During the electrolysis of copper (II) tetraoxosulphate (VI) solution, a steady current of 4.0 x 10\(^2\) A flowing for one hour liberated 0.48 g of copper. Calculate the mass of copper liberated by one coulomb of charge.
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Correct Answer: Option n
Explanation:
M = Zit
Z = \(\frac{M}{It}\)
= \(\frac{0.48 \times 10^{-3}}{4 \times 10^2 \times 3600}\) = 3.3 x 10\(^{-10}\)kgc\(^{-1}\)
M = Zit
Z = \(\frac{M}{It}\)
= \(\frac{0.48 \times 10^{-3}}{4 \times 10^2 \times 3600}\) = 3.3 x 10\(^{-10}\)kgc\(^{-1}\)