An electric current is passed through two voltameters connected in series and containing `CuSO_4` and `AgNO_3` solutions respectively. The masses of copper and silver deposited are 0.424 g and 1.44 g respectively. Find the equivalent mass of silver if that of copper is 31.75.

Answer

To solve the problem, we will use Faraday's second law of electrolysis, which states that the mass of a substance deposited during electrolysis is directly proportional to the equivalent weight of that substance and the amount of electric charge passed through the electrolyte.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of copper deposited (m_Cu) = 0.424 g
- Mass of silver deposited (m_Ag) = 1.44 g
- Equivalent mass of copper (E_Cu) = 31.75 g

2. **Apply Faraday's Second Law:**
According to Faraday's second law:
\[ \frac{m_{Cu}}{m_{Ag}} = \frac{E_{Cu}}{E_{Ag}} \]
where \(E_{Ag}\) is the equivalent mass of silver that we need to find.

3. **Substituting the Values:**
Substitute the known values into the equation:
\[ \frac{0.424 \, \text{g}}{1.44 \, \text{g}} = \frac{31.75 \, \text{g}}{E_{Ag}} \]

4. **Cross-Multiply to Solve for \(E_{Ag}\):**
Cross-multiplying gives:
\[ 0.424 \, \text{g} \cdot E_{Ag} = 1.44 \, \text{g} \cdot 31.75 \, \text{g} \]

5. **Calculate the Right Side:**
Calculate \(1.44 \times 31.75\):
\[ 1.44 \times 31.75 = 45.72 \, \text{g} \]

6. **Rearranging the Equation:**
Now, we can rearrange the equation to find \(E_{Ag}\):
\[ E_{Ag} = \frac{45.72 \, \text{g}}{0.424 \, \text{g}} \]

7. **Perform the Division:**
Calculate \(E_{Ag}\):
\[ E_{Ag} = \frac{45.72}{0.424} \approx 107.8 \, \text{g} \]

8. **Conclusion:**
The equivalent mass of silver is approximately 107.8 g.

### Final Answer:
The equivalent mass of silver is **107.8 g**.