The freezing point of an aqueous solution of `KCN` containing `0.1892 mol kg^(-1)` was `-0.704^(circ)C`. On adding `0.45` mole of `Hg(CN)_(2)`, the freezing point of the solution was `=0.620^(circ)C`. If whole of `Hg(CN)_(2)` is used in complex formation according to the equation,
`Hg(CN)_(2)+mKCN rarrK_(m)[Hg(CN)_(m+2)]`
what is the formula of the complex ? Assume `[Hg(CN)_(m+2)]^(m-)` is not ionised and the complex molecule is `100%` ionised. `(K_(f)(H_(2)O)` is `1.86 kg mol^(-1) K`.)

Answer

`K_(2)[Hg(CN)_4)]`

Explanation

`DeltaT_(f)` (`KCN` solution) `=0.704^(@)`
`M'` (molality of `KCN` solution) `=0.1892 "mol"kg^(-1)`
`i=(1+x)`
`DeltaT_(f)=k_(f)M'i`
`i` (complex) `=(1+x)=(DeltaT_(f))/(k_(f)M')`
`=(0.704)/(1.86xx0.1892)=2.0`
This gives `x=1`, indicating `100%` ionisation of `KCN`
`DeltaT_(f)`(of the complex) `=0.530^(@)`
`M'[Hg(CN)_(2)]=0.095 "mol" kg^(-1)`
`k_(m)[Hg(CN)_(m+2)]hArrmK^(+)+[Hg(CN)_(m+2)]^(m-)`
`therefore i` (complex)=1+(y-1)x=1+(m+1-1)x`
`=1+mx=(1+m)`
`i=(1+m)=(DeltaT_(f))/(K_(f()M')=(0.530)/(1.86xx0.095)=3`
`therefore m=2`
Thus, complex is `K_(f)[Hg(CN)_(4)]`