Question

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?

Answer 1

I got `0.5385`.

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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)`