One litre of 1M solution of ${\mathrm{CoCl}}_3.4{\mathrm{NH}}_3$ require ‘V’ litres of 2M ${\mathrm{AgNO}}_3$ solution to precipitate all the chloride ions. Value of ‘V’ is

$$\begin{gathered} \mathrm{Modern} \mathrm{formula} \mathrm{of} {\mathrm{CoCl}}_3.4{\mathrm{NH}}_3 \mathrm{is} \left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{Cl}}_2\right]\mathrm{Cl}; \\ \left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{Cl}}_2\right]\mathrm{Cl} \rightleftharpoons {\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{Cl}}_2\right]}^{+}+{\mathrm{Cl}}^{-}; \\ \mathrm{Valency} \mathrm{of} \mathrm{the} \mathrm{complex} \mathrm{is} \mathrm{one}; \\ \mathrm{Valency} \mathrm{of} \mathrm{silver} \mathrm{nitrate} \mathrm{is} \mathrm{one}; \\ {\mathrm{V}}_1{\mathrm{N}}_1={\mathrm{V}}_2{\mathrm{N}}_2; \\ 1 \left(1\mathrm{M}\right) = \mathrm{V} \left(2\mathrm{M}\right); \\ {\mathrm{V}}_{{\mathrm{AgNO}}_3}= 0.5 \mathrm{litre}; \end{gathered}$$