Search any question & find its solution
Question:
Answered & Verified by Expert
In producing chlorine through electrolysis $100 \mathrm{~W}$ power at $125 \mathrm{~V}$ is being consumed. How much chlor ine per minute is leberated? (ECE of chlor ine is $0.367 \times 10^{-6} \mathrm{~kg} / \mathrm{C}$ )
Options:
Solution:
2810 Upvotes
Verified Answer
The correct answer is:
$17.6 \mathrm{mg}$
$$
\begin{array}{l}
\mathrm{P}=\mathrm{VI} \Rightarrow \mathrm{I}=\frac{P}{V}=\frac{100}{125} \mathrm{~A} \\
\text { Mass of chlorine liberated = zit } \\
=0.367 \times 10^{-6} \times \frac{100}{125} \times 60=17.6 \mathrm{mg}
\end{array}
$$
\begin{array}{l}
\mathrm{P}=\mathrm{VI} \Rightarrow \mathrm{I}=\frac{P}{V}=\frac{100}{125} \mathrm{~A} \\
\text { Mass of chlorine liberated = zit } \\
=0.367 \times 10^{-6} \times \frac{100}{125} \times 60=17.6 \mathrm{mg}
\end{array}
$$
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.