1.

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 :

Answer»

0.56
1.12 K
`-0.56 K`
`01.12 K`

Solution :`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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