Step-by-Step Solution

1. Calculate the initial number of moles of NaCl. The problem provides the molarity and volume of the NaCl solution.

   • Molarity (M) = 4.0 mol/L
   • Volume (V) = 500 mL = 0.5 L

   The number of moles of solute (NaCl) can be calculated using the formula: $\text{Moles} = \text{Molarity} \times \text{Volume (in L)}$ $\text{Moles of NaCl} = 4.0 \, \text{mol/L} \times 0.5 \, \text{L} = 2.0 \, \text{mol}$

2. Determine the number of moles of chloride ions ($Cl^-$). Sodium chloride (NaCl) is a strong electrolyte and dissociates completely in water according to the equation: $NaCl(aq) \rightarrow Na^+(aq) + Cl^-(aq)$ From the stoichiometry of this dissociation, 1 mole of NaCl produces 1 mole of chloride ions ($Cl^-$). Therefore, the number of moles of $Cl^-$ ions in the solution is equal to the number of moles of NaCl. $\text{Moles of } Cl^- = \text{Moles of NaCl} = 2.0 \, \text{mol}$

3. Analyze the electrolysis process and write the relevant half-reaction. During the electrolysis of aqueous NaCl, chlorine gas ($Cl_2$) is evolved at the anode. The half-reaction for this oxidation process is: $2Cl^-(aq) \rightarrow Cl_2(g) + 2e^-$

4. Calculate the total number of moles of chlorine gas ($Cl_2$) evolved. The question asks for the total number of moles of chlorine gas evolved, which implies that the electrolysis is carried out until all the available chloride ions are consumed. From the stoichiometry of the anode half-reaction, 2 moles of chloride ions ($Cl^-$) produce 1 mole of chlorine gas ($Cl_2$). We can use this ratio to find the moles of $Cl_2$ produced from 2.0 moles of $Cl^-$. $\text{Moles of } Cl_2 = \text{Moles of } Cl^- \times \frac{1 \, \text{mol } Cl_2}{2 \, \text{mol } Cl^-}$ $\text{Moles of } Cl_2 = 2.0 \, \text{mol} \times \frac{1}{2} = 1.0 \, \text{mol}$ Thus, a total of 1.0 mole of chlorine gas is evolved.

5. Conclusion The calculated number of moles of chlorine gas evolved is 1.0. Comparing this with the given options: A: 0.5 B: 1.0 C: 2.0 D: 3.0

   The correct option is B.