One Faraday of electricity liberates \(x×10^{−1 }\)gram atom of copper from copper sulphate, x is ______.
Correct Answer: 5
Solution and Explanation
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]
$2 \, \text{Faraday} \rightarrow 1 \, \text{mol Cu}$
$1 \, \text{Faraday} \rightarrow 0.5 \, \text{mol Cu deposit}$
$0.5 \, \text{mol} = 0.5 \, \text{g atom} = 5 \times 10^{-1}$
\[ x = 5 \]
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