1. Write the neutralisation reaction

Oxalic acid is dibasic: $\mathrm{H_2C_2O_4 + 2NaOH \rightarrow Na_2C_2O_4 + 2H_2O}$

So, 1 mole of oxalic acid reacts with 2 moles of NaOH.

2. Calculate moles of oxalic acid used

Given:

• Volume of oxalic acid $= 50\,\mathrm{mL} = 0.050\,\mathrm{L}$
• Molarity of oxalic acid $= 0.5\,\mathrm{M}$

Moles of oxalic acid: $n = M \times V = 0.5 \times 0.050 = 0.025$

So, moles of oxalic acid $= 0.025$ mol.

3. Find moles of NaOH neutralised

From the stoichiometry: $1\text{ mol oxalic acid} : 2\text{ mol NaOH}$

Therefore, $\text{moles of NaOH} = 2 \times 0.025 = 0.050\text{ mol}$

This amount of NaOH is present in $25\,\mathrm{mL}$ of the NaOH solution.

4. Find moles of NaOH in $50\,\mathrm{mL}$

If $25\,\mathrm{mL}$ contains $0.050$ mol, then $50\,\mathrm{mL}$ contains: $0.050 \times 2 = 0.100\text{ mol}$

5. Convert moles into mass

Molar mass of NaOH: $23 + 16 + 1 = 40\,\mathrm{g/mol}$

Mass of NaOH in $50\,\mathrm{mL}$: $m = n \times M = 0.100 \times 40 = 4\,\mathrm{g}$

6. Final answer

The amount of $\mathrm{NaOH}$ in $50\,\mathrm{mL}$ of the solution is: $\boxed{4\,\mathrm{g}}$

7. Comparison with stored answer

Stored correct answer = $4$

Our derived answer also = $4$, so they agree.