1. Write the balanced redox reaction in acidic medium

$2\,\mathrm{MnO_4^-} + 5\,\mathrm{H_2C_2O_4} + 6\,\mathrm{H^+} \rightarrow 2\,\mathrm{Mn^{2+}} + 10\,\mathrm{CO_2} + 8\,\mathrm{H_2O}$

So, the mole ratio is:

$\mathrm{KMnO_4} : \mathrm{H_2C_2O_4} = 2:5$

2. Calculate moles of oxalic acid used

Given:

• Volume of oxalic acid $= 20\,\mathrm{mL} = 0.020\,\mathrm{L}$
• Molarity of oxalic acid $= 2\,\mathrm{M}$

Hence,

$n(\mathrm{H_2C_2O_4}) = M \times V = 2 \times 0.020 = 0.040\,\mathrm{mol}$

3. Use stoichiometric ratio to find moles of $\mathrm{KMnO_4}$

From the balanced equation,

$\frac{n(\mathrm{KMnO_4})}{n(\mathrm{H_2C_2O_4})} = \frac{2}{5}$

So,

$n(\mathrm{KMnO_4}) = \frac{2}{5} \times 0.040 = 0.016\,\mathrm{mol}$

4. Calculate molarity of $\mathrm{KMnO_4}$ solution

Given volume of $\mathrm{KMnO_4}$ used:

$2\,\mathrm{mL} = 0.002\,\mathrm{L}$

Therefore,

$M = \frac{n}{V} = \frac{0.016}{0.002} = 8\,\mathrm{M}$

5. Final answer

$\boxed{8}$

This matches the stored correct answer.