Question

At ${\mathbf{20}}^{\mathbf{0}}\mathit{C}$, the vapor pressure of toluene is 0.0289atmand the vapor pressure of benzene is n .Equal numbers of moles of toluene and benzene are mixed and form an ideal solution. Compute the mole fraction of benzene in the vapor in equilibrium with this solution.

Short answer

The mole fraction of benzene in the vapor in equilibrium with a solution is $0.773$.

Step-by-step solution

Given data

The vapor pressure of toluene is $0.0289\text{atm}$ and the vapor pressure of benzene is n.

Concept of Raoult's law

According to Raoult's law, the partial vapor pressure of a solvent in a solution (or mixture) is equal to or the same as the vapor pressure of the pure solvent times the mole fraction of the solvent present in the solution.

The formula of Raoult's law is as follows:

${\mathit{p}}_{\mathbf{i}}\mathbf{=}{\mathit{X}}_{\mathbf{i}}\mathbf{\times }{\mathit{p}}_{\mathbf{i}}^{\mathbf{*}}$

Calculate the total pressure of the solution

The total pressure is calculated by Dalton’s law, $P_{\text{total}}=P_{C_8{H}_8}+P_{{\text{C}}_8{\text{H}}_5{\text{CH}}_3}$

$$\begin{gathered} P_{\text{total}}=(0.5)(0.0987\text{atm})+(0.5)(0.0289\text{atm}) \\ {P}_{\text{total}}=0.0638\text{atm} \end{gathered}$$

Calculate the mole fraction of benzene

The mole fraction of benzene is obtained as

$$\begin{gathered} y{\text{C}}_8{\text{H}}_8=\frac{P_{{\text{Cs}}_8{\text{H}}_8}}{P_{\text{totsl}}} \\ =\frac{X_{{\text{C}}_8{\text{H}}_8}{P}_{{\text{C}}_8{\text{H}}_8}^{\ne }}{P_{\text{total}}} \\ {y}_8{\text{H}}_8=\frac{0.5\times 0.0987\text{atm}}{0.0638\text{atm}} \\ =0.773 \\ \backslash \mathrm{endgathered}\backslash \mathrm{begingathered}y{\text{C}}_8{\text{H}}_8=\frac{P_{{\text{Cs}}_8{\text{H}}_8}}{P_{\text{totsl}}} \\ =\frac{X_{{\text{C}}_8{\text{H}}_8}{P}_{{\text{C}}_8{\text{H}}_8}^{\ne }}{P_{\text{total}}} \\ {y}_8{\text{H}}_8=\frac{0.5\times 0.0987\text{atm}}{0.0638\text{atm}} \\ =0.773 \end{gathered}$$