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?
1 Answer
I got
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) = 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
#=> 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)#
