A solution is 25% ethanol ans 50% acetic acid by mass. Calculate the mole fraction of ethanol and acetic acid in the solution.

Explanation

Let us start with 100 g of the solutioon in which
Mass of water =25g
Mass of ethanol=25g
Mass of acetic acid = 50g
`n_(H_(2)O)=((25g))/((18gmol^(-1)))=1.388mol`
`n_(C_(2)H_(5)OH)=((25g))/((46gmol^(-1)))=0.543 mol`
`n_(CH_(3)COOH)=((50g))/((60gmol^(-1)))=0.833mol`
Total number of moles =(1,388+0.543+0.833)=2.764 mole
`x_(C_(2)COOH)" "=n_(C_(2)H_(5)OH)=((0.543 mol))/((2.764 mol))=0.196`
`x_(CH_(3)COOH)" "n_(CH_(3)COOH)=((0.833 mol))/((2.764 mol))=0.301`