The “proof” of an alcoholic beverage is twice the volume percent of ethyl alcohol, C2H5OH, in solution. For an 80-proof rum, what is the mole fraction of ethyl alcohol in the rum? Take the densities of the ethyl alcohol and water to be 0.789g/mL and 1.00g/mL, respectively.
An 80-proof rum has 40% ethyl alcohol by volume, which means that there is 40 mL of ethyl alcohol in every 100 mL of rum. We will first find the mass of the ethyl alcohol in the solution and then use the molar masses to find the mole fraction.
First, let's find the mass of the ethyl alcohol in 100 mL of solution:
Mass_ethyl_alcohol = volume_ethyl_alcohol * density_ethyl_alcohol
Mass_ethyl_alcohol = 40 mL * 0.789 g/mL = 31.56 g
Now, let's find the mass of water in the solution:
volume_water = volume_solution - volume_ethyl_alcohol
volume_water = 100 mL - 40 mL = 60 mL
Mass_water = volume_water * density_water
Mass_water = 60 mL * 1.00 g/mL = 60 g
Now we need to find the moles of ethyl alcohol (C2H5OH) and water (H2O) in the solution. We can use their molar masses (C2H5OH has a molar mass of 46.07 g/mol, and H2O has a molar mass of 18.02 g/mol):
Moles_ethyl_alcohol = Mass_ethyl_alcohol / Molar_mass_ethyl_alcohol
Moles_ethyl_alcohol = 31.56 g / 46.07 g/mol = 0.685 mol
Moles_water = Mass_water / Molar_mass_water
Moles_water = 60 g / 18.02 g/mol = 3.33 mol
Finally, we can find the mole fraction of ethyl alcohol in the rum (mole fraction = moles of a component / total moles of the mixture):
Mole_fraction_ethyl_alcohol = Moles_ethyl_alcohol / (Moles_ethyl_alcohol + Moles_water)
Mole_fraction_ethyl_alcohol = 0.685 mol / (0.685 mol + 3.33 mol) ≈ 0.171
The mole fraction of ethyl alcohol in the 80-proof rum is approximately 0.171.