What is the mol fraction of benzene in the vapour phase if equal mols of benzene and toluene are dissolved in the same solution? The pure vapor pressures of benzene and toluene are #"22 torr"# and #"75 torr"# respectively, at this temperature.

#chi_("benz"(g)) = 0.773#

The important thing to note is that the mole fraction in the vapor is asked for, not the mole fraction in the liquid (although that is readily found to be #0.50#).

Recall Raoult's law for one component in solution with a second:

#P_j = chi_(j(l))P_j^"*"##" "" "" "" "bb((1))#

where:

• #P_j# is the vapor pressure above the solution of component #j# in solution.
• #P_j^"*"# is the pure vapor pressure in the absence of the solvent.
• #chi_(j(l))# is the mol fraction of #j# in the solution (liquid phase).

From Dalton's law of partial pressures, we assume with ideal solutions that, upon plugging in #bb((1))# for each component in solution, we have:

#P_"tot" = P_i + P_j#

#= chi_(i(l))P_i^"*" + chi_(j(l))P_j^"*"# #" "" "" "" "bb((2))#

Let #i# be toluene and #j# be benzene. Then, we can get the total pressure (#"1 mm Hg = 1 torr"#), knowing that the solution is equimolar for both components:

#color(green)(P_"tot") = overbrace((n_i)/(n_i + n_i)("22 torr"))^(P_i) + overbrace((n_i)/(n_i + n_i)("75 torr"))^(P_j) = color(green)("48.5 torr")#

If we then focus on each component, we have two equations by relating partial pressures to the total pressure via mol fractions in the vapor phase:

#chi_(i(l))P_i^"*" = chi_(i(g))P_"tot"# #" "" "" "" "bb((3a))#
#chi_(j(l))P_j^"*" = chi_(j(g))P_"tot"##" "" "" "" "bb((3b))#

We could choose either one, really. Since we want benzene's mole fraction above the solution, plug in the mole fraction of benzene in the liquid phase, its pure vapor pressure, and the total pressure (previously found from #bb((2))#) to get:

#color(blue)(chi_(j(g))) = (chi_(j(l))P_j^"*")/P_"tot"#

#= color(blue)(0.773)#