Given the following, how do you calculate the mole fractions of water and alcohol in commercial vodka?

Commercial vodka is typically sold at a concentration of 40.0% alcohol by volume (ABV), which is equivalent to 31.2% by mass ethanol($C {H}_{3} C {H}_{2} O H$) in water. The vapor pressure of pure water and ethanol at 20 °C are 17.5 torr and 44.6 torr respectively.

Jun 13, 2017

Well, let's take a $100 \cdot m L$ volume of vodka..........

$\chi \text{EtOH} = 0.151$; $\chi \text{H"_2"O} = 0.849$.

Explanation:

And this represents a $31.2 \cdot g$ mass of $E t O H$, and a $68.8 \cdot g$ mass of water......

And the $\text{mole fraction}$ ${\chi}_{\text{EtOH"="Moles of EtOH"/"Moles of EtOH+moles of water}}$

$\text{Moles of EtOH} = \frac{31.2 \cdot g}{46.07 \cdot g \cdot m o {l}^{-} 1} = 0.677 \cdot m o l$.

$\text{Moles of water} = \frac{68.8 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1} = 3.82 \cdot m o l$.

$\chi \text{EtOH} = \frac{0.677 \cdot m o l}{0.677 \cdot m o l + 3.82 \cdot m o l} = 0.151$

Now of course, in a binary solution, ${\chi}_{\text{other component}} = 1 - 0.151$, but we might as well go thru the motions......

$\chi \text{H"_2"O} = \frac{3.82 \cdot m o l}{0.677 \cdot m o l + 3.82 \cdot m o l} = 0.849$.

And $\chi \text{EtOH"+chi"H"_2} O = 1$ as required.......

The vapour pressure of each component in solution will be proportional to their mole fractions.......and the vapour pressure of the solution will be this sum.........

${P}_{\text{EtOH"=0.151xx44.6*"Torr"=6.7*"Torr}}$

${P}_{\text{water"=0.849xx17.5*"Torr"=14.9*"Torr}}$

${P}_{\text{solution"=P_"EtOH"+P_"water"=(14.9+6.7)*"Torr}}$

$= 21.6 \cdot \text{Torr}$

As is typical of the vapour of such solutions; by comparison to the vapour pressures of the pure solvents, the vapour is ENRICHED with respect to the MORE VOLATILE component, which here is ethanol.