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 `

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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 `

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`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

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`

4 views Answered 2023-02-05 15:29