$100$ ml of $KMn{O}_4$ solution is exactly reduced by $100$ ml of $0.5$ M oxalic acid under acidic condition. The molarity of $KMn{O}_4$ solution is:

The correct option is B $0.2$ M

$2KMn{O}_4+5H_2{C}_2{O}_4+6H^{+}\to 2M{n}^{2+}+10C{O}_2+8H_{2}O$

$5\times M_{KMn{O}_4}\times V_{KMn{O}_4}=2\times M_{H_2{C}_2{O}_4}\times V_{H_2{C}_2{O}_4}$

$5\times M_{KMn{O}_4}\times \text{100 mL}=2\times \text{0.5 M}\times \text{100 mL}$

$M_{KMn{O}_4}=0.2$ M.

The molarity of $KMn{O}_4$ solution is $0.2$ M.