An excess of AgNO_3 is added to 100 ~mL of a 0.01 M solution of...

The formula of dichlorotetraqua chromium (III) chloride is $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl}$. On ionisation it generates only one $\mathrm{Cl}^{-}$ ion. $\left[\mathrm{Cr} \left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl} \longrightarrow\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right]^{+}+ \mathrm{Cl}^{-}Initial 100 \times0.01 0 \mathrm{mmol}=1 \mathrm{mmol} After ionisation 0 1 \mathrm{mmol} \quad1 \mathrm{mmol} One mole of \mathrm{Cl}^{-}ions react with only 1 mole of \mathrm{AgNO}_3 molecule to produce 1 mole of \mathrm{AgCl}$. $\therefore 1$ mmol or $1 \times10^{-3}$ mole reacts with $\mathrm{AgNO}_3$ to give \mathrm{AgCl}=\frac{1 \times 1 \times 10^{-3}}{1}=10^{-3} \text{ or }0.001 \mathrm{ mol} \mathrm{AgCl}$$