137. Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure, in bar of a mixture of 26.0 g of heptane and 35.0 g of octane? 

Answer:
We have given,

Molar mass of heptane,
C₇H₁₆ = 100 g mol^–1
Molar mass of octane,
C₈H₁₈ = 114 g mol^–1
Moles of heptane

$= \frac{\mathrm{Wt}. \mathrm{of} \mathrm{heptane}}{\mathrm{Mol}. \mathrm{mass} \mathrm{of} \mathrm{heptane}}$

$=\frac{26}{100}=0.26$

Similarly, Moles of octane

$=\frac{35}{114}=0.31$

Mole fraction heptane

$=\frac{{\mathrm{n}}_{\mathrm{A}}}{{\mathrm{n}}_{\mathrm{A}}+{\mathrm{n}}_{\mathrm{B}}}$

$= \frac{0.26}{0.26+0.31}=0.456$

Mole fraction of octane

$= \frac{0.31}{0.26+0.31}=0.543$

Partial vapour pressure = Mole fraction x Vap.
Pressure of pure component.
Partial vapour pressure of heptane

= 0.456 x 105.2 = 47.97 kPa

Partial vapour pressure of octane
= 0.543 x 46.8 = 25.4 kPa

Total vapour pressure of solution = 73.08 kPa.