153.

Heptane and Octane form an ideal solution at 373 K. The vapour pressures of the pure liquids at this temperature are 105.2 K Pa and 46.8 K Pa respectively. If the solution contains 25 g of heptane and 28.5 g of octane, calculate
(i) Vapour pressure exerted by heptane.
(ii) Vapour pressure exerted by solution.
(iii) Mole fraction of octane in the vapour phase.

Components A (Heptane, ${\mathrm{C}}_7{\mathrm{H}}_{16}$)
$$\begin{gathered} {{\mathrm{p}}^{\mathrm{o}}}_{\mathrm{A}} = 105.2 \mathrm{kpa} \\ {\mathrm{W}}_{\mathrm{A}} = 25 \mathrm{g} \\ {\mathrm{M}}_{\mathrm{A}} = (12\times 7) + (16\times 1) = 100 \end{gathered}$$
Components of B $(\mathrm{octane}, {\mathrm{C}}_8{\mathrm{H}}_{18})$
           $$\begin{gathered} {{\mathrm{p}}^{\mathrm{o}}}_{\mathrm{B}} = 46.8 \mathrm{kpa} \\ {\mathrm{W}}_{\mathrm{A}} = 28.5 \mathrm{g} \\ {\mathrm{M}}_{\mathrm{A}} = (12\times 8) + (18\times 1) = 114 \end{gathered}$$
No. of moles of heptane,
              ${\mathrm{n}}_{\mathrm{A}} = \frac{25}{100}=0.25$
No. of moles of octane,
                ${\mathrm{n}}_{\mathrm{B}} =\frac{{\mathrm{W}}_{\mathrm{B}}}{{\mathrm{M}}_{\mathrm{B}}}=\frac{28.5}{114}=0.25$
Total moles in solution,
               ${\mathrm{n}}_{\mathrm{A}}+{\mathrm{n}}_{\mathrm{B}} = 0.25+0.25 = 0.50$
(i) Vapour pressure exerted by heptane
                 ${\mathrm{p}}_{\mathrm{A}} = {{\mathrm{p}}^{\mathrm{o}}}_{\mathrm{A}\times \mathrm{A}}$
where ${\mathrm{X}}_{\mathrm{A}}$ is mole fraction of component A
                   $$\begin{gathered} {\mathrm{p}}_{\mathrm{A}} = 105.2\times \frac{{\mathrm{n}}_{\mathrm{A}}}{{\mathrm{n}}_{\mathrm{A}}+{\mathrm{n}}_{\mathrm{B}}} = \frac{105.2\times 0.25}{0.50} \\ = 52.6 \mathrm{k} \mathrm{pa} \end{gathered}$$
(ii) Vapour pressure of octane
              $$\begin{gathered} {\mathrm{p}}_{\mathrm{B}} = {{\mathrm{p}}^{\mathrm{o}}}_{\mathrm{B}}\times \mathrm{B} \\ {\mathrm{p}}_{\mathrm{B}} = 46.8 \times \frac{{\mathrm{n}}_{\mathrm{B}}}{{\mathrm{n}}_{\mathrm{A}}+{\mathrm{n}}_{\mathrm{B}}} = \frac{46.8\times 0.25}{0.50} \\ = 23.4 \mathrm{kpa} \end{gathered}$$
   Total vapour pressure of solution
                  $\mathrm{p} ={\mathrm{p}}_{\mathrm{A}}+{\mathrm{p}}_{\mathrm{B}} = 52.6+23.4 = 76 \mathrm{kpa}$
(iii) Mole fraction of octane
                              $= \frac{{\mathrm{n}}_{\mathrm{B}}}{{\mathrm{n}}_{\mathrm{A}}+{\mathrm{n}}_{\mathrm{B}}}=\frac{0.25}{0.50}=0.5.$