On passing 0.1 faraday of electricity through fused sodium chloride,
the amount of chlorine liberated is (At. Mass of `Cl=35.45`)

Options

A. 35.45g
B. 70.9g
C. 3.545g
D. 17.77g

Explanation

To solve the problem of determining the amount of chlorine liberated when 0.1 Faraday of electricity is passed through fused sodium chloride, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- When sodium chloride (NaCl) is fused, it dissociates into sodium ions (Na⁺) and chloride ions (Cl⁻).
- The relevant half-reaction for the liberation of chlorine gas (Cl₂) from chloride ions is:
\[ 2 \text{Cl}^- \rightarrow \text{Cl}_2 + 2 \text{e}^- \]
- This indicates that 2 moles of Cl⁻ are required to produce 1 mole of Cl₂.

2. **Calculating Moles of Electrons**:
- One Faraday (1 F) of electricity corresponds to the transfer of 1 mole of electrons (approximately 96500 coulombs).
- Therefore, 0.1 Faraday corresponds to:
\[ 0.1 \text{ F} = 0.1 \text{ moles of electrons} \]

3. **Relating Moles of Electrons to Moles of Chlorine**:
- From the half-reaction, we see that 2 moles of electrons are required to produce 1 mole of Cl₂.
- Thus, 0.1 moles of electrons will produce:
\[ \text{Moles of Cl}_2 = \frac{0.1 \text{ moles of electrons}}{2} = 0.05 \text{ moles of Cl}_2 \]

4. **Calculating Mass of Chlorine**:
- The molar mass of Cl₂ (chlorine gas) is:
\[ \text{Molar mass of Cl}_2 = 2 \times 35.45 \text{ g/mol} = 70.90 \text{ g/mol} \]
- Therefore, the mass of chlorine liberated can be calculated using the number of moles:
\[ \text{Mass of Cl}_2 = \text{Moles of Cl}_2 \times \text{Molar mass of Cl}_2 = 0.05 \text{ moles} \times 70.90 \text{ g/mol} = 3.545 \text{ g} \]

5. **Final Answer**:
- The amount of chlorine liberated when 0.1 Faraday of electricity is passed through fused sodium chloride is **3.545 grams**.