Mole fraction of ethanol in ethanol water mixture is 0.25. Hence, the percentage concentration of ethanol by weight of mixture is

Explanation

Mole fraction of ethanol `(X_(1))=0.25`
Mole fraction of `H_(2)O(X_(2))=0.75`
`X_(1)=(n_(1))/(n_(1)+n_(2)), X_(2)=(n_(2))/(n_(1)+n_(2))`
`therefore" "(X_(1))/(X_(2))=(n_(1))/(n_(2))=(w_(1))/(M_(1))(M_(2))/(w_(2))=(1)/(3)rArr (w_(1))/(46)xx(18)/(w_(2))=(1)/(3)`
`therefore" "(w_(1))/(w_(2))=(46)/(54)`
Thus, if (ethanol) is 46 g then `w_(2)` (water) is 54 g in `(w_(1)+w_(2))=100g` solution.
Percentage concentration of ethanol `=(46)/(100)xx100` in solution `=46%`.