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

`{:(HA,rarr,H^(+),+,A^(-)),(1-alpha,,alpha,,alpha),(1-0.3,,0.3,,0.3):}`
`therefore i=(1-0.3)+0.3+0.3=1.3`
`DeltaT_(f)=iK_(f)m=1.3xx1.86xx0.1=0.2418`
`T_(f)=0-0.2418^(@)C=-0.2418^(@)C~=-0.24^(@)C`.