What is the time (in sec) required for depositing all the silver present in 125 ml of 1M AgNO₃ solution by passing a current of 241.25 amperes ? (1F = 96500 coulombs)

$\mathrm{Moles} \mathrm{of} {\mathrm{AgNO}}_3= \mathrm{Molarity} \times \frac{{\mathrm{V}}_{\left(\mathrm{ml}\right)}}{1000}=1\times \frac{125}{1000}=0.125 \mathrm{moles}$

${\mathrm{Ag}}^{+}+ {\mathrm{e}}^{-}\to \mathrm{Ag}$

1 mole Ag requires 1 mole of e^-.

0.125 moles of Ag requires = 0.125 moles of  electrons = 0.125 X 96500 c 

since , Q= i X t 

0.125 X 96500 = t X 241.25

t= 50 sec.