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The freezing point of an aqueous solutio...

The freezing point of an aqueous solution of `KCN` containing `0.1892 mol Kg^(-1)`, the freezing point of the solution was found to be `-0.530^(@)C`. If the complex formation takes place according to the following equation:
`Hg(CN)_(2)`, the freezing point of the solution was found to be `-0.530^(@)C`. If the complex formation takes place according to the following equation:
`Hg(CN)_2 +nKCN hArr K_n[Hg (CN)_(n+2)]``
what is the formula of the complex? [`K_(f)(H_(2)O)` is `1.86 K kg mol^(-1)`]

Text Solution

Verified by Experts

For `KCN`,`i=1+alpha`, where `alpha`= degree of dissociation
`DeltaT_(f)=K_(f)xxixx"molarity"`
`0.704=1.86xxixx0.1892`
`i=(0.704)/(1.86xx0.1892)=2`
or `1+alpha=2`
`:.alpha=1` indicates `100%` ionization of `KCN`.
Now, `DeltaT_(f)`=(of the complex)=`0.530^(@)C`
Molality of `Hg(CN)_(2)=0.095 "mol" kg^(-1) =0.095 m`

Here, `i=(1-alpha)+nalpha+alpha=1+nalpha`
`because alpha=1`
`:. i=1+n`
`DeltaT_(f)=K_(f)xxixx"molality"`
`rArr0.503=1.86xxixx0.095`
`rArr i~~3`
`1+n=3` or `n=2`
Hence, the complex is `K_(2)[Hg(CN)]`.
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