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NEET UGChemistryModerateRe-NEET 2026

In an acidic medium, 10 mL of 0.25 M oxalic acid is titrated with KMnO4KMnO_4 solution. If the volume of KMnO4KMnO_4 solution required to reach end point is 10 mL, the strength of the KMnO4KMnO_4 solution is

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KMnO4+H2C2O4→H+Mn2++2CO2+H2OKMnO_4 + H_2C_2O_4 \xrightarrow{H^+} Mn^{2+} + 2CO_2 + H_2O

No. of equivalent of KMnO4KMnO_4 = No. of equivalent of H2C2O4H_2C_2O_4

10×5×M=10×0.25×210 \times 5 \times M = 10 \times 0.25 \times 2

M=550M = \frac{5}{50}

M=110=0.10M = \frac{1}{10} = 0.10 M

Source: geekyrakshit/indian-exam-questions / reneet2026-code50

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