20 ml of 0.02 M ${\mathrm{KMnO}}_4$ was required to completely oxidise 10 ml of oxalic acid solution. What is the molarity of the oxalic acid solution?

The concerned reaction is 

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

[For ${\mathrm{KMnO}}_4$]                         [For oxalic acid]

${\mathrm{V}}_1=20\mathrm{ml}$                              ${\mathrm{V}}_2=10\mathrm{ml}$

${\mathrm{M}}_1=0.02\mathrm{M}$                           ${\mathrm{M}}_2=?$

${\mathrm{n}}_1=2, {\mathrm{n}}_2=5$

$\frac{{\mathrm{M}}_1{\mathrm{V}}_1}{{\mathrm{n}}_1}=\frac{{\mathrm{M}}_2{\mathrm{V}}_2}{{\mathrm{n}}_2}$

$\frac{0.02\times 20}{2}=\frac{{\mathrm{M}}_2\times 10}{5}$

${\mathrm{M}}_2=\frac{0.02\times 20\times 5}{10\times 2}=0.1\mathrm{M}$