Question

In a mixture of volatile substances, the vapor pressure of the solution depends on the vapor pressure of both substances (solute and solvent). Each component can evaporate, but its vapor pressure is lowered by the presence of the other substance following Rault's law. The vapor pressure of the solvent is calculated using Psolvent = Xsolvent*Posolvent and the vapor pressure of the solute is calculated using Psolute = Xsolute*Posolute. The total vapor pressure can be calculated by Dalton's law of partial pressures, Ptotal = Psolvent + Psolute. At 25oC, the vapor pressure of pure benzene (C6H6, 78.11 g/mol) is 96 torr. At the same temperature, the vapor pressure of pure toluene (C7H8, 92.14 g/mol) is 14 torr. Consider a solution containing 3.71 mol of benzene and 5.52 mol of toluene. Calculate the vapor pressure above the solution. Enter your answer in units of torr to three significant figures.

Answer 1

Vapour pressure of the solution is 30.5 torr.

Step-by-step explanation:

`\eta _(Benzene)= 2.15`

`\eta _(Toulene)= 8.53`

`\chi _(Benzene)= (\eta_(Benzene))/(\eta_(Toulene))= (2.15)/((8.53+2.15))=0.2013`

`\chi _(Total)= (8.53)/((8.53+2.15))=0.7987`

`\rho _(Benzene)= \chi_(Benzene)+\rho^(o)_(Benene)= 0.2013* 96torr=19.32torr`

`\rho _(total)= \chi_(Total)+\rho^(o)_(Total)= 0.7987* 14torr=11.18torr`

`\rho _(total)= \rho_(Benzene)+\rho_(Total)=19.32+11.18=30.5torr`

Therefore, Vapour pressure of the solution is 30.5 torr.