Question

What is the time (in sec) required for depositing all the silver present in $125 \,mL$ of $1 \,M\, AgNO _{3}$ solution by passing a current of $241.25\, A ?$
$(1\, F =96500$ coulombs $)$

• A. 10
• B. 50
• C. 1000
• D. 100

Answer 1

Answer: (B)

Given $125\, mL$ of $1 \,M \,AgNO _{3}$ solution. It means that
$\because 1000\, mL$ of $AgNO _{3}$ solution contains
$=108\, g\, Ag$
$\therefore 125\, mL$ of $AgNO _{3}$ solution contains
$=\frac{108 \times 125}{1000}\, g\, Ag$
$=13.5\, g \,Ag$
$\because 108\, g$ of $Ag$ is deposited by $96500 \,C$
$\therefore 13.5 \,g$ of $Ag$ is deposited by $=\frac{96500}{108} \times 13.5$
or $=12062.5 C$
$Q=i t$
$t=\frac{Q}{i}=\frac{12062.5}{241.25}=50$