If benzene and toluene have pure vapor pressures of #"700 torr"# and #"600 torr"# respectively at a certain temperature, then upon mixing equal mols of both together, what is the mol fraction of benzene above the solution phase?

I got #0.5385#.

Benzene and toluene are very similar (can you look up their structures?), so they form an essentially ideal solution.

Raoult's law for ideal solutions would work well here.

#P_A = chi_(A(l))P_A^"*"#

where:

• #A# is benzene and #B# is toluene.
• #chi_(k(l))# is the mol fraction of #k# in the solution phase.
• #P_A# is the partial vapor pressure of #A# above the solution.
• #"*"# indicates the pure substance.

We also know from Dalton's law of partial pressures for ideal gases that the total pressure above the combined solution, #P_(AB)#, is described by:

#P_(AB) = P_A + P_B#

The partial pressure above the solution is precisely the thing described by Raoult's law. So:

#P_(AB) = chi_(A(l))P_A^"*" + chi_(B(l))P_B^"*"#

It may seem like not enough information is provided, but we know that equal mols of both were mixed, so #chi_(A(l)) = chi_(B(l)) = 0.5#.

#=> P_(AB) = 0.5P_A^"*" + 0.5P_B^"*"#

#= 0.5(P_A^"*" + P_B^"*")#

#= 0.5("700 torr" + "600 torr")#

#=# #"650 torr"#

Now, we know the definition of partial pressure is that:

#P_A = chi_(A(v))P_(AB)#,

where #chi_(A(v))# is the mol fraction of #A# in the vapor phase.

Notice the distinction here... we are now regarding the components in the vapor phase above the solution, NOT the components in the solution.

Therefore:

#color(blue)(chi_(A(v))) = P_A/P_(AB)#

#= (chi_(A(l))P_A^"*")/(P_(AB))#

#= ((0.5)("700 torr"))/("650 torr")#

#= color(blue)(0.5385)#