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0.5 molal aqueous solution of a weak aci...

0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If `K_(f)` for water is 1.86 K kg `mol^(-1)`, the lowering in freezing point of solution will be :

A

0.56

B

1.12 K

C

`-0.56 K`

D

`01.12 K`

Text Solution

Verified by Experts

The correct Answer is:
b
`HXhArrH^(+)+X^(-)`
For dessociation of electrolyte,
`alpha=((i-1))/((n-1))or0.2=((i-1))/((2-1))`
i=1+0.2=1.2
`DeltaT_(f)=I K_(f)m`
`=(1.2)xx(1.86 K m^(-1))xx(0.5m)=1.12 K`
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