When 0.1 mol CoCl₃(NH₃)₅ is treated with excess of AgNO₃, 0.2 mol of AgCl are obtained. The conductivity of solution will correspond to

When 0.1 mole of CoCl₃ (NH₃)₅ was reacted with excess of AgNO₃, we get 0.2 moles of AgCl. So, there are two chloride ions that are free and not part of the complex. The formula for complex has to be [Co(NH₃)₅ Cl] Cl₂.

$\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5\mathrm{Cl}\right]{\mathrm{Cl}}_2\to {\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5\mathrm{Cl}\right]}^{2+}+2{\mathrm{Cl}}^{-}$

Therefore, the conductivity of the solution will be 1 : 2 electrolyte.