108 g fairly concentrated solution of `AgNO_(3)` is electrolysed by using 0.1 F charge the mass of resulting solution is
(a)94g
(b)11.6g
(c)96.4g
(d)None of these

Options

A. 94g
B. 11.6g
C. 96.4g
D. None of these

Answer

To solve the problem of determining the mass of the resulting solution after electrolyzing 108 g of a fairly concentrated solution of AgNO₃ using 0.1 F charge, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Electrolysis Reaction**:
- At the cathode, silver ions (Ag⁺) are reduced to solid silver (Ag).
- At the anode, water is oxidized, producing oxygen gas (O₂) and hydrogen ions (H⁺).

2. **Calculate Moles of Silver Deposited**:
- Given that 0.1 Faraday of charge is used, we can determine how many moles of silver are deposited.
- The reaction for the deposition of silver is:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]
- 1 mole of electrons (1 Faraday) deposits 1 mole of silver. Therefore, 0.1 Faraday will deposit 0.1 moles of silver.

3. **Calculate the Mass of Silver Deposited**:
- The molar mass of silver (Ag) is approximately 108 g/mol.
- For 0.1 moles of silver:
\[ \text{Mass of Ag} = 0.1 \, \text{moles} \times 108 \, \text{g/mol} = 10.8 \, \text{g} \]

4. **Calculate Moles of Oxygen Produced**:
- The reaction at the anode can be simplified as:
\[ 2 \text{H}_2\text{O} \rightarrow \text{O}_2 + 4 \text{H}^+ + 4 e^- \]
- This indicates that 4 moles of electrons produce 1 mole of O₂. Therefore, 0.1 Faraday (which corresponds to 0.1 moles of electrons) will produce:
\[ \text{Moles of O}_2 = \frac{0.1}{4} = 0.025 \, \text{moles} \]

5. **Calculate the Mass of Oxygen Produced**:
- The molar mass of oxygen (O₂) is approximately 32 g/mol.
- For 0.025 moles of oxygen:
\[ \text{Mass of O}_2 = 0.025 \, \text{moles} \times 32 \, \text{g/mol} = 0.8 \, \text{g} \]

6. **Calculate the Total Mass of Substances Removed**:
- The total mass of substances removed from the solution (mass of silver deposited + mass of oxygen produced):
\[ \text{Total mass removed} = 10.8 \, \text{g} + 0.8 \, \text{g} = 11.6 \, \text{g} \]

7. **Calculate the Mass of the Resulting Solution**:
- The initial mass of the solution was 108 g. Therefore, the mass of the resulting solution after electrolysis is:
\[ \text{Mass of resulting solution} = 108 \, \text{g} - 11.6 \, \text{g} = 96.4 \, \text{g} \]

### Final Answer:
The mass of the resulting solution is **96.4 g** (Option c).