The ratio of amounts of ${\mathrm{H}}_2\mathrm{S}$ needed to precipitate all the metal ions from 100 ml of 1 M ${\mathrm{AgNO}}_3\mathrm{and}100\mathrm{ml}\mathrm{of}1\mathrm{M}{\mathrm{CuSO}}_4\mathrm{will}\mathrm{be}$

$\begin{array}{l}{\mathrm{AgNO}}_3\equiv 2{\mathrm{Ag}}^{+}+\underset{\left({\mathrm{H}}_2\mathrm{S}\right)}{{\mathrm{S}}^{2-}}\to {\mathrm{Ag}}_2\mathrm{S}\\ \because 2\mathrm{moles}\to 1\mathrm{mole}\left[100\times 1=100\mathrm{millimoles}\right]\\ \therefore 100\mathrm{millimoles}\to 50\mathrm{millimoles}{\mathrm{H}}_2\mathrm{S}\mathrm{required}\\ {\mathrm{CuSO}}_4\equiv {\mathrm{Cu}}^{+2}+\underset{\left({\mathrm{H}}_2\mathrm{S}\right)}{{\mathrm{S}}^{2-}}\to \mathrm{CuS}\\ \because 1\mathrm{mole}\to 1\mathrm{mole}\left[100\times 1=100\mathrm{millimoles}\right]\\ \therefore 100\mathrm{millimoles}\to 100\mathrm{millimoles}{\mathrm{H}}_2\mathrm{S}\mathrm{required}\\ \mathrm{Ratio}\frac{50}{100}=\frac{1}{2}.\end{array}$