Question

Question: Refer to the data of Problem 62. Calculate the mole fraction of toluene in a mixture of benzene and toluene that boils at 90° C under atmospheric pressure.

Short answer

The mole fraction of toluene in the vapor is 0.473.

Step-by-step solution

Concept of Mole fraction

The mole fraction or molar fraction $\left(x_i\right)$is defined as unit of theamountof a constituent (expressed inmoles),${\mathit{n}}_{\mathbf{i}}$ divided by the total amount of all constituents in a mixture (also expressed in moles),${\mathit{n}}_{\mathbf{\text{tot}}}$ .This expression is given below:

${\mathit{x}}_{\mathbf{i}}\mathbf{=}\frac{{\mathbf{n}}_{\mathbf{i}}}{{\mathbf{n}}_{\mathbf{\text{tot}}}}$

Calculation of the mole fraction of each component

The mole fraction of benzene can be calculated as:

$$\begin{gathered} X_{\text{benzene}}=\frac{n_{\text{benzene}}}{n_{\text{benzene}}+n_{\text{toluene}}} \\ =\frac{0.400}{0.400+0.900} \\ =0.308 \end{gathered}$$

The mole fraction of toluene can be calculated as:

$$\begin{gathered} X_{\text{toluene}}=\frac{n_{\text{toluene}}}{n_{\text{benzene}}+n_{\text{toluene}}} \\ =\frac{0.900}{0.400+0.900} \\ =0.692 \end{gathered}$$

Partial pressure of each gas

The pressure of the benzene molecules is calculated as:

$$\begin{gathered} P_{\text{benzene}}=X_{\text{benzene}}{P}_{\text{benzene}}^0 \\ =0.308\times 1.34\text{atm} \\ =0.413\text{atm} \end{gathered}$$

Pressure of the toluene molecules is calculated as:

$$\begin{gathered} P_{\text{toluene}}=X_{\text{toluene}}{P}_{\text{toluene}}^0 \\ =0.692\times 0.534\text{atm} \\ =0.370\text{atm} \end{gathered}$$

Therefore, mole fraction of toluene in vapor is calculated as shown below:

$$\begin{gathered} X_{\text{toluene}}=\frac{P_{\text{toluene}}}{P_{\text{benzene}}+P_{\text{toluene}}} \\ =\frac{0.370\text{atm}}{0.413\text{atm}+0.370\text{atm}} \\ =0.473 \end{gathered}$$

Thus, the mole fraction of toluene in the vapor is .