`50ml` of `0.2M` solution of a compound with emprical formula `CoCl_(3).4NH_(3)` treatment with excess of `AgNO_(3)(aq)` yields `1.435gm` of `AgCl`. Ammonia is not removed by treatment with concenrated `H_(2)SO_(4)`. The formula of compound is:

Options

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

Answer

C

Explanation

No. of moles of `AgCl=(1.435 gm)/(M_(AgCl))=0.01` moles
No. of moles of `CoCl_(3).4NH_(3)=0.01` moles
`:.` 1 ionizable `Cl` per mole of `CoCl_(3).4NH_(3)`