When a certain amount of benzene and toluene are mixed, the total vapor pressure above the solution is #"760 torr"#. Assuming an ideal solution is formed, and knowing the pure vapor pressures, what is the mol fraction of benzene in the solution phase?

The pure vapor pressures of benzene and toluene at this temperature are #"900 torr"# and #"360 torr"#, respectively.

2 Answers
Nov 19, 2017

#chi_(i(l)) = 0.7407#

What is the mol fraction of toluene in the solution phase?


This asks you to apply Raoult's law

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

where

  • #P_j# is the vapor pressure of component #j# in solution,
  • #P_j^"*"# is the vapor pressure of the pure component #j# (i.e. no component #i#!),
  • #chi_(j(l))# is the mol fraction of component #j# in the liquid phase (in the solution, not above it),

and the notion of partial pressures:

#P = P_i + P_j#

where #P# is the total pressure and the #P_k# are the partial pressures of the #k#th component.

The total pressure above the solution is #"1 atm"#, or #"760 torr"#. Use Raoult's law for ideal solutions to get:

#P = P_i + P_j#

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

Note that we have not yet specified which species is which. Let benzene be component #i# and toluene be component #j#. Then we have:

#P = chi_(i(l))P_i^"*" + (1 - chi_(i(l))) P_j^"*"#

#= chi_(i(l))P_i^"*" + P_j^"*" - chi_(i(l)) P_j^"*"#

#= P_j^"*" + chi_(i(l))(P_i^"*" - P_j^"*")#

where we have used the fact that the mol fractions of all components in solution MUST add up to #1#.

Thus, the mol fraction of benzene in the solution phase is:

#color(blue)(chi_(i(l))) = (P - P_j^"*")/(P_i^"*" - P_j^"*")#

#= ("760 torr" - "360 torr")/("900 torr" - "360 torr")#

#= color(blue)(0.7407)#

Nov 19, 2017

Answer:

The mole fraction of benzene is 0.741.

Explanation:

This is a problem involving Raoult's Law, which states that the partial vapor pressure of each component in an ideal mixture of liquids is equal to the vapor pressure of the pure component multiplied by its mole fraction.

If we let component 1 be benzene and component 2 be toluene,
we can express Raoult's Law as

#color(blue)(bar(ul(|color(white)(a/a)chi_1p_1^@ + chi_2p_2^@ = p_text(tot)color(white)(a/a)|)))" "#

where

#chi_1 = # the mole fraction of benzene
#p_1^@ = # the partial vapour pressure of pure benzene
#chi_2 = # the mole fraction of toluene
#p_2^@ = # the partial vapour pressure of pure toluene
#p_text(tot) =# the total vapour pressure over the solution

Since there are only two components in the mixture,

#chi_1 + chi_2 = 1#

Let #chi_1 = x#; then #"chi_2 = 1 - x#

#x × 900 color(red)(cancel(color(black)("torr"))) + (1 - x) × 360 color(red)(cancel(color(black)("torr"))) = 760 color(red)(cancel(color(black)("torr")))#

#900x + 360 - 360x = 760#

#540x = 400#

#x = 400/540 = 0.741#

#chi_1 = 0.741#

The mole fraction of benzene is 0.741.