A solution containing $$2.675$$ g of $$\text{CoCl}_3 \cdot 6\text{NH}_3$$ (molar mass $$= 267.5$$ g mol$$^{-1}$$) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of $$\text{AgNO}_3$$ to give $$4.78$$ g of AgCl (molar mass $$= 143.5$$ g mol$$^{-1}$$). The formula of the complex is (At. Mass of Ag $$= 108$$ u)

Number of moles of the complex taken
$$n_{\text{complex}} = \frac{2.675\ \text{g}}{267.5\ \text{g mol}^{-1}} = 0.010\ \text{mol}$$

After the solution is passed through a cation-exchange resin, only the free (ionisable) $$\text{Cl}^-$$ ions remain in solution. These chloride ions are precipitated as AgCl.

Number of moles of AgCl obtained
$$n_{\text{AgCl}} = \frac{4.78\ \text{g}}{143.5\ \text{g mol}^{-1}} = 0.0333\ \text{mol}$$

The reaction $$\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl}\downarrow$$ shows a 1 : 1 ratio; hence

$$n_{\text{Cl}^-} = n_{\text{AgCl}} = 0.0333\ \text{mol}$$

Ionisable chloride ions per mole of the complex
$$\text{Cl}^- \text{ per complex} = \frac{0.0333\ \text{mol}}{0.010\ \text{mol}} \approx 3$$

Therefore, all three chloride ions present in $$\text{CoCl}_3\cdot6\text{NH}_3$$ are outside the coordination sphere, while the six $$\text{NH}_3$$ molecules act as ligands around cobalt.

Hence, the coordination entity and complete formula are
$$[\text{Co}(\text{NH}_3)_6]^{3+}\;3\text{Cl}^- \;=\;[\text{Co}(\text{NH}_3)_6]\text{Cl}_3$$

Option A which is: $$[\text{Co}(\text{NH}_3)_6]\text{Cl}_3$$