The redox reaction between oxalic acid ($H_{2}C_{2}O_{4}$) and potassium permanganate ($KMnO_{4}$) in an acidic medium is: $$2KMnO_{4} + 5H_{2}C_{2}O_{4} + 3H_{2}SO_{4} \rightarrow K_{2}SO_{4} + 2MnSO_{4} + 10CO_{2} + 8H_{2}O$$
From the balanced equation, 2 moles of $KMnO_{4}$ react with 5 moles of oxalic acid. Therefore, the mole ratio is: $$\frac{n_{KMnO_{4}}}{n_{oxalic}} = \frac{2}{5}$$
Given volume $V_{ox} = 10\text{ mL} = 0.01\text{ L}$ and molarity $M_{ox} = 0.25\text{ M}$. $$n_{oxalic} = M \times V = 0.25 \times 0.01 = 0.0025\text{ moles}$$
Using the stoichiometry: $$n_{KMnO_{4}} = \frac{2}{5} \times n_{oxalic} = \frac{2}{5} \times 0.0025 = 0.001\text{ moles}$$ Given volume of $KMnO_{4}$ is $10\text{ mL} = 0.01\text{ L}$. $$M_{KMnO_{4}} = \frac{n}{V} = \frac{0.001}{0.01} = 0.10\text{ M}$$
Final Answer: 0.10 M
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