1.

0.1m solution of KCI and `BaCI_(2)` are preparred. The freezing point of KCI solution is found to be- `4.0^(@)C`. What will be the freezing point of `BaCI_(2)` solution assuming that both KCI and `BaCI_(2)` solution are sompletely ionised in solution ?A. `-3^(@)C`B. `-4^(@)C`C. `-5^(@)C`D. `-6^(@)C`

Answer» Correct Answer - a
`"For" KCI,DeltaT_(f)=iK_(f)xxm`
`DeltaT_(f)=(0.0.4^(@)C)=4^(@)C`
`4K=2xxK_(f)xxm`
`"For " BaCI_(2),DeltaT_(f)=iK_(f)xxm`
`=3xxK_(f)xxm`
Both `K_(f)` and m are same for the solute,
`DeltaT_(f)=(BaCI_(2))=3`
Freezing point of `BaCI_(2)=0-3=-3^(@)C`


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