1. Find moles of NaCl present

Given:

• Concentration of NaCl solution $= 4.0\,\text{M}$
• Volume electrolysed $= 500\,\text{mL} = 0.5\,\text{L}$

So, moles of NaCl are: $n = M \times V = 4.0 \times 0.5 = 2.0\,\text{mol}$

Thus, the solution contains:

• $2.0$ mol $\text{Na}^+$
• $2.0$ mol $\text{Cl}^-$

2. Electrode reactions during electrolysis

At the anode, chloride ions are oxidized: $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-$

At the mercury cathode, sodium ions are reduced and dissolve in Hg to form sodium amalgam: $\text{Na}^+ + e^- + \text{Hg} \rightarrow \text{Na(Hg amalgam)}$

Each mole of $\text{Na}^+$ requires $1$ mole of electrons.

Since there are $2.0$ mol of $\text{Na}^+$ available, at maximum all can be converted into amalgam.

3. Amount of amalgam formed

If all $2.0$ mol of sodium are discharged, then sodium formed in amalgam = $2.0$ mol.

Mass of sodium deposited: $m_{\text{Na}} = 2.0 \times 23 = 46\,\text{g}$

Now amalgam contains sodium dissolved in mercury. In such questions, amalgam is taken as the compound-like combination of Na with Hg in molar ratio $1:1$.

So for $2.0$ mol Na, mercury needed = $2.0$ mol Hg.

Mass of Hg: $m_{\text{Hg}} = 2.0 \times 200 = 400\,\text{g}$

Hence total mass of amalgam formed: $m_{\text{amalgam}} = 46 + 400 = 446\,\text{g}$

4. Check options

• A: $200$ g
• B: $225$ g
• C: $400$ g
• D: $446$ g

Therefore, the correct option is: $\boxed{D\; (446\,\text{g})}$

5. Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

So the derived answer agrees with the stored answer.