# A solution, that we may consider to be ideal, is composed of a 100*g masses EACH of ethanol and methanol....what is the vapour pressure of the solution?

## A solution, that we may consider to be ideal, is composed of a $100 \cdot g$ masses EACH of ethanol and methanol. If the vapour pressures of each PURE solvent at a given temperature are $44.5 \cdot m m \cdot H g$ and $88.7 \cdot m m \cdot H g$ respectively, what is the vapour pressure of each component, and the vapour pressure of the solution?

Nov 23, 2016

${P}_{\text{solution}} \cong 60 \cdot m m \cdot H g$

#### Explanation:

The partial pressure of each component in solution is proportional to its mole fraction.

Thus ${P}_{\text{MeOH"=chi_"MeOH"xxP_"pure MeOH}}$

${\chi}_{\text{MeOH"="Moles of MeOH"/"Moles of MeOH + Moles of EtOH}}$

$=$ $\frac{\frac{100 \cdot g}{32.04 \cdot g \cdot m o {l}^{-} 1}}{\left(\frac{100 \cdot g}{32.04 \cdot g \cdot m o {l}^{-} 1}\right) + \left(\frac{100 \cdot g}{46.07 \cdot g \cdot m o {l}^{-} 1}\right)}$

$=$ $\frac{3.12 \cdot m o l}{3.12 \cdot m o l + 2.17 \cdot m o l} = 0.590$. $\chi$ is dimensionless; why?

We could do the same thing to get ${\chi}_{\text{EtOH}}$, however this is a binary solution, and the mole fractions of each component must sum to $1$.

Thus ${\chi}_{\text{EtOH"=1-chi_"MeOH}} = 1 - 0.590 = 0.410$

And thus ${P}_{\text{solution"=P_"MeOH"+P_"EtOH}}$

$=$ ${\chi}_{\text{MeOH"xx88.7*mm*Hg+chi_"EtOH}} \times 44.5 \cdot m m \cdot H g$

$=$ $0.410 \times 88.7 \cdot m m \cdot H g + 0.590 \times 44.5 \cdot m m \cdot H g$

$=$ ??*mm*Hg

As is typical in these solutions, the vapour is enriched with respect to the more volatile component, here methyl alcohol.