Search any question & find its solution
Question:
Answered & Verified by Expert
What is the time (in sec) required for depositing all the silver present in $125 \mathrm{~mL}$ of $1 \mathrm{M} \mathrm{AgNO}{ }_3$ solution by passing a current of $241.25 \mathrm{~A}$ ? (1F $=96500$ coulombs $)$
Options:
Solution:
2541 Upvotes
Verified Answer
The correct answer is:
50
Given $125 \mathrm{~mL}$ of $1 \mathrm{M} \mathrm{AgNO}_3$ solution. It means that
$\because 1000 \mathrm{~mL}$ of $\mathrm{AgNO}_3$ solution contains
$=108 \mathrm{~g} \mathrm{Ag}$
$\therefore 125 \mathrm{~mL}$ of $\mathrm{AgNO}_3$ solution contains
$=\frac{108 \times 125}{1000} \mathrm{~g} \mathrm{Ag}$
$=13.5 \mathrm{~g} \mathrm{Ag}$
$\because 108 \mathrm{~g}$ of $\mathrm{Ag}$ is deposited by $96500 \mathrm{C}$
$\therefore 13.5 \mathrm{~g}$ of $\mathrm{Ag}$ is deposited by $=\frac{96500}{108} \times 13.5$
$=12062.5 \mathrm{C}$
$Q=i t$
$t=\frac{Q}{i}=\frac{12062.5}{241.25}=50$
$\because 1000 \mathrm{~mL}$ of $\mathrm{AgNO}_3$ solution contains
$=108 \mathrm{~g} \mathrm{Ag}$
$\therefore 125 \mathrm{~mL}$ of $\mathrm{AgNO}_3$ solution contains
$=\frac{108 \times 125}{1000} \mathrm{~g} \mathrm{Ag}$
$=13.5 \mathrm{~g} \mathrm{Ag}$
$\because 108 \mathrm{~g}$ of $\mathrm{Ag}$ is deposited by $96500 \mathrm{C}$
$\therefore 13.5 \mathrm{~g}$ of $\mathrm{Ag}$ is deposited by $=\frac{96500}{108} \times 13.5$
$=12062.5 \mathrm{C}$
$Q=i t$
$t=\frac{Q}{i}=\frac{12062.5}{241.25}=50$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.