How many mols are there of methanol and ethanol above the solution if the partial vapor pressures of methanol and ethanol are "2.619 kPa" and "4.556 kPa" respectively?

1 Answer
Jun 1, 2017

Methanol:

${\chi}_{\text{MeOH} \left(v\right)} = 0.3650$

Ethanol:

${\chi}_{\text{EtOH} \left(v\right)} = 0.6350$

As with many other problems about ideal solutions and volatile solutes and solvents, we refer to Raoult's law for ideal solutions and Dalton's law of partial pressures for ideal gases, respectively:

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

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

where:

• ${\chi}_{A \left(l\right)}$ is the mol fraction of methanol in the solution phase. $B$ indicates ethanol.
• ${P}_{A}$ is the vapor pressure of methanol above the solution, in the context of the solution.
• $\text{*}$ indicates that it's for the pure substance.
• ${P}_{A B}$ is the total pressure above the solution.

It is not entirely obvious, but it's likely we are given the vapor pressures in the context of the solution (rather than the pure substances), because you wrote "partial vapor pressures":

${P}_{A} = \text{2.619 kPa}$
${P}_{B} = \text{4.556 kPa}$

(Otherwise, we would not be able to solve for the mol fractions, because we do not know the total pressure.)

In this case then, the total pressure from Dalton's law is:

${P}_{A B} = {P}_{A} + {P}_{B} = \text{7.175 kPa}$

This means we can now solve for the mol fraction of one substance of choice in the vapor phase. Recall that ${P}_{A}$ and ${P}_{B}$ are known as partial vapor pressures. So, by considering the definition of partial pressures...

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

where $\left(v\right)$ indicates the vapor phase...

...we can then get the mol fractions of each one in the vapor phase.

$\implies \textcolor{b l u e}{{\chi}_{A \left(v\right)}} = {P}_{A} / {P}_{A B} = \text{2.619 kPa"/"7.175 kPa}$

$= \textcolor{b l u e}{0.3650}$

$\implies \textcolor{b l u e}{{\chi}_{B \left(v\right)}} = 1 - {\chi}_{A \left(v\right)} = \textcolor{b l u e}{0.6350}$

However, since mol fractions are always relative, without knowing exactly what quantity of one substance was put in, we cannot know the actual $\text{mol}$s of it.

So, apparently, we would have to leave it at knowing the mol fractions...