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

Explanation

Let us suppose that the total mass of the solution is 100 g. Then
Amount of water = 25 g , `" "` Amount of ethanol = 25 g
`"25 g of water "=("25 g ")/("18 g mol"^(-1))="1.388 moles "[because" Malar mass of water "="18 g mol"^(-1)]`
`"50 g of ethanol "=("25 g")/("46 g mol"^(-1))="0.543 moles "[because " Molar mass of ethanol "(C_(2)H_(5)OH)="46 g mol"^(-1)]`
`"50 g of acetic acid "=("50 g")/("60 g mol"^(-1))="0.833 moles"`
`" "[because" Molar mass of acetic acid "(CH_(3)COOH)="60 g mol"^(-1)]`

`therefore" Mole fraction of water "=(1.388)/(1.388+0.543+0.833)=(1.388)/(2.764)=0.503`
Mole fraction of ethanol `=(0.543)/(2.764)=0.196" and mole fraction of acetic acid "=(0.833)/(2.764)=0.301.`