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 =965000\, C)$

• A. 10
• B. 50
• C. 1000
• D. 100

Answer 1

The Correct Option is 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 $=\frac{108 \times 125}{1000} g Ag$ $= 13/5 \,g\,Ag$ $108\, g$ of $Ag$ is deposited by $=96500\, C$ $\therefore 13.5\, g$ of $Ag$ is deposited by $=\frac{96500}{108} \times 13.5$ $=12062.5\, C$ $Q=i t$ or $t=\frac{Q}{i}=\frac{12062.5}{241.25}=50$