# 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)#