Question

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(`CH_3CH_2OH`) in water. The vapor pressure of pure water and ethanol at 20 °C are 17.5 torr and 44.6 torr respectively.

Answer 1

Well, let's take a `100*mL` volume of vodka..........

`chi"EtOH"=0.151`; `chi"H"_2"O"=0.849`.

Explanation:

And this represents a `31.2*g` mass of `EtOH`, and a `68.8*g` mass of water......

And the `"mole fraction"` `chi_"EtOH"="Moles of EtOH"/"Moles of EtOH+moles of water"`

`"Moles of EtOH"=(31.2*g)/(46.07*g*mol^-1)=0.677*mol`.

`"Moles of water"=(68.8*g)/(18.01*g*mol^-1)=3.82*mol`.

`chi"EtOH"=(0.677*mol)/(0.677*mol+3.82*mol)=0.151`

Now of course, in a binary solution, `chi_"other component"=1-0.151`, but we might as well go thru the motions......

`chi"H"_2"O"=(3.82*mol)/(0.677*mol+3.82*mol)=0.849`.

And `chi"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_"EtOH"=0.151xx44.6*"Torr"=6.7*"Torr"`

`P_"water"=0.849xx17.5*"Torr"=14.9*"Torr"`

`P_"solution"=P_"EtOH"+P_"water"=(14.9+6.7)*"Torr"`

`=21.6*"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.