$100\, \mathrm{mL}, 0.1 \,\mathrm{M}$ $1.2\, \times 10^{22} \text { ion }$

$\text {moles }\;\; 0.01 \,\mathrm{mol}$   $\frac{1.2 \times 10^{22}}{6 \times 10^{23}}=0.02\, \mathrm{mol}$

since, $0.01\, \mathrm{mol}$ of $\mathrm{CoCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ produce $0.02\,\mathrm{mol}$ of $\mathrm{AgCl}$

Therefore, $1\,mol$ of $\mathrm{CoCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ will produce $2\, \mathrm{mol}$ of $\mathrm{AgCl}$

Thus, the complex is $\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}$