A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed . The leads to the evolution of chlorine gas at one the electrodes (atomic masses : Na = 23 , Hg = 200, 1 Faraday = 96500 coulombs ). The total number of moles of chlorine gas evolved is

Options

A. 0.5
B. `1.0`
C. `2.0`
D. `3.0`

Answer

To solve the problem, we need to determine the total number of moles of chlorine gas (Cl₂) evolved during the electrolysis of a 4.0 M NaCl solution when 500 mL of it is electrolyzed.

### Step 1: Calculate the number of moles of NaCl in the solution.
To find the number of moles, we use the formula:

\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (in L)} \]

Given:
- Molarity of NaCl = 4.0 M
- Volume of solution = 500 mL = 0.500 L

\[ \text{Number of moles of NaCl} = 4.0 \, \text{mol/L} \times 0.500 \, \text{L} = 2.0 \, \text{mol} \]

### Step 2: Determine the number of moles of Cl⁻ ions.
When NaCl dissociates in solution, it produces Na⁺ and Cl⁻ ions:

\[ \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \]

From 2.0 moles of NaCl, we get 2.0 moles of Cl⁻ ions.

### Step 3: Calculate the moles of Cl₂ produced.
During electrolysis, chlorine gas (Cl₂) is produced at the anode from the oxidation of Cl⁻ ions:

\[ 2 \text{Cl}^- \rightarrow \text{Cl}_2 + 2 \text{e}^- \]

This reaction shows that 2 moles of Cl⁻ produce 1 mole of Cl₂. Therefore, the number of moles of Cl₂ produced from 2.0 moles of Cl⁻ is:

\[ \text{Moles of Cl}_2 = \frac{\text{Moles of Cl}^-}{2} = \frac{2.0 \, \text{mol}}{2} = 1.0 \, \text{mol} \]

### Final Answer:
The total number of moles of chlorine gas evolved is **1.0 mol**.

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