What is the time (in sec) required for depositing all the silver present in 125 ml of 1M $AgN{O}_3$ solution by passing a current of 241.25 amperes? (1F = 96500 coulombs)

$$\begin{gathered} \text{ Moles of }{\mathrm{AgNO}}_3=\mathrm{M}\times \mathrm{V} \\ =1\times \frac{125}{1000}=0.125\text{ mole } \\ {\mathrm{Ag}}^{+}+{\mathrm{e}}^{-}\to \mathrm{Ag} \\ 1 \mathrm{mole} \mathrm{Ag} \mathrm{and} 1 \mathrm{mole} {\mathrm{e}}^{-} \mathrm{are} \mathrm{needed}. \\ \mathrm{Requires} 0.125 \mathrm{mole} \mathrm{of} \mathrm{Ag} \mathrm{and} 0.125 \mathrm{mole} \mathrm{of} {\mathrm{e}}^{-}. \\ \text{ Total Charge }=0.125\times 96500 \\ 0.125\times 96500=\mathrm{t}\times 241.25 \\ \mathrm{t}=50 \sec \end{gathered}$$

Therefore, the correct option is B