5H₂O₂ + 2MnO₄^- + 6H⁺ $\\ \to$ 2Mn²⁺ + 5O₂ + 8H₂O
Moles of KMnO₄ = ${{0.316} \over {158}}$ = 2 $\times$ 10^-3
Equivalents of H₂O₂ = Equivalent of KMnO₄
= 2 × 10^–3 × 5 = 0.01
Moles of H₂O₂ = ${{0.01} \over 2}$ = 0.005
Mass of pure H₂O₂ = 0.005 × 34 = 0.170 gm
Percentage purity = ${{0.17} \over {0.2}} \times$ 100 = 85 %