The ratio of amounts of H₂S needed to precipitate all the metal ions from 100 ml of 1 M AgNO₃ and 100 ml of 1 M CUSO₄ will be

${\mathrm{AgNO}}_3\equiv 2{\mathrm{Ag}}^{+}+\underset{\left({\mathrm{H}}_2\mathrm{S}\right)}{{\mathrm{S}}^{2-}}\to {\mathrm{Ag}}_2\mathrm{S}$

$\because 2\mathrm{mol}\to 1\mathrm{mol}[100\times 1=100\text{ millimole }]$

$\therefore 100\text{ millimole }\to 50\text{ millimole }{\mathrm{H}}_2\mathrm{S}\text{ required }$

${\mathrm{CuSO}}_4\equiv {\mathrm{Cu}}^{+2}+\underset{\left({\mathrm{H}}_2\mathrm{S}\right)}{{\mathrm{S}}^2}\to \mathrm{CuS}$

$\because 1\mathrm{mol}\to 1\mathrm{mol}[100\times 1=100\text{ millimole }]$

$\therefore 100\text{ millimole }\to 100\text{ millimole }{\mathrm{H}}_2\mathrm{S}\text{ required }$

$\text{ Ratio }\frac{50}{100}=\frac{1}{2}\text{. }$