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

In an acidic medium, the reaction of KMnO$_4$ with oxalic acid involves the following changes in oxidation states:

For KMnO$_4$, Mn$^{7+} \rightarrow$ Mn$^{2+}$, so the n-factor is $5$.

For oxalic acid (H$_2$C$_2$O$_4$), C$^{3+} \rightarrow$ C$^{4+}$, so the n-factor is $2 \times 1 = 2$.

At the equivalence point, the number of equivalents of KMnO$_4$ equals the number of equivalents of oxalic acid:

$N_1 V_1 = N_2 V_2$

$(M_1 \times n_1) V_1 = (M_2 \times n_2) V_2$

Substituting the given values:

$M_1 \times 5 \times 10 = 0.25 \times 2 \times 10$

$50 M_1 = 5$

$M_1 = 0.10$ M

The strength of the KMnO$_4$ solution is $0.10$ M.