Question

Impure copper containing $Fe,Au,Ag$ as impurities is electrolytically refined. A current of $140A$ for $482.5s$ decreased the mass of the anode by $22.26g$ and increased the mass of cathode by $22.011g$. Percentage of iron in impure copper is :
(Given molar mass $Fe=55.5g{mol}^{-1}$, molar mass $Cu=63.54g{mol}^{-1}$)

• A. $0.85$
• B. $0.90$
• C. $0.95$
• D. $0.97$

Answer 1

The correct option is B $0.90$

$22.26-22.011=0.249$ g of impurity is present.
A current of 140 A for 482.5 s corresponds to $140\times 482.5=67550$ C of electricity.
Equivalent weight of Cu is 31.77 g.
The mass of Cu that should be deposited $=\frac{67550}{96500}\times 31.77=22.239$ g
Increase in mass of anode $=22.011$ g.
$22.239-22.011=0.228$ g corresponds to the mass of Cu that is not deposited. This can be equated to the mass of iron that passed into solution.
The equivalent mass of Fe is 27.75 g.
The mass of Fe $=0.228\times 27.7531.77=0.199$ g.
The percent of Fe $=\frac{0.199\times 100}{22.26}=0.90$%
Hence, the option (B) is the correct answer.