The volume (in mL ) of 0 . 1   M   AgNO 3 required for complete precipitation of chloride ions…

$\left(\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\mathrm{Cl}\right){\mathrm{Cl}}_2⟶\left(\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\mathrm{Cl}\right)+2{\mathrm{Cl}}^{-}$

Number of ionisable ${\mathrm{Cl}}^{-}$ in $\left(\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\mathrm{Cl}\right){\mathrm{Cl}}_2$​ is $2.$

Molarity of ${\mathrm{Cl}}^{–}$ ions $=0.01\times 2=0.02 \mathrm{M}$

$\therefore$ Millimoles of ${\mathrm{Cl}}^{–}$ ions in the given solution

$=20\times 0.02$

$\therefore$ Millimoles of ${\mathrm{AgNO}}_3$ required $=20\times 0.2$

$\mathrm{V}\times 0.1=20\times 0.02$

$\mathrm{V}=4 \mathrm{mL}$