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Q.
What is the time required (in seconds) for depositing the silver present in $125 ml$ of $1 M AgNO _{3}$ solution passing a current of $241.25 A [1 F =96500 C ] ?$
Electrochemistry
Solution:
$1000\, ml$ of $AgNO _{3}$ solution contains $108 \, g \, Ag$
$\therefore 125\, ml$ of $AgNO _{3}$ solution will contain
$=\frac{108 \times 125}{1000}=13.5 \, g \, Ag$
$Ag ^{+}+ e ^{-} \rightarrow Ag $
(1 F =96500) 108 g
$108 g$ of $Ag$ is deposited by $96500 C$
$\therefore 13.5$ of $Ag$ is deposited $by =\frac{96500}{108} \times 13.5=12062.5 C$
$Q = It $
$t =\frac{ Q }{ I }=\frac{12062.5}{241.25}=50$ sec