A current liberates 0.50 8 of hydrogen in 2 hours. The weight of copper (At. wt. = 635) deposited at the same time by the same current through copper sulphate solution is :

Options

A. `63*5 g`
B. `31*8 g`
C. `15*9 g`
D. `15*5 g`

Explanation

To solve the problem, we will use Faraday's second law of electrolysis, which states that the weights of substances deposited or liberated at an electrode during electrolysis are proportional to their equivalent weights.

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Weight of hydrogen (H2) liberated = 0.508 g
- Time = 2 hours (not needed for calculations)
- Atomic weight of copper (Cu) = 63.5 g/mol

2. **Determine the Equivalent Weight of Hydrogen**:
- The molecular weight of hydrogen (H2) = 2 g/mol.
- The n-factor for hydrogen (H2) = 2 (since 2 moles of electrons are required to liberate 1 mole of H2).
- Therefore, the equivalent weight of hydrogen = Molecular weight / n-factor = 2 g/mol / 2 = 1 g/equiv.

3. **Determine the Equivalent Weight of Copper**:
- The n-factor for copper (Cu) = 2 (since 2 moles of electrons are required to deposit 1 mole of Cu).
- Therefore, the equivalent weight of copper = Molecular weight / n-factor = 63.5 g/mol / 2 = 31.75 g/equiv.

4. **Set Up the Proportionality According to Faraday's Second Law**:
\[ \frac{\text{Weight of H2}}{\text{Equivalent weight of H2}} = \frac{\text{Weight of Cu}}{\text{Equivalent weight of Cu}} \]
Substituting the known values:
\[ \frac{0.508 \text{ g}}{1 \text{ g/equiv}} = \frac{\text{Weight of Cu}}{31.75 \text{ g/equiv}} \]

5. **Cross-Multiply to Solve for Weight of Copper**:
\[ 0.508 \text{ g} \times 31.75 \text{ g/equiv} = \text{Weight of Cu} \]
\[ \text{Weight of Cu} = 0.508 \times 31.75 = 16.15 \text{ g} \]

6. **Final Calculation**:
- The weight of copper deposited is approximately 16.15 g.

### Conclusion:
The weight of copper deposited at the same time by the same current through copper sulphate solution is approximately **16.15 grams**.