The quantity of electricity required to deposit `1*15g` of sodium from molten `NaCl (Na = 23, Cl = 35*5)` is

Options

A. `1 F`
B. `0*5 F`
C. `0*05 F`
D. `1*5 F`

Explanation

To solve the problem of calculating the quantity of electricity required to deposit 1.15 g of sodium from molten NaCl, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Molar Mass of Sodium (Na)**:
- The atomic mass of sodium (Na) is given as 23 g/mol.

2. **Determine the Number of Moles of Sodium**:
- To find the number of moles of sodium in 1.15 g, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
- Substituting the values:
\[ \text{Number of moles of Na} = \frac{1.15 \, \text{g}}{23 \, \text{g/mol}} = 0.05 \, \text{mol} \]

3. **Understand the Electrochemical Reaction**:
- In the electrolysis of molten NaCl, sodium ions (Na⁺) are reduced to sodium metal (Na) by gaining one electron (e⁻):
\[ \text{Na}^+ + e^- \rightarrow \text{Na} \]
- This means that 1 mole of Na requires 1 mole of electrons for reduction.

4. **Calculate the Number of Faradays Required**:
- According to Faraday's laws of electrolysis, 1 mole of electrons corresponds to 1 Faraday (F).
- Therefore, the number of Faradays required to deposit 0.05 moles of sodium is:
\[ \text{Faradays required} = \text{Number of moles of Na} = 0.05 \, \text{F} \]

5. **Final Answer**:
- The quantity of electricity required to deposit 1.15 g of sodium from molten NaCl is **0.05 Faraday**.