On mixing, heptane and octane form an ideal solution. At $373 \mathrm{~K}$, the vapour pressures of the two…
On mixing, heptane and octane form an ideal solution. At $373 \mathrm{~K}$, the vapour pressures of the two liquid components (heptane and octane) are $105 \mathrm{kPa}$ and $45 \mathrm{kPa}$ respectively. Vapour pressure of the solution obtained by mixing $25.0 \mathrm{~g}$ of heptane and $35 \mathrm{~g}$ of octane will be (molar mass of heptane $=100 \mathrm{~g} \mathrm{~mol}^{-1}$ an dof octane $=114 \mathrm{~g} \mathrm{~mol}^{-1}$ ).
$72.0 \mathrm{kPa}$
$36.1 \mathrm{kPa}$
$96.2 \mathrm{kPa}$
$144.5 \mathrm{kPa}$
Solution
Mole fraction of Heptane $=\frac{25 / 100}{\frac{25}{100}+\frac{35}{114}}=\frac{0.25}{0.557}=0.45$
$\mathrm{X}_{\text {Hep tane }}=0.45$.
$\therefore$ Mole fraction of octane $=0.55=\mathrm{X}_{\text {octane }}$
Total pressure $=\sum X_i P_i^0$
$=(105 \times 0.45)+(45 \times 0.55) \mathrm{kP}_{\mathrm{a}}$
$=72.0 \mathrm{KPa}$