Question

At a given temperature, you have a mixture of benzene (vapor pressure of pure benzene = 745 torr) and toluene (vapor pressure of pure toluene = 290 torr). The mole fraction of benzene in the vapor above the solution is 0.590. Assuming ideal behavior, calculate the mole fraction of toluene in the solution.

Answer 1

Explanation:

The given data is as follows.

     Vapor pressure of pure benzene (y) = 745 torr

     Vapor pressure of pure toluene (z) = 290 torr

      Mole fraction of benzene = 0.590

      Mole fraction of toulene = ?

Let $X_{a}$ and $X_{b}$ be mole fractions of benzene and toulene in its solution.

And, let $Y_{a}$ and $Y_{b}$ be mole fractions of benzene and toulene in its vapour.

$p_{a}_{o}$ and $p_{b}_{o}$ are vapour pressures of pure benzene and pure toulene.

$p_{a}$ and $p_{b}$ are partial pressures of vapors of benzene and toulene.

and, P is total pressure.  

Then, $Y_{a} = \frac{p_{a}}{P}$  

                  = $\frac{X_{a} p_{a}_{o}}{P}$ ............... (1)

            $Y_{b} = \frac{p_{b}}{P}$

                      = $\frac{X_{b} p_{b}_{o}}{P}$ ............... (2)

Now, divide equation (1) by (2) as follows.

      $\frac{Y_{a}}{Y_{b}}$ = $\frac{X_{a} p_{a}_{o}}{X_{b} p_{b}_{o}}$

It is given that, $Y_{a}$ = 0.590, then $Y_{b}$ = 1 - 0.590 = 0.410.

And, $p_{a}_{o}$ = 745 torr , $p_{b}_{o}$ = 290 torr

and, $X_{a} + X_{b}$ = 1

Therefore, $\frac{0.590}{0.41}$ = $\frac{(1 - X_{b}) \times 745}{X_{b} \times 290}$

               1.44 = $\frac{745 - 745 X_{b}}{X_{b} \times 290}$

                $417.6 X_{b} = 745 - 745X_{b}$

                         $X_{b} = \frac{745}{1162.6}$

                                    = 0.641

Hence, we can conclude that the mole fraction of toluene in the solution is 0.641.