On treating 100 mL of 0.1 M aqueous solution of the complex \(\mathrm{CrCl}_3 \cdot 6 \mathrm{H}_2 \mathrm{O}\) with excess of \(\mathrm{AgNO}_3, 2.86 \mathrm{~g}\) of AgCl was obtained. The complex is

Given, Molarity of the complex \(=0.1 \mathrm{M}\)
Volume of the complex \(=100 \mathrm{~mL}\)
Mass of AgCl obtained \(=2.86 \mathrm{~g}\)
Now, no. of moles in \(\mathrm{CrCl}_3 \cdot 6 \mathrm{H}_2 \mathrm{O}\)
\(=\frac{\text { molarity } \times \text { volume }}{1000}=\frac{0.1 \times 100}{1000}=0.01 \mathrm{moles}\)
No. of moles of \(\mathrm{AgCl}=\frac{2.86}{143}=0.02\) moles
Number of \(\mathrm{Cl}^{-}\)ions present in the ionisation sphere
\(=\frac{\text { Moles of ions precipitated with excess } \mathrm{AgNO}_3}{\text { Moles of complex }}\)
\(=\frac{0.02}{0.01}=2\)
It means two chlorine ions must be present in the solution. So, the complex will be
\(\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}^{-}\right] \mathrm{Cl}_2 \cdot \mathrm{H}_2 \mathrm{O}\)