One Faraday of electricity liberates $$x \times 10^{-1}$$ gram atom of copper from copper sulphate, $$x$$ is ______.

Correct Answer: 5

$$Cu^{2+} + 2e^- \rightarrow Cu$$. One Faraday = 1 mol electrons.

1 Faraday deposits $$\frac{1}{2}$$ mol Cu = 0.5 gram atom = 5 × 10⁻¹ gram atom. x = 5.

The answer is $$\boxed{5}$$.

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