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

Explanation

Let the total mass of solution = 100 g
Mass of water = 25 g
Mass of ethanol = 25 g
Mass of acetic acid = 50 g
Moles of water ` = (25)/(18) = 1.388 " "(because" Molar mass of " H_(2)O = 18)`
Moles of ethanol `= (25)/(46) = 0.543 " "(because "Molar mass of " C_(2)H_(5)OH = 46)`
Moles of acetic acid `= (50)/(60) = 0.833 " "(because" 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 water ` = (0.543)/(2.764) = 0.502`
Mole fraction of acetic acid ` = (0.833)/(2.764) = 0.302`