A certain current liberates `0.504g` of `H_(2)(g)` in 2 hours. The weight of `Cu` deposited by same current flowing for the same time in `CuSO_(4)` solution is `……………………..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 gas (H₂) liberated, \( w_1 = 0.504 \, \text{g} \)
- Time of electrolysis, \( t = 2 \, \text{hours} \) (not directly needed for calculations)
- Equivalent weight of hydrogen, \( e_1 = 1 \, \text{g/equiv} \) (since H₂ has a molar mass of 2 g/mol and valency is 2)
- Equivalent weight of copper (Cu), \( e_2 = \frac{63.5}{2} = 31.75 \, \text{g/equiv} \)

2. **Use Faraday’s Second Law of Electrolysis:**
\[ \frac{w_1}{w_2} = \frac{e_1}{e_2} \]
where \( w_2 \) is the weight of copper deposited.

3. **Substitute the Known Values:**
\[ \frac{0.504}{w_2} = \frac{1}{31.75} \]

4. **Cross Multiply to Solve for \( w_2 \):**
\[ 0.504 \cdot 31.75 = 1 \cdot w_2 \]
\[ w_2 = 0.504 \cdot 31.75 \]

5. **Calculate \( w_2 \):**
\[ w_2 = 0.504 \cdot 31.75 = 16.007 \, \text{g} \]

6. **Final Answer:**
The weight of copper deposited by the same current flowing for the same time in copper sulfate solution is approximately \( 16.007 \, \text{g} \).