A complex is represented as `CoCl_(3) . XNH_(3)`. Its `0.1` molal solution in aqueous solution shows `Delta T_(f) = 0.558^(circ). (K_(f)` for `H_(2)O` is `1.86 K "molality"^(-1))` Assuming `100%` ionisation of complex and co-ordination number of `Co` as six, calculate formula of complex.

Answer

`[Co(NH_(3))_(5)Cl]Cl_(2)`

Explanation

`Delta"T"_("f")=[1+("n"-1)alpha]"K"_("f")xx"Molality"`
`0.558=[1+("n"-1)xx1]xx1.86xx0.1`
`"n"=(0.558)/(1.86xx0.1)`
`"n"=3`
`therefore "formula of complex"=["CO"("NH"_(3))_(5)"Cl"]"Cl"_(2)`