Q. The amount of charge required to liberate $9\ gm$ of aluminium (atomic weight =27 and valency = 3) in the process of electrolysis is (Faraday’s number =96500 $coulombs/gm$ equivalent)
A) $321660\ coulombs$
B) $69500\ coulombs$
C) $289500\ coulomb$
D) $96500\ coulomb$
Explanation
(d) Equivalent weight of aluminium $= \frac{27}{3} = 9$
So $1\ faraday = 96500\ C$ are required to liberate $9\ gm$ of $Al$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.