Q: 10 ml of a solution of H_{2}O_{2} labelled ’10 volume’ just decolorises 100 ml of potassium permanganate solution acidified with dilute H_{2}SO_{4}. Calculate the amount of potassium permanganate in the given solution.

(A) 0.1563 gm

(B) 0.563 gm

(C) 5.63 gm

(D) 0.256 gm

Sol: 2H_{2}O_{2} → H_{2}O + O_{2}

22400 ml of O_{2} evolved from 68 gm of H_{2}O_{2}

∴ 10 ml of O_{2} is evolved from $\frac{680}{22400}$ gm of H_{2}O_{2}

Hence 1 ml of H_{2}O_{2} contain $ \frac{0.0303}{34} mol $ = 0.00178 equivalent 10 ml of H_{2}O_{2} will 0.0178 equivalents which will be present in 100 ml of KMnO_{4} solution.

Amount of KMnO_{4} in given sample $=\frac{158}{5} \times 0.00178 $ = 0.563 gm

Ans: (B)