Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Open MARKS Web Download App
Search any question & find its solution
Question: Answered & Verified by Expert
On passing 'C ampere of current for time 't' sec through 1 L of 2 (M) $\mathrm{CuSO}_{4}$ solution (atomic weight of $\mathrm{Cu}=63.5$ ), the amount ${ }^{\prime} \mathrm{m}^{\prime}$ of Cu (in gram) deposited on cathode will be
ChemistryElectrochemistryWBJEEWBJEE 2012
Options:
  • A $m=C t /(63.5 \times 96500)$
  • B $m=C t /(31.25 \times 96500)$
  • C $m=(C \times 96500) /(31.25 \times t)$
  • D $m=(31.25 \times C \times t) / 96500$
Solution:
1021 Upvotes Verified Answer
The correct answer is: $m=(31.25 \times C \times t) / 96500$
$m=\frac{E G}{F}$ $=\frac{\frac{63.5}{2} \times C \times t}{96500}=\frac{31.75 \times C \times t}{96500}$

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.

Use on iOS or Desktop Download from Google Play