108 g solution of `AgNO_(3)` is electrolysed using Pt electrodes by passing a charge of 0.1 F. The mass of resultant solution left is :

Options

A. 98g
B. 107.2g
C. 11.6g
D. `96.4g`

Explanation

To solve the problem step by step, we need to analyze the electrolysis of the silver nitrate (AgNO3) solution and calculate the mass of the resultant solution left after passing a charge of 0.1 Faraday.

### Step 1: Determine the amount of Ag deposited
When AgNO3 is electrolyzed, silver ions (Ag⁺) are reduced at the cathode to form solid silver (Ag). The reaction at the cathode can be represented as:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]

From the stoichiometry of the reaction, we know that:
- 1 mole of Ag (which has a molar mass of 108 g) requires 1 mole of electrons (1 Faraday).

### Step 2: Calculate the mass of Ag deposited using the charge passed
Given that we are passing a charge of 0.1 Faraday:
- The amount of silver deposited can be calculated as follows:
\[ \text{Mass of Ag deposited} = \left( \frac{0.1 \text{ F}}{1 \text{ F}} \right) \times 108 \text{ g} = 10.8 \text{ g} \]

### Step 3: Determine the amount of O2 evolved
At the anode, the oxidation of water occurs, producing oxygen gas (O2). The reaction can be represented as:
\[ 2 \text{H}_2\text{O} \rightarrow \text{O}_2 + 4 \text{H}^+ + 4 e^- \]

From the stoichiometry of this reaction:
- 4 Faraday produces 1 mole of O2 (32 g).
- Therefore, 1 Faraday produces 8 g of O2.

Since we are passing 0.1 Faraday:
\[ \text{Mass of O2 evolved} = \left( \frac{0.1 \text{ F}}{4 \text{ F}} \right) \times 32 \text{ g} = 0.8 \text{ g} \]

### Step 4: Calculate the total mass loss
The total mass loss from the solution due to the deposition of silver and the evolution of oxygen is:
\[ \text{Total mass loss} = \text{Mass of Ag deposited} + \text{Mass of O2 evolved} \]
\[ \text{Total mass loss} = 10.8 \text{ g} + 0.8 \text{ g} = 11.6 \text{ g} \]

### Step 5: Calculate the mass of the resultant solution left
The initial mass of the solution is 108 g. Therefore, the mass of the resultant solution left after electrolysis is:
\[ \text{Mass of resultant solution} = \text{Initial mass} - \text{Total mass loss} \]
\[ \text{Mass of resultant solution} = 108 \text{ g} - 11.6 \text{ g} = 96.4 \text{ g} \]

### Final Answer
The mass of the resultant solution left is **96.4 g**.

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