Question

The concentration of grain alcohol (C2H5OH) in whisky is given in "degrees proof", which is twice the percent alcohol by volume (v/v). What are the mole fraction and molality of C2H5OH in 85°proof vodka? Assume that vodka is a solution of only C2H5OH and water and that the volumes are additive. The density of C2H5OH is 0.79 g/mL.

Answer 1

The mole fraction and molality of alcohol, C₂H₅OH, in 85°proof vodka is 0.186 and 12.70 mol/kg respectively.

Data provided:

• Molality = moles of solute/kilogram of solvent
• Mole fraction = number of moles of solute/ total number of moles
• Density = mass/volume
• number of moles = mass/molar mass
• molar mass alcohol, C₂H₅OH = 46 g/mol
• molar mass of water = 18 g/mol
• density of water = 1 g/mL
• density of alcohol = 0.79 g/mL

Calculating moles of solute

85° proof vodka = 1/2 * 85% v/v

85° proof vodka = 42.5 % alcohol

In 100 mL vodka:

Volume of alcohol = 42.5 mL

volume of water = 57.5 mL

mass of alcohol, C₂H₅OH = density * volume

mass of alcohol, C₂H₅OH = 0.79 * 42.5

mass of alcohol, C₂H₅OH = 33.575 g

moles of alcohol = 33.575 g / 46 g/mol

moles of alcohol = 0.730 moles

Calculating mole fraction

mass of water = 57.5 ml * 1 g/mL

mass of water = 57.5 g

moles of water = 57.5 g/ 18 g/mL

moles of water = 3.194 moles

Total moles = 3.194 + 0.730

Total moles = 3.924 moles

Mole fraction of alcohol = 0.730/3.924

Mole fraction of alcohol = 0.186

Calculating molality of alcohol

mass of solvent = 57.5 = 0.0575 kg

Molality of alcohol = 0.730 / 0.0575

Molality of alcohol = 12.70 mol/kg

Therefore, the mole fraction and molality of alcohol, C₂H₅OH, in 85°proof vodka is 0.186 and 12.70 mol/kg respectively.

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