Question

At \(20^{\circ} \mathrm{C}\) the vapor pressure of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) is 75 torr, and that of toluene \(\left(\mathrm{C}_{7} \mathrm{H}_{8}\right)\) is 22 torr. Assume that benzene and toluene form an ideal solution. (a) What is the composition in mole fractions of a solution that has a vapor pressure of 35 torr 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

The composition of an ideal solution with a vapor pressure of 35 torr at 20°C has mole fractions of benzene (C₆H₆) and toluene (C₇H₈) in the solution as 0.245 and 0.755, respectively. The mole fraction of benzene in the vapor above the solution is 0.721.

Step-by-step solution

(a) Calculate the mole fractions of benzene and toluene in the solution.

According to Raoult's law, the partial vapor pressure of a component in an ideal solution is equal to the product of its mole fraction in the solution and its vapor pressure. We are given the vapor pressures of benzene and toluene at 20°C and the total vapor pressure of the solution, and asked to find the mole fractions of benzene and toluene in the solution. Let x₁ and x₂ represent the mole fractions of benzene and toluene in the solution, respectively. The total pressure above the solution is given by: \(P = x_1P_1^0 + x_2P_2^0\) where \(P_1^0\) and \(P_2^0\) are the vapor pressures of benzene and toluene, respectively. We also know that: \(x_1 + x_2 = 1\) Now, plugging in the given values: \(35\,\text{torr} = x_1(75\,\text{torr}) + x_2(22\,\text{torr})\) We can solve for x₁ and x₂ using these two equations.

(a) Solving for x₁ and x₂ using the equations.

We can rearrange the equation for mole fractions: \(x_2 = 1 - x_1\) Now substitute this expression of x₂ into the equation for total pressure: \(35\,\text{torr} = x_1(75\,\text{torr}) + (1 - x_1)(22\,\text{torr})\) Expanding and simplifying this equation: \(35\,\text{torr} = 75x_1\,\text{torr} + 22\,\text{torr} - 22x_1\,\text{torr}\) \(13\,\text{torr} = 53x_1\,\text{torr}\) So, \(x_1= \frac{13\,\text{torr}}{53\,\text{torr}} = 0.245\) Now, we can find x₂: \(x_2 = 1-0.245=0.755\) Therefore, the mole fractions of benzene and toluene in the solution are 0.245 and 0.755, respectively.

(b) Calculate the mole fraction of benzene in the vapor.

To calculate the mole fraction of benzene in the vapor, we can use the following relation between the mole fractions in the solution and the vapor: \(\frac{y_1}{y_2} = \frac{x_1P_1^0}{x_2P_2^0}\) Plugging in the given values and the values of x₁ and x₂ that we just calculated: \(\frac{y_1}{y_2} = \frac{(0.245)(75\,\text{torr})}{(0.755)(22\,\text{torr})}\) \( \frac{y_1}{y_2} = 2.587 \) Now, we can find the mole fraction of benzene and toluene in the vapor using: \(y_1 + y_2 = 1\) \(y_1 = 1 - y_2\) Substitute this expression of y₁ into the equation for the ratio of mole fractions: \(\frac{1 - y_2}{y_2} = 2.587\) Solve for y₂: \(y_2 = \frac{1}{3.587} = 0.279\) Now, find y₁: \(y_1 = 1 - 0.279 = 0.721\) Therefore, the mole fraction of benzene in the vapor above the solution is 0.721.