A solution has ` 25% `of water,` 25% `ethanol and ` 50% `acetic acid by mass. Calculate the mol e fraction of each component.

Explanation

Let the total mass of solution = 100g
mass of water = 25 g
mass of ethanol = 25 g
Mass of acetic acid = 50 g
moles of water `=25/18 = 1.388`
(`therefore` Molar mass of `H_(2)O=18`)
Moles of ethanol `=25/46 =0.543`
(`therefore` Molar mass of `C_(2)H_(5)OH = 46`)
moles of acetic acid `=50/60 =0.833`
`(therefore` Molar mass of `CH_(3)COOH=60`)
Total number of moles = 1.388 + 0.543 + 0.833
=2.764
mole fraction of water `=(1.388)/(2.764) = 0.502`
Mole fraction of ethanol `=(0.543)/(2.764) = 0.302`