Question

Passage of $10800C$ of electricity through the electrolyte deposited $2.977g$ of metal with atomic mass $106.4gmo{l}^{-1}$. The charge on the metalcation is

• A. +4
• B. +3
• C. +2
• D. +1

Answer 1

Answer: (A)

Let the charge be $n^{+}$
$\therefore M^{n+}+n{e}^{-}\to M$
$106.4g$ of metal require charge $=n\times 96500C$
$2.977g$ of metal require charge
$=(\frac{n\times 96500}{106.4}\times 2.977)$
$\therefore \frac{n\times 96500}{106.4}\times 2.977$
$=10800$
$n=\frac{10800\times 106.4}{96500\times 2.977}\approx 4$