In the electrolysis of ${\mathrm{CuSO}}_4$ solution, $6.35 \mathrm{g}$ of copper was deposited in a coulometer. If the same amount of electricity is passed through silver nitrate solution, the mass of $\mathrm{Ag}$ deposited at cathode is

The reaction is${\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to \mathrm{Cu}\left(\mathrm{s}\right)$, i.e. $1 \mathrm{mol}$ of $\mathrm{Cu}$is deposited by 2mol of electrons. For the deposition of$6.35 \mathrm{g} \mathrm{Cu} (= 0.1 \mathrm{mol} \mathrm{Cu}), 0.2 \mathrm{mol}$ of electrons will be needed. The deposition of $\mathrm{Ag}\left(\mathrm{s}\right)$involves the reaction .${\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}\to \mathrm{Ag}\left(\mathrm{s}\right)$ Thus, for $0.2 \mathrm{mol}$of electrons, $0.2 \mathrm{mol}$ of$\mathrm{Ag}$ will be deposited which is $0.2 \mathrm{X} 108 \mathrm{g}$i.e. $21.6 \mathrm{g}$