A complex is represented as `CoCl_(3)Xnh_(3)` . Its `0.1` molal solution in water `DeltaT_(f)=0.588 K. K_(f)` for `H_(2)O` is `1.86K" molality"^(-1)`. Assuming `100%` ionisation of complex and co`-` ordination number of `Co` is six calculate formula of complex.

Answer

`CoCl_(3),xNH_(3)rarrCoCl_((3-n)),NH_(3)+nCl^(-)`
`{:(1,0,0),((1-alpha),alpha,alpha):}`
Assume `n Cl^(-)` are attached with `Co` through primary valencles which undergo ionisation. All the `NH_(3)` molecules are attached with `Co` through secondary valencies. `0.558=1.86xx0.1xx(1+n) (because alpha=1)`
Thus, complex is `=[CoCl.xNH_(3)]. Cl_(2)`. Since co-ordination number of `Co` is six, thus complex is `[Co(NH_(3)_(5)Ci]Cl_(2)`.