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

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.