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.

Options

A. `[Co(NH_(3))_(5)]Cl_(3)`
B. `[Co(NH_(3))_(5)Cl]Cl_(2)`
C. `[Co(NH_(3))_(4)Cl_(2)]Cl`
D. None of these

Answer

2

Explanation

`DeltaT_(f)=K_(f)xx`molality `xxi,`
`0.558=1.86xx0.1xxi,`
or `" "i=3`
Thus complex is `[CoCl.xNH_(3)].Cl_(2).` Since co- ordination number of Co is six. Thus complex is `[Co(NH_(3))_(3)Cl]Cl_(2)`