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 $241.25 \, A$? ($IF = 96500$ coulombs)

• A. 10
• B. 50
• C. 1000
• D. 100

Answer 1

The Correct Option is B

Given $125 \,mL$ of $1 \,M \,Ag NO _{3}$ solution. It means that $\because \,1000 \,ML$ of $AgNO _{3}$ solution contains $=108\, g \,Ag$ $\therefore 125 \,mL$ of $AgNO _{3}$ solution contains $=\frac{108 \times 125}{1000} \,g \,A g=13.5 \,Ag$ $\because 108\, g$ of Ag is deposited by $96500\, C$ $\therefore 13.5 \,g$ of $Ag$ is deposited by $=\frac{96500}{108} \times 13.5=12062.5 C$ $\because Q=i t$ $\therefore t=\frac{Q}{i}=\frac{12062.5}{241.25}=50$