The freezing point of a solution contaning `0.3 g` of acetic acid in `43 g` of benzene reduces by `0.3^(@)`. Calculate the Van's Hoff factor
"(`K_(f)` for benzene =`5.12 K kg mol^(-1)`)"

Explanation

Let us first calculate observed molar mass
`M_(B) = (k_(f) xx w_(B) xx 1000)/(Delta T_(f) xx w_(A))`
`w_(B) = 0.3 g , w_(A) = 30.0 g , Delta T_(f) = 0.45^(@) , K_(f) = 5.12 K kg mol^(-1)`
`therefore " " M_(B) = (5.12 xx 0.3 xx 100)/(0.45 xx 30) = 113.8`
Normal molar mass = 60 .
Now `" "i=("Normal molar mass")/("Oberved molar mass") = (60)/(113.8) = 0.527`