A) 0.529 g
B) 10.623 g
C) 0.0529 g
D) 1.2708 g
Correct Answer: C
Solution :
[c] Using Faraday's second law of electrolysis, \[\frac{Weight\,of\,Cu\,deposoited}{Weight\,of\,Ag\,deposited}=\frac{Equ.\,wt.\,of\,Cu}{Equ.\,wt.\,of\,Ag}\] \[\Rightarrow \frac{{{w}_{Cu}}}{0.18}=\frac{63.5}{2}\times \frac{1}{108}\] \[\Rightarrow {{w}_{Cu}}=\frac{63.5\times 18}{2\times 108\times 100}=0.0529g.\]You need to login to perform this action.
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