A colloidal sol is formed by moving AgNO₃ and KI in 2. 1 molar ratio This colloidal sol is to be coagulated

I. NaI            II. MgSO₄            III. AlPO₄

Molar ratio of these required to coagulate 1 mole colloidal sol is

$$\begin{gathered} {\mathrm{AgNO}}_3+\mathrm{KI}⟶\mathrm{AgI}+{\mathrm{KNO}}_3 \\ \mathrm{AgI}+\underset{\mathrm{excess}}{{\mathrm{AgNO}}_3}⟶\underset{\underset{\left(\mathrm{colloidal}\right)}{\mathrm{Fixed} \mathrm{layer}}}{\left[\mathrm{AgI}\right]{\mathrm{Ag}}^{+}}⋮{\mathrm{NO}}_3^{-} \end{gathered}$$

By ${\text{ [AgI] Ag }}^{+}$ can be coagulated by negatively charged sol I^- from Na, $\left.{\mathrm{SO}}_4^{2-}\left(\text{ from }{\mathrm{MgSO}}_4\right)\text{ and }{\mathrm{PO}}_4^{3-}\text{ (from }{\mathrm{AlPO}}_4\right)$. Greater the charge on anion, greater the coagulating power. (Hardy-Schulze rule)

$\begin{array}{l}\left[\mathrm{AgI}\right]{\mathrm{Ag}}^{+}+{\mathrm{I}}^{-}⟶\mathrm{AgI}+\mathrm{AgI}\\ 2\left[\mathrm{AgI}\right]{\mathrm{Ag}}^{+}+{\mathrm{SO}}_4^{2-}⟶2\mathrm{AgI}+{\mathrm{Ag}}_2{\mathrm{SO}}_4\\ 3\left[\mathrm{AgI}\right]{\mathrm{Ag}}^{+}+{\mathrm{PO}}_4^{3-}⟶3\mathrm{AgI}+{\mathrm{Ag}}_3{\mathrm{PO}}_4\end{array}$

Thus, I^- required by 1 mole colloidal sol = 1 mole

${\mathrm{SO}}_4^{2-}\text{ required by }1\text{ mole colloidal sol }=\frac{1}{2}\text{ mole }$

${\mathrm{PO}}_4^{3-}\text{ required by }1\text{ mole colloidal }\mathrm{sol}=\frac{1}{3}\text{ mole }$

$\text{ Ratio }=1:\frac{1}{2}:\frac{1}{3}=6:3:2$