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) solutionassuming that both KCI and BaCI_(2) solution are sompletely ionised in solution ?

Answer»

`-3^(@)C`
`-4^(@)C`
`-5^(@)C`
`-6^(@)C`

SOLUTION :`"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`


Discussion

No Comment Found

Related InterviewSolutions