Question

What is the time (in sec) required for depositing all the silver present in $125mL$ of $1MAgN{O}_3$ solution by passing a current of $241.25A?$
$(1F=96500$ coulombs $)$

• A. 10
• B. 50
• C. 1000
• D. 100

Answer 1

Answer: (B)

Given $125mL$ of $1MAgN{O}_3$ solution. It means that
$\because 1000mL$ of $AgN{O}_3$ solution contains
$=108gAg$
$\therefore 125mL$ of $AgN{O}_3$ solution contains
$=\frac{108\times 125}{1000}gAg$
$=13.5gAg$
$\because 108g$ of $Ag$ is deposited by $96500C$
$\therefore 13.5g$ of $Ag$ is deposited by $=\frac{96500}{108}\times 13.5$
or $=12062.5C$
$Q=it$
$t=\frac{Q}{i}=\frac{12062.5}{241.25}=50$