have the information that the mole fractions of benzene and toluene in the vapor phase are equal: \(\frac{P_{A}}{P^{o}_{A}} = \frac{P_{B}}{P^{o}_{B}}\) The vapor pressures of benzene and toluene at 25°C are given as 95 torr and 28 torr, respectively. So, \(\frac{P_{A}}{95} = \frac{P_{B}}{28}\)
03

Express mole fraction relationship for benzene and toluene

Since there are only two components in the solution, their mole fractions must add up to 1: \(x_{A} + x_{B} = 1\) Hence, we can write the mole fraction of toluene as: \(x_{B} = 1 - x_{A}\)
04

Substitute mole fraction of toluene and solve for partial pressures

Replace \(x_{B}\) in Raoult's law for toluene: \(P_{B} = (1 - x_{A}) \cdot 28\) Substituting this expression for \(P_{B}\) in the equality of mole fractions from Step 2: \(\frac{P_{A}}{95} = \frac{(1 - x_{A}) \cdot 28}{28}\) We can now solve for \(P_{A}\): \(P_{A} = 95 (1 - x_{A})\)
05

Substitute the expression for partial pressure of benzene

Replace \(P_{A}\) in Raoult's law for benzene: \(95 (1 - x_{A}) = x_{A} \cdot 95\)
06

Solve for the mole fraction of benzene

We have: \(95 - 95x_{A} = 95x_{A}\) Adding \(95x_{A}\) to both sides: \(95 = 190x_{A}\) Dividing both sides by 190: \(x_{A} = \frac{95}{190}\) Calculating the mole fraction of benzene: \(x_{A} = 0.5\)
07

Determine the composition of the solution

From Step 3, we can find the mole fraction of toluene: \(x_{B} = 1 - 0.5 = 0.5\) Our composition of the solution is 50% benzene and 50% toluene (by mole fraction).

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