Ten grams of a fairly concentrated solution of cupric sulphate is electrolysed using 0.01 faraday of electricity. Calculate (i) the weight of the resulting solution and (ii) the number of equivalent of acid or alkali in the solution. (Cu = 63.5, F = 96500 coulombs)

Explanation

To solve the problem step by step, we will first determine the weight of copper deposited during the electrolysis and then calculate the weight of the resulting solution. Finally, we will find the number of equivalents of acid or alkali in the solution.

### Step 1: Calculate the amount of copper deposited
1. **Understanding Faraday's Law**: According to Faraday's law of electrolysis, 1 Faraday (F) deposits 1 mole of monovalent ions or 0.5 moles of divalent ions. Copper (Cu) is a divalent ion (Cu²⁺).
2. **Calculate moles of copper deposited**:
\[ \text{Moles of Cu deposited} = \text{Faraday used} \times \frac{0.5 \text{ moles}}{1 \text{ Faraday}} = 0.01 \text{ F} \times 0.5 = 0.005 \text{ moles} \]
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