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.

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

Sol:  2H2O2 → H­2O + O2

22400 ml of O2 evolved from 68 gm of H2O2

∴ 10 ml of O­2 is evolved from $\frac{680}{22400}$ gm of H2O2

Hence 1 ml  of H2O2 contain  $ \frac{0.0303}{34} 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 $=\frac{158}{5} \times 0.00178 $  = 0.563 gm

Ans: (B)