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#.

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.......

Each component in the solution will have a vapour pressure that is proportionate to its mole fraction; the total vapour pressure of the solution will be this amount.

#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, the vapour is ENRICHED with respect to the MORE VOLATILE component, in this case, ethanol, when compared to the vapour pressures of the pure solvents.

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Answer 2

To calculate the mole fractions of water and alcohol in commercial vodka, you need to know the mole fractions of water and alcohol in the solution, which can be determined from the composition of the vodka. You can use the following formula:

Mole fraction of water = (moles of water) / (total moles of water and alcohol)

Mole fraction of alcohol = (moles of alcohol) / (total moles of water and alcohol)

To find the moles of each component, you can use the given mass of water and alcohol and their respective molar masses, then convert them to moles using the formula:

moles = mass / molar mass

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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