Solubility of AgCl is $1.435\times {10}^{-3}\mathrm{g}/\mathrm{litre}$  at 298 K. Solubility product of AgCl will be (GFW of AgCl = 143.5g)

 Solubility of AgCl       = $\frac{1.435\times {10}^{-3}}{143.5}\mathrm{mole}/\mathrm{litre}$

                                  $= {10}^{-5} \mathrm{M}$

                             $\begin{array}{l}\mathrm{AgCl}\underset{ }{\underset{ }{ }}\underset{\mathrm{S}}{{\mathrm{Ag}}^{+}}+\underset{\mathrm{S}}{{\mathrm{Cl}}^{-}}\\ {\mathrm{K}}_{\mathrm{sp}}={\mathrm{s}}^2\\ {\mathrm{K}}_{\mathrm{sp}}={\left({10}^{-5}\right)}^2{\mathrm{M}}^2\\ ={10}^{-10}{\mathrm{M}}^2\end{array}$