Same amount of electric current is passed through the solutions of `AgNO_3` and HCI. If 1.08 g of silver is obtained from `AgNO_3` solution. The amount of hydrogen liberted at STP will be

Options

A. 1.008 g
B. 11.2 g
C. 0.01 g
D. 1.1 g

Explanation

To solve the problem, we need to use the concept of electrolysis and the relationship between the amount of substance deposited or liberated during electrolysis and the amount of electric current passed through the solutions.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of silver (Ag) deposited = 1.08 g
- Molar mass of silver (Ag) = 108 g/mol
- Molar mass of hydrogen (H₂) = 2 g/mol
- The same amount of electric current is passed through both solutions.

2. **Calculate the Number of Moles of Silver Deposited:**
\[ \text{Number of moles of Ag} = \frac{\text{Mass of Ag}}{\text{Molar mass of Ag}} = \frac{1.08 \, \text{g}}{108 \, \text{g/mol}} = 0.01 \, \text{mol} \]

3. **Determine the Number of Equivalents of Silver:**
- The n-factor for silver (Ag) is 1 (since it gains 1 electron to form Ag).
\[ \text{Number of equivalents of Ag} = \text{Number of moles of Ag} = 0.01 \, \text{mol} \]

4. **Relate the Equivalents of Silver to Hydrogen:**
- Since the same amount of electric current is passed, the number of equivalents of hydrogen (H₂) liberated will be equal to the number of equivalents of silver deposited.
\[ \text{Number of equivalents of H₂} = 0.01 \, \text{mol} \]

5. **Calculate the Mass of Hydrogen Liberated:**
- The n-factor for hydrogen (H₂) is 2 (since it requires 2 electrons to form 1 mole of H₂).
\[ \text{Mass of H₂} = \text{Number of equivalents of H₂} \times \text{Equivalent weight of H₂} \]
- Equivalent weight of H₂ = \(\frac{\text{Molar mass of H₂}}{\text{n-factor}} = \frac{2 \, \text{g/mol}}{2} = 1 \, \text{g/equiv}\)
\[ \text{Mass of H₂} = 0.01 \, \text{mol} \times 1 \, \text{g/equiv} = 0.01 \, \text{g} \]

### Final Answer:
The amount of hydrogen liberated at STP is **0.01 g**.