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

Options

A. `25%`
B. `75%`
C. `45%`
D. `54%`

Answer

C

Explanation

`chi_(2) = (n_(2))/(n_(1) + n_(2)) = 0.25` (ethanol) (solute) `(MW_(2) = 46g), " "chi_(1) = (n_(1))/(n_(1) + n_(2)) = 0.75` (water)(solvent) `(MW_(1) = 18g)`
`(chi_(1))/(chi_(2)) = (n_(1))/(n_(2)) = (3)/(2) rArr (W_(1) xx MW_(2))/(MW_(1) xx W_(2)) = (3)/(1), " "(W_(1) xx 46)/(18 xx W_(2)) = 3, " "(W_(1))/(W_(2)) = 1.17 :. (W_(2))/(W_(1) + W_(2)) = 0.46`