A certain current jdeposits `0.50` g of hydrogen gas in 2 hrs. The amount of copper liberated from a solution of copper sulphate by the same current flowing for the same time would be:

Options

A. `31.8` g
B. `63.6` g
C. `15.9` g
D. `6.36` g

Answer

To solve the problem, we will use Faraday's second law of electrolysis, which states that the amount of substance deposited or liberated during electrolysis is directly proportional to the equivalent weight of the substance and the quantity of electricity passed.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of hydrogen deposited (m_H) = 0.50 g
- Time (t) = 2 hours (not directly needed for calculation)
- Equivalent weight of hydrogen (E_H) = Molecular weight of H / n-factor = 1 g/mol (since n-factor for H is 1).

2. **Calculate the Equivalent Weight of Copper:**
- Molecular weight of copper (Cu) = 63.5 g/mol
- n-factor for copper (in CuSO₄) = 2 (since Cu²⁺ ion is formed).
- Therefore, Equivalent weight of copper (E_Cu) = Molecular weight of Cu / n-factor = 63.5 g / 2 = 31.75 g.

3. **Using Faraday's Second Law of Electrolysis:**
- According to Faraday's law, the ratio of the masses deposited is given by:
\[ \frac{m_H}{m_{Cu}} = \frac{E_H}{E_{Cu}} \]
- Substituting the known values:
\[ \frac{0.50 \, \text{g}}{m_{Cu}} = \frac{1}{31.75} \]

4. **Cross-Multiply to Find the Mass of Copper (m_Cu):**
- Rearranging the equation gives:
\[ m_{Cu} = 0.50 \, \text{g} \times \frac{31.75}{1} \]
- Calculating this gives:
\[ m_{Cu} = 0.50 \times 31.75 = 15.875 \, \text{g} \]

5. **Final Answer:**
- The amount of copper liberated from the solution of copper sulfate is approximately **15.9 g**.