Question

A solution of benzene (C6H6) and toluene (C7H8) is 25.0 % benzene by mass. The vapor pressures of pure benzene and pure toluene at 25∘C are 94.2 and 28.4 torr, respectively.

1.)Assuming ideal behavior, calculate the vapor pressure of benzene in the mixture.

Express the pressure to three significant figures and include the appropriate units.

2.)Assuming ideal behavior, calculate the vapor pressure of toluene in the mixture.

Answer 1

Molar mass of benzene (C₆H₆) = 78.11 g/mol

Molar mass of toluene (C₇H₈) = 92.14 g/mol

Therefore,

Moles of benzene in the solution= 25 g / (78.11 g/mol) = 0.32 moles

Moles of toluene in the solution = 75 g / (92.14 g/mol) = 0.8139 moles

Now,

Vapour pressure of benzene = (Mole fraction of benzene) x (Vapour pressure of pure benzene)

                                                   = [0.32/(0.32 + 0.8139)] x 94.2 = 26.58 torr

Vapour pressure of toluene = (Mole fraction of toluene) x (Vapour pressure of pure toluene)

                                                   = [0.8139/(0.32 + 0.8139)] x 28.4 = 20.38 torr