1.

Equimolar solutions of NaCl and BaCl_(2) are prepared . If the freezing point of NaCl is - 2^(@)C, the freezing point of BaCl_(2) is expected to be

Answer»

`-2^(@)C`
`-3^(@)C`
`-1.5^(@)C`
`-1.66^(@)C`

Solution :`DeltaT_(f)=iK_(f)xxm`
For NACL, `DeltaT_(f)=2, i=2`
`2=2xxK_(f)xxm`
For `BaCl_(2), DeltaT_(f)=?, i=3`
`DeltaT_(f)=3xxK_(f)xxm`
Since `K_(f)` and m are same for equimolar solutions,
`(DeltaT_(f))/(2)=3/4` or `DeltaT_(f)=3`
`:.` Freezing POINT of `BaCl_(2)` solution `=0-3=-3^(@)C`


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