Question

A solution of 75.0g of benzene (C6H6) and 75.0g of toluene (C7H8) has a total vapor pressure of 80.9 mmHg at 303 K. Another Solution of 100.0g of benzene and 50.0g of toluene has a total vapor pressure of 93.3 mmHg at this temperature. Find the vapor pressure of pure benzene and toluene at 303K.

Answer 1

Mol. wt of benzene = 78 and that of toluene = 92

No. of moles in 75.0 g of benzene = 75/ 78 = 0.96

No. of moles in 75.0 g of toluene = 75/ 92 = 0.82

Therefore mole fraction of benzene in the mixture is 0.96/(0.96+0.82) = 0.54

Similarily,

No. of moles in 100.0 g of benzene = 100/ 78 = 1.28

No. of moles in 50.0 g of toluene = 50/ 92 = 0.54

Therefore mole fraction of benzene in the mixture is 1.28/(1.28+0.54) = 0.70

From the graph we find that the vap. pressure is 119 when the mole fraction of benzene is 1.0 i.e. of pure benzene

and the same is 39 when the mole fraction of benzene is 0, i.e, of pure tolune.