`10^(5)` coloumb charge liberated 1 gm silver (Ag). If now charge is doubled then the amount of liberated Ag will be :

Options

A. a. 1 gm
B. b. 2 gm
C. c. 3 gm
D. d. 4 gm

Explanation

To solve the problem, we will use the relationship between the mass of a substance liberated during electrolysis and the charge passed through the electrolyte. According to Faraday's laws of electrolysis, the mass (m) of a substance liberated is directly proportional to the quantity of charge (Q) passed through the electrolyte.

### Step-by-Step Solution:

1. **Identify the initial conditions**:
- Given: \( Q_1 = 10^5 \) coulombs (initial charge)
- Mass of silver liberated with this charge: \( m_1 = 1 \) gram

2. **Understand the relationship**:
- According to Faraday's law, the mass liberated is directly proportional to the charge:
\[ m \propto Q \]
- This can be expressed as:
\[ \frac{m_1}{m_2} = \frac{Q_1}{Q_2} \]

3. **Determine the new charge**:
- The problem states that the charge is doubled:
\[ Q_2 = 2 \times Q_1 = 2 \times 10^5 \text{ coulombs} \]

4. **Set up the equation**:
- Using the proportionality relationship:
\[ \frac{m_1}{m_2} = \frac{Q_1}{Q_2} \]
- Substituting the known values:
\[ \frac{1 \text{ gram}}{m_2} = \frac{10^5}{2 \times 10^5} \]

5. **Simplify the equation**:
- This simplifies to:
\[ \frac{1}{m_2} = \frac{1}{2} \]

6. **Solve for \( m_2 \)**:
- Cross-multiplying gives:
\[ m_2 = 2 \text{ grams} \]

### Final Answer:
The amount of liberated silver (Ag) when the charge is doubled is **2 grams**.