Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

A solution of  \mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3  is electroly zed for \mathrm{X~min} with a current of  1.8 A to deposit 0.6247g of Fe. Find the value of X.

Option: 1

2523 Sec


Option: 2

1279 Sec


Option: 3

1182 Sec


Option: 4

1794 Sec


Answers (1)

best_answer

\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3 \rightarrow 2 \mathrm{Fe} \\
\mathrm{ \omega=\text { zit } }

\mathrm{ \omega=\left(\frac{E}{96500}\right) i t }
\mathrm{ 0.6247=\left(\frac{56}{3 \times 96500}\right) \times 1.8 \times t \\ }

\mathrm{ t=1794 \mathrm{sec} \\ }.
 

Posted by

Pankaj Sanodiya

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026 application correction begins on jeemain.nta.nic.in; edit details by December 2
anam-khan

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus
anam-khan

JEE Main 2026 correction window opens tomorrow; NTA allows exam city change

Latest Question