During the electrolysis of molten NaCl solution, 230 g of sodium metal is deposited on the cathode, then how many moles of chlorine will be obtained at anode?

Options

A. `10.0`
B. `5.0`
C. `35.5`
D. `17.0`

Explanation

To solve the problem of how many moles of chlorine will be obtained at the anode during the electrolysis of molten NaCl when 230 g of sodium metal is deposited at the cathode, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Molar Mass of Sodium (Na)**:
The molar mass of sodium (Na) is approximately 23 g/mol.

2. **Calculate the Number of Moles of Sodium Deposited**:
To find the number of moles of sodium deposited, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
Substituting the values:
\[ \text{Number of moles of Na} = \frac{230 \, \text{g}}{23 \, \text{g/mol}} = 10 \, \text{moles} \]

3. **Understanding the Electrolysis Reaction**:
During electrolysis, sodium ions (Na⁺) are reduced at the cathode to form sodium metal (Na), while chloride ions (Cl⁻) are oxidized at the anode to form chlorine gas (Cl₂). The half-reaction at the anode can be represented as:
\[ 2 \, \text{Cl}^- \rightarrow \text{Cl}_2 + 2 \, e^- \]
This means that 2 moles of chloride ions produce 1 mole of chlorine gas.

4. **Calculate the Moles of Chlorine Produced**:
From the stoichiometry of the reaction, we know that for every 2 moles of Cl⁻, 1 mole of Cl₂ is produced. Since each mole of Na deposited corresponds to 1 mole of Cl⁻ consumed, we can calculate the moles of chlorine produced from the moles of sodium deposited:
\[ \text{Moles of Cl}_2 = \frac{\text{Moles of Na}}{2} = \frac{10 \, \text{moles}}{2} = 5 \, \text{moles} \]

### Final Answer:
Therefore, the number of moles of chlorine gas (Cl₂) obtained at the anode is **5 moles**.

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