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. `-0.1^(@)C`
B. `-0.54^(@)C`
C. `-0.3^(@)C`
D. `-0.24^(@)C`

Answer

D

Explanation

`{:(H_(A),rarr,H^(+),+,A'),((1-alpha),,(alpha),,(alpha)):}`
Total number of particles after dissociation `=1+alpha`
`alpha = (30)/(100)=0.3`
As we know, `i=1+alpha=1.3`
`Delta T_(f)=ixxk_(f)xxm=1.30xx1.86xx0.1`
`=0.2418`
Freezing point `= -0.2418`