When $0.1 \ mol$ of $CoCl_3(NH_3)_5$ is treated with an excess of $AgNO_3$,$0.2 \ mol$ of $AgCl$ are obtained. The formula of the compound is:

  • A
    $[CoCl_3(NH_3)_3]NH_3$
  • B
    $[CoCl(NH_3)_5]Cl_2$
  • C
    $[CoCl_2(NH_3)_3]Cl \cdot NH_3$
  • D
    $[Co(NH_3)_6]Cl_3$

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