Question

At ${\mathbf{90}}^{\mathbf{0}}\mathit{C}$ , the vapor pressure of toluene is 0.0534 and the vapor pressure of benzene is 1.34 atm . Benzene (0.400 mol)is mixed with toluene(0.900 mol)to 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 the solution is 0.773 .

Step-by-step solution

Given data

$P_i^{\ne }$(benzene)$=1.34\text{atm}$

$P_i^{\ne }($toluene$)=0.534\text{atm}$

${\mathbf{n}}_{\text{benzene}}=0.400\text{mol}$

${\mathbf{n}}_{\text{toluene}}=0.900\text{mol}$

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 follow

Calculate the mole fraction

The mole fraction is obtained as, ${\mathbf{x}}_{\text{benzene}}=\frac{\text{moles of benzene}}{\text{total moles}}$

$P_i=x_i\times P_i^{*}$

$$\begin{gathered} {\mathbf{x}}_{\text{benzene}}=\frac{0.400}{(0.400+0.900)} \\ {\mathbf{x}}_{\text{benzene}}=0.308 \end{gathered}$$

Now, mole fraction of toluene is obtained by subtraction of mole fraction of benzene from total mole fraction.

$$\begin{gathered} {\text{x}}_{\text{toluene}}=1-molefractionofbenzene \\ =1-0.308 \\ =0.692 \end{gathered}$$

Evaluate vapour pressure of benzene and toluene

The vapour pressure of benzene is obtained by Raoult’s law.

$$\begin{gathered} P_{\text{benzene}}=x_{benzene}\times P_{\text{benzene}}^{\backslash \mathrm{ddag}} \\ {P}_{\text{benzene}}=0.308\times 1.34\text{atm} \\ =0.413\text{atm} \end{gathered}$$

Similarly, the vapour pressure of toluene is obtained by Raoult’s law.

$$\begin{gathered} P_{\text{toluene}}=x_{\text{toluene}}\times P_{\text{toluene}}^{*} \\ {P}_{\text{toluene}}=0.692\times 0.534\text{atm} \\ =0.370\text{atm} \end{gathered}$$

Calculate mole fraction in vapour

Mole fraction of benzene in vapor is given by

$Molefractionofbenzeneinvapor=\frac{\text{vapor pressure of benzene}}{\text{total pressure}}$

$$\begin{gathered} =\frac{0.413\text{atm}}{(0.413+0.370)\text{atm}} \\ =0.527 \end{gathered}$$