What time in sec is required for depositing all the silver present in $125 \mathrm{ml}$ of $1 \mathrm{M} {\mathrm{AgNO}}_3$ solution by passing a current of $241.25$ A (Given $1 \mathrm{F}= 96500 \mathrm{C}$)

Given, $125 \mathrm{ml}$ of $1 \mathrm{M} {\mathrm{AgNO}}_3$

which is equal to $13.5 \mathrm{g}$ of $\mathrm{Ag}$

$\because 1000 \mathrm{ml}$ of ${\mathrm{AgNO}}_3=108 \mathrm{g}$ of $\mathrm{Ag}$

$\therefore 125 \mathrm{ml}$ of ${\mathrm{AgNO}}_3=\frac{108\times 125}{1000}=13.5 \mathrm{g}$

Also, $\because 108 \mathrm{g}$ of $\mathrm{Ag}$ is deposited by $=96500 \mathrm{C}$

$\therefore 13.5 \mathrm{g}$ of $\mathrm{Ag}$ is deposited by $=\frac{96500\times 13.5}{108}=12062.5 \mathrm{C}$

$\because \mathrm{Q}=\mathrm{it}$

$\therefore \mathrm{t}$ (time) $=\mathrm{Q}/\mathrm{i}=\frac{12062.50}{241.25}, \mathrm{t}=50 \sec$