The volume (in mL ) of 0.1 M AgNO_3 required for complete...

\mathrm{mmol}$of complex$=30 \times 0.01=0.3$Also, 1 mole of complex \left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2$ gives only two moles of chloride ion when dissolved in solution $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_{2}$ \longrightarrow\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_5 \mathrm{Cl}^{2+}+2 \mathrm{Cl}^{-}\right.\Rightarrow \mathrm{mmol}$of$\mathrm{Cl}^{-}ion produced from its 0.3 \mathrm{mmol}=0.6$Hence,$0.6 \mathrm{mmol}$of$\mathrm{Ag}^{+}$would be required for precipitation.$$\begin{aligned} & \Rightarrow 0.60 \mathrm{mmol} \text{\text{ of } } \mathrm{Ag}^{+}=0.1 \mathrm{M} \times \mathrm{V}(\text{ in } \mathrm{mL})\\ & \Rightarrow \quad V=6 \mathrm{ mL} . \end{aligned}$$$