If 0.5 A current is passed through acidified silver nitrate solution for 10 minutes. The mass of silver deposited on cathode, is (eq. wt. of silver nitrate = 108)

$\begin{array}{l}\text{Given }\mathrm{I}=0.5\text{ amp}\\ \mathrm{t}=10\times 60\sec \\ \mathrm{Q}=\mathrm{I}\times \mathrm{t}=0.5\times 10\times 60=300\mathrm{C}\\ {\mathrm{AgNO}}_3\to {\mathrm{Ag}}^{+}+{\mathrm{NO}}_3^{-}\\ \text{Anode: }{\mathrm{NO}}_3^{-}\to 1{\mathrm{e}}^{-}+{\mathrm{NO}}_2+\frac{1}{2}{\mathrm{O}}_2\\ 1\mathrm{F}\to 1\text{ mole of Ag}\\ 1\mathrm{F}\to 108\text{ g of Ag}\\ \text{96500 C}\to \text{108 g of Ag}\\ \text{300 C }\to \frac{300\times 108}{96500}\text{g of AgO}\\ =0.336 \mathrm{gm}\end{array}$