Q: 10 ml of a solution of H2O2 labelled ’10 volume’ just decolorises 100 ml of potassium permanganate solution acidified with dilute H2SO4. 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
2H2O2 → H2O + O2
22400 ml of O2 evolved from 68 gm of H2O2
∴ 10 ml of O2 is evolved from gm of H2O2
Hence 1 ml of H2O2 contain mol = 0.00178 equivalent 10 ml of H2O2 will 0.0178 equivalents which will be present in 100 ml of KMnO4 solution.
Amount of KMnO4 in given sample = = 0.563 gm