10.0 mL of 0.05 M ${\mathrm{KMnO}}_4$ solution was consumed in a titration with 10.0 mL of given oxalic acid dehydrate solution. The strength of given oxalic acid solution is ___ × ${10}^{-2} \mathrm{g}/\mathrm{L}$.

$$\begin{gathered} {\mathrm{n}}_{\mathrm{eq}}{\mathrm{KMnO}}_4={\mathrm{n}}_{\mathrm{eq}}{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4·2{\mathrm{H}}_2\mathrm{O} \\ {\mathrm{n}}_{\mathrm{eq}}=\frac{\mathrm{V}\left(\mathrm{ml}\right)\mathrm{M}}{1000}(\mathrm{n}-\mathrm{factor}) \end{gathered}$$
Or, $\frac{10\times 0.05}{1000}\times 5=\frac{10\times M}{1000}\times 2$
Conc. of oxalic acid solution =0.125 M = 0.125$\times 126 \mathrm{g}/\mathrm{L}=15.75 \mathrm{g}/\mathrm{L}$
                                             =$1575\times {10}^{-2} \mathrm{g}/\mathrm{L}$

Hence, the correct answer is 1575 X 10^-2 g / L.