# 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?

May 31, 2017

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.

${P}_{A} = {\chi}_{A \left(l\right)} {P}_{A}^{\text{*}}$

where:

• $A$ is benzene and $B$ is toluene.
• ${\chi}_{k \left(l\right)}$ is the mol fraction of $k$ in the solution phase.
• ${P}_{A}$ is the partial vapor pressure of $A$ above the solution.
• $\text{*}$ 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, ${P}_{A B}$, is described by:

${P}_{A B} = {P}_{A} + {P}_{B}$

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

${P}_{A B} = {\chi}_{A \left(l\right)} {P}_{A}^{\text{*" + chi_(B(l))P_B^"*}}$

It may seem like not enough information is provided, but we know that equal mols of both were mixed, so ${\chi}_{A \left(l\right)} = {\chi}_{B \left(l\right)} = 0.5$.

$\implies {P}_{A B} = 0.5 {P}_{A}^{\text{*" + 0.5P_B^"*}}$

$= 0.5 \left({P}_{A}^{\text{*" + P_B^"*}}\right)$

$= 0.5 \left(\text{700 torr" + "600 torr}\right)$

$=$ $\text{650 torr}$

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

${P}_{A} = {\chi}_{A \left(v\right)} {P}_{A B}$,

where ${\chi}_{A \left(v\right)}$ 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:

$\textcolor{b l u e}{{\chi}_{A \left(v\right)}} = {P}_{A} / {P}_{A B}$

$= \frac{{\chi}_{A \left(l\right)} {P}_{A}^{\text{*}}}{{P}_{A B}}$

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

$= \textcolor{b l u e}{0.5385}$