To find the value of x when one Faraday of electricity liberates x times $10^{-1}$ gram atom of copper from copper sulphate, we need to understand how electricity interacts with copper ions in solution. The key reaction is:

$$\mathrm{Cu}^{2+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}$$

This shows that copper ions (Cu²⁺) gain two electrons (e^-) to become copper metal (Cu). From electrochemistry, we know that:

• 2 Faradays of electricity are required to deposit 1 mole (or 1 gram atom) of copper.
• Therefore, 1 Faraday will deposit half of that amount, which is 0.5 mole, or in other terms, 0.5 gram atom of copper.
• Expressing 0.5 in the form of x times $10^{-1}$ gives us $5 \times 10^{-1}$.

This means x equals 5.