Question

100.0 g of benzene is mixed with 100.0 g of toluene to form a solution that can be considered ideal. What is the total vapor pressure (in Torr) of this solution at 293 K? The molar masses of benzene and toluene are 78.11 g/mol and 92.14 g/mol respectively. At 293 K the vapor pressures of pure benzene and pure toluene are 74.7 Torr and 22.3 Torr respectively. Report your result with 1 decimal place.

Answer 1

Mol of benzene,

n(benzene) = mass of benzene / molar mass of benzene

= 100.0 g / 78.11 g/mol

= 1.280 mol

n(toluene) = mass of toluene / molar mass of toluene

= 100.0 g / 92.14 g/mol

= 1.085 mol

mole fraction of benzene,

X(benzene) = n(benzene) / total mol

= 1.280 mol / (1.280 mol + 1.085 mol)

= 0.541

X(toluene) = 1- X(benzene)

= 1-0.541

= 0.459

Now use:

P = po(benzene)*X(benzene) + po(toluene)*X(toluene)

= 74.7*0.541 + 22.3*0.459

= 50.6 torr

Answer: 50.6 torr