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Q.
Passage of $10800 \,C$ of electricity through the electrolyte deposited $2.977\, g$ of metal with atomic mass $106.4 \,g \,mol^{-1}$. The charge on the metalcation is
Electrochemistry
Solution:
Let the charge be $n^+$
$\therefore M^{n+} + ne^- \rightarrow M$
$106.4 \,g$ of metal require charge $ = n \times 96500\, C$
$2.977\, g$ of metal require charge
$ = (\frac{n\times 96500}{106.4} \times2.977)$
$\therefore \frac{n \times 96500}{106.4} \times 2.977$
$ = 10800$
$ n = \frac{10800\times 106.4}{96500\times 2.977} \approx 4$