The weight of KMnO₄ that can oxidise 100 ml of. 0.2M oxalic acid in acidic medium is [Mol. wt of KMnO₄ = 158]

$2KMn{O}_4 + 5 H_2{C}_2{O}_4 + 6HCl \to 2MnC{l}_2 + 10C{O}_2 + 8 H_{2}O$

${\mathrm{n}}_{\mathrm{oxalic} \mathrm{acid}}= 0.2\times \frac{100}{1000}=0.02$

From (1)

$\frac{5}{2}mol of H_2{C}_2{O}_4 require 1 mol KMn{O}_4 = 158 g KMn{O}_4$

$$\begin{gathered} 0.02 mol of H_2{C}_2{O}_4 require 1 mol of KMn{O}_4 = \frac{158}{5}\times 0.02 \\ = 1.26 g of KMn{O}_4 \end{gathered}$$

Hence, correct option is: (B) 1.264g