Question

The vapour pressures of benzene and toluene at 20°C are 75 mm and 22 mm Hg respectively. 23.4 g of benzene and 64.4 g of toluene are mixed. If the two form an ideal solution, calculate the mole fraction of benzene in vapour phase when vapours are in equilibrium with the liquid phase.

Answer 1

Given: Vapour pressure of benzene `P_"benzene"^circ` = 75 mm Hg

Vapour pressure of toluene `P_"toluene"^circ` = 22 mm Hg

Mass of benzene = 23.4 g

Mass of toluene = 64.4 g

The solution is ideal, and vapours are in equilibrium with liquid.

Temperature = 20°C

Molar mass of benzene (C₆H₆) = 78 g/mol

Molar mass of toluene (C₇H₈) = 92 g/mol

Moles of benzene = `23.4/78` = 0.3 mol

Moles of toluene = `64.4/92` = 0.7 mol

Total moles = 0.3 + 0.7 = 1.0 mol

`chi_"benzene" = 0.3/1.0` = 0.3

`P_"benzene" = chi_"benzene" xx P_"benzene"^circ`

= 0.3 × 75

= 22.5 mm Hg

`P_"toluene" = chi_"toluene" xx P_"toluene"^circ`

= 0.7 × 22

= 15.4 mm Hg

P_total = 22.5 + 15.4 = 37.9 mm Hg

Using Dalton’s law,

`Y_"benzene" = (P_"benzene")/(P_"total")`

= `22.5/37.9`

= 0.5937

∴ The mole fraction of benzene in the vapour phase is approximately 0.5937.