1.

The freezing point (in .^(@)C) of a solution containing 0.1 g of K_(3)[Fe(CN)_(6)[ (Mol. Wt. 329) in 100 g of water (K_(f)="1.86 K kg mol"^(-1)) is

Answer»

`-2.3xx10^(-2)`
`-5.7xx10^(-2)`
`-5.7xx10^(-3)`
`-1.2xx10^(-2)`

SOLUTION :`K_(3)[Fe(CN)_(6)]HARR 3K^(+)+[Fe(CN)_(6)]^(3-)`
`therefore I = 4`
`DeltaT_(F)=iK_(f)m=ixxK_(f)xx(w_(2))/(M)xx(1)/(w_(1))xx1000`
`=4xx1.86xx(0.1)/(329)xx(1)/(100)xx1000`
`=0.023 = 2.3xx10^(-2).^(@)C or K`
`T_(f)=0-2.3xx10^(-2).^(@)C=-2.3xx10^(-2).^(@)C`


Discussion

No Comment Found

Related InterviewSolutions