A `0.1` molal aqueous solution of a weak acid is `30%` ionized. If `K_(f)` for water is `1.86^(@)C//m`, the freezing point of the solution will be.

Options

A. `-0.18^(@)C`
B. `-0.54^(@)C`
C. `-0.36^(@)C`
D. `-0.24^(@)C`

Answer

D

Explanation

`Delta T_(f) = i xx K_(f) xx m`
Here I is van't Hoff factor.
ifor weak adic is `1 + alpha`
Hera `alpha` is degree of dissociation i.e. 30%
`i.e. 30/100=0.3`
`:. i=1 + alpha=1 + 0.3 = 1.3`
`Delta T_(f) = i xx K_(f) xx m`
`= 1.3 xx 1.86 xx 0.1 =0.24`
Freezing point = - 0.24