50 ml of 0.2 M solution of a compound with empirical formula `CoCl_(3). 4NH_(3)` on treatment with excess of `AgNO_(3)(aq)` yields 1.435 g of AgCl. Ammonia is not removed by treatment with concentrated `H_(2)SO_(4)`. The formula of the compound is

Options

A. `Co(NH_(3))_(4)Cl_(3)`
B. `[Co(NH_(3))_(4)Cl_(2)]Cl`
C. `[Co(NH_(3))_(4)]Cl_(3)`
D. `[CoCl_(3)(NH_(3))]NH_(3)`

Answer

B

Explanation

`n_(AgCl)=1.435/143.5=0.01`
=10 m mol,
m mol of compound =50`xx`0.2=10
`rArr` Only 1 `Cl^(-)` permolecule is ionisable