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 on the cathode, we can follow these steps:

### Step 1: Determine the number of moles of sodium deposited
To find the number of moles of sodium (Na) deposited, we can use the formula:

\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

The molar mass of sodium (Na) is approximately 23 g/mol. Given that 230 g of sodium is deposited:

\[ \text{Number of moles of Na} = \frac{230 \, \text{g}}{23 \, \text{g/mol}} = 10 \, \text{moles} \]

### Step 2: Relate moles of sodium to moles of chlorine
During the electrolysis of NaCl, the following reactions occur:

- At the cathode (reduction):
\[ \text{Na}^+ + e^- \rightarrow \text{Na} \]

- At the anode (oxidation):
\[ 2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^- \]

From the cathode reaction, we see that 1 mole of Na is produced for every 1 mole of electrons (1 Farad).

From the anode reaction, we see that 1 mole of Cl2 is produced for every 2 moles of electrons.

### Step 3: Calculate the moles of chlorine produced
Since 10 moles of sodium were deposited, this means that 10 moles of electrons were used (as 1 mole of Na corresponds to 1 mole of electrons).

Now, to find out how many moles of Cl2 are produced from the 10 moles of electrons:

\[ \text{Moles of Cl}_2 = \frac{\text{Moles of electrons}}{2} = \frac{10}{2} = 5 \, \text{moles} \]

### Conclusion
Therefore, the number of moles of chlorine gas (Cl2) produced at the anode is **5 moles**.

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