Explanation

(b)
Given 125mL of 1 M $AgNO_{3}$solution. It means that
∵ 1000mL of $AgNO_{3}$ solution contains
= 108 g Ag
∴ 125mL of $AgNO_{3}$ solution contains
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ \frac{108\ \times 125}{1000}\ gAg$
= 13.5 g Ag
∵ 108 g of Ag is deposited by 96500 C
∴ 13.5 g of Ag is deposited by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ \frac{96500}{108}\ \times 13.5$
= 12062.5 C
$\mathcal{\ \ \ \ \ \ \ \ \ \ \ \ Q =}it$
Or $t\ = \ \frac{\mathcal{Q}}{i}\ = \ \ \frac{12062.5}{241.25}\ = \ 50$