$50 \, mL$ of $0.2 \, M$ solution of a compound with empirical formula $CoCl_3 \cdot 4NH_3$ on treatment with excess of $AgNO_{3(aq)}$ yields $1.435 \, g$ of $AgCl$. Ammonia is not removed by treatment with concentrated $H_2SO_4$. The formula of the compound is:
- A$Co(NH_3)_4Cl_3$
- B$[Co(NH_3)_4Cl_2]Cl$
- C$[Co(NH_3)_4Cl_3]$
- D$[CoCl_3(NH_3)]NH_3$
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