What is the total vapor pressure above an ideal binary mixture consisting of #"2 mols"# of benzene and #"1 mol"# of toluene? The pure vapor pressures are #"953 torr"# for benzene and #"378 torr"# for toluene.

Raoult's Law relates the partial pressure of a solvent of a mixture and its mole fraction to the colligative property of the solution.

#P_s = chi_iP_0#

Benzene is the solvent in this case, so we'll only worry about its vapor pressure. Hence,

#P_s = (2mol)/(3mol) * 953t o rr approx 635t o rr#

is the vapor pressure of this gaseous solution in this case.

I got #P = "761 torr"# for the mixture.

You are given the info:

#P_"ben"^"*" = "953 torr"#
#P_"tol"^"*" = "378 torr"#

That is, when they are by themselves. These obey Raoult's law well, as they mix to form an essentially ideal solution:

#color(green)(P_j) = color(darkblue)(chi_(j(l))P_j^"*")#

where:

• #P_j# is the partial pressure of component #j# in the context of the solution.
• #"*"# indicates pure solvent.
• #chi_(j(l))# is the mol fraction of component #j# in the solution phase.

The total pressure #P# (what we're looking for!) is then the sum of the partial pressures (in the solution context) for ideal gas mixtures:

#P = color(green)(P_"ben" + P_"tol")#

#= color(darkblue)(chi_"ben(l)"P_"ben"^"*" + chi_"tol(l)"P_"tol"^"*")#

where we have used Raoult's law to rewrite the partial pressures.

Since we want the vapor pressure of the mixture, we want #P#. I assume you are saying there are #"2 mols"# of benzene and #"1 mol"# of toluene.

If so,

#chi_"ben(l)" = "2 mol benzene"/("2 mol benzene" + "1 mol toluene") = 0.bar(66)#

#=> chi_"tol(l)" = 0.bar(33)#

This gives a vapor pressure of:

#color(blue)(P) = 2/3 cdot "953 torr" + 1/3 cdot "378 torr" = color(blue)("761 torr")#