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In an acidic medium, 10 mL of 0.25 M oxalic acid is titrated with $KMnO_4$ solution. If the volume of $KMnO_4$ solution required to reach end point is 10 mL, the strength of the $KMnO_4$ solution is
Detailed Solution
Equivalents of oxalic acid = equivalents of $KMnO_4$: $M_1\times n_1\times V_1=M_2\times n_2\times V_2 \Rightarrow 0.25\times2\times10=M_2\times5\times10 \Rightarrow M_2=0.10$ M.