Q. The amount of silver deposited by passing 241.25C of current through silver nitrate solution is
A) 2.7 g
B) 2.7 mg
C) 0.27 g
D) 0.54 g
Explanation
(c)
Given, current = 241.25 C
We know that 1 C electricity will deposit $1.118\ \times \ 10^{- 3}g$ of silver.
∴ 241.25 C electricity will deposit
$= \ (1.118\ \times \ 10^{- 3})\ \times \ 241.25$
= 0.27 g of silver.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.