Acidified ${\mathrm{KMnO}}_4$ oxidizes oxalic acid to ${\mathrm{CO}}_2$. What is the volume (in ml) of ${10}^{-1} M$ ${\mathrm{KMnO}}_4$ required to completely oxidize 0.5 liters of ${10}^{-2}$ M oxalic acid in the acid medium?

The above mentioned reaction gives,

$2{\mathrm{KMnO}}_4+3{\mathrm{H}}_2{\mathrm{SO}}_4+5\underset{\mathrm{COOH}}{\mathrm{COOH}}\to {\mathrm{K}}_2{\mathrm{SO}}_4+2{\mathrm{MnSO}}_4+10{\mathrm{CO}}_2+8{\mathrm{H}}_2\mathrm{O}$

$\frac{{\mathrm{M}}_1{ \mathrm{V}}_1}{{\mathrm{n}}_1}\left({\mathrm{KMnO}}_4\right)=\frac{{\mathrm{M}}_2{ \mathrm{V}}_2}{{\mathrm{n}}_2}\left(\begin{array}{c}\mathrm{COOH}\\ \mathrm{COOH}\end{array}\right)$

From the formula,

$\frac{{10}^{-4}\times V_1}{2}=\frac{{10}^{-2}\times 0.5}{5}$

Therefore,

${ \mathrm{V}}_1=20 \mathrm{L}.$

Hence, option D is the correct answer