In the redox reaction,  $2{\mathrm{KMnO}}_4+3{\mathrm{H}}_2{\mathrm{SO}}_4+5{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\to$${\mathrm{K}}_2{\mathrm{SO}}_4+2{\mathrm{MnSO}}_4+10{\mathrm{CO}}_2+8{\mathrm{H}}_2\mathrm{O}\text{, }$

the volume of $0.1M {\mathrm{KMnO}}_4$ required to oxidize 25ml of $0.25M H_2{\mathrm{C}}_2{\mathrm{O}}_4$ solution is.

The provided reaction is:

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

According to the molarity equation

${\left(\frac{M_1{V}_1}{n_1}\right)}_{KMn{O}_4}={\left(\frac{M_2{V}_2}{n_2}\right)}_{\text{oxalic acid }}$

or $\frac{0.1\times V_1}{2}=\frac{0.25\times 25}{5}$

$V_1=\frac{2\times 0.25\times 25}{5\times 0.1}=\frac{12.5}{0.5}$

$V_1=25\mathrm{ml}$

Hence, option(c) is the answer.