RE-NEET 2026 Chemistry Titration MCQ Question

In a redox titration involving oxalic acid and KMnO₄, the balanced reaction shows that 5 moles of oxalic acid react with 2 moles of KMnO₄.

First, calculate the moles of oxalic acid:

$$\text{Moles of oxalic acid} = \text{Volume (L)} \times \text{Molarity} = 0.010 \, \text{L} \times 0.25 \, \text{M} = 0.0025 \, \text{mol}$$

Using the stoichiometry of the reaction:

$$\text{Moles of KMnO₄} = \frac{2}{5} \times \text{Moles of oxalic acid} = \frac{2}{5} \times 0.0025 = 0.001 \, \text{mol}$$

Now, calculate the molarity of KMnO₄:

$$\text{Molarity of KMnO₄} = \frac{\text{Moles}}{\text{Volume (L)}} = \frac{0.001 \, \text{mol}}{0.010 \, \text{L}} = 0.10 \, \text{M}$$

Thus, the strength of the KMnO₄ solution is 0.10 M. Other options fail because they do not align with the stoichiometric calculations based on the reaction.