Vapour pressure of C6H6 and C7H8 mixture at 50 degree C is given P(mm Hg)= 180Xb+90, where Xb is the mole fraction of C6H6. A solution is prepared by mixing 936 gm benzene and 736 gm toluene and if vapours over this solution are removed and condensed into liquid and again brought to temperature 50 degree C, what would be mole fraction of C6H6 in the vapour state. (At. wt. of C=12, H=1)

Given: mass of C₆H₆ (w_b)= 936 g
​​            mass of C₇H₈ (w_a) = 736 g
​            pressure of mixture, p = 180 X_b + 90 mmHg
To calculate: mole fraction of benzene in vapour state
Solve: Mole fraction of benzene in liquid state (Xb) is given by
${\mathrm{X}}_{\mathrm{b}}=\frac{{\mathrm{n}}_{\mathrm{b}}}{{\mathrm{n}}_{\mathrm{b}} +\hspace{0.17em}{\mathrm{n}}_{\mathrm{a}}}$
where n_a and n_b are number of moles of toluene and benzene respectively.
Number of moles is given by
$n= \frac{w}{M}$
where w is the given mass and M is molar mass of the substance.

Molar mass of Benzene (M_b) = 78 g mol^-1
Molar mass of Toluene (M_a) = 92 g mol^-1

$$\begin{gathered} {\mathrm{n}}_{\mathrm{b}} = \frac{{\mathrm{w}}_{\mathrm{b}}}{{\mathrm{M}}_{\mathrm{b}}} \\ {\mathrm{n}}_{\mathrm{b}} = \frac{936 \mathrm{g}}{78 \mathrm{g} {\mathrm{mol}}^{-1}} \\ {\mathrm{n}}_{\mathrm{b}} = 12 \mathrm{mol} \end{gathered}$$

 $$\begin{gathered} {\mathrm{n}}_{\mathrm{a}} = \frac{{\mathrm{w}}_{\mathrm{a}}}{{\mathrm{M}}_{\mathrm{a}}} \\ {\mathrm{n}}_{\mathrm{a}} = \frac{736 \mathrm{g} }{92 \mathrm{g} {\mathrm{mol}}^{-1}} \\ {\mathrm{n}}_{\mathrm{a}} = 8 \mathrm{mol} \end{gathered}$$ 
$$\begin{gathered} {\mathrm{X}}_{\mathrm{b}} = \frac{{\mathrm{n}}_{\mathrm{b}}}{{\mathrm{n}}_{\mathrm{b}}+ {\mathrm{n}}_{\mathrm{a}}} \\ {\mathrm{X}}_{\mathrm{b}} = \frac{12}{8+12} \\ {\mathrm{X}}_{\mathrm{b}} = 0.6 \end{gathered}$$

and X_a = 1 - X_b
​              = 1 - 0.6
            =0.4
Thus substituting the value of X_b in the given equation of pressure

p = 180 X_b + 90 mmHg

p = 180 $\times$ 0.6 +90 mmHg

p = 198 ​mmHg

Comparing the given equation of pressure
p = (180 X_b + 90 ) mmHg 
with the standard equation of 
$$\begin{gathered} p= p_a^0+ (p_b^0-p_a^0)X_b \\ \mathrm{we} \mathrm{get} {\mathrm{p}}_{\mathrm{a}}^0 = 90 \mathrm{mmHg} \\ \mathrm{and} ({\mathrm{p}}_{\mathrm{b}}^0-{\mathrm{p}}_{\mathrm{a}}^0)=180 \mathrm{mmHg} \\ \mathrm{Thus} {\mathrm{p}}_{\mathrm{b}}^0 = 90 \mathrm{mmHg} \end{gathered}$$


mole fraction of benzene in the vapour state is given by
 $$\begin{gathered} {\mathrm{y}}_{\mathrm{b}}=\frac{{\mathrm{p}}_{\mathrm{b}}}{\mathrm{p}} \\ \mathrm{and} {\mathrm{p}}_{\mathrm{b}} ={\mathrm{p}}_{\mathrm{b}}^0 {\mathrm{x}}_{\mathrm{b}} \\ \mathrm{where} {\mathrm{p}}_{\mathrm{b}}^0 \mathrm{is} \mathrm{the} \mathrm{partial} \mathrm{pressure} \mathrm{of} \mathrm{pure} \mathrm{benzene} \mathrm{at} {50}^0 \mathrm{C}. \end{gathered}$$
p_b= 90 $\times$0.6
      = 54 mmHg
Thus
$$\begin{gathered} y_b= \frac{54}{198} \\ = \mathbf{0}\mathbf{.}\mathbf{27} \end{gathered}$$
Therefore, the mole fraction of benzene in vapour state is 0.27.