NTA Abhyas JEE Main2020ChemistryRedox ReactionsPractice
To a 25 mL H 2 O 2 solution, excess of acidified solution of potassium iodide was added. The iodine liberated required 20 mL of 0.3 N sodium thiosulphate solution. Calculate the volume strength of H 2 O 2 solution and report your answer by multiplying it with 1000.
Correct answer
1334
Step-by-step solution
Meq of H 2 O 2 = Meq of I 2 = Meq of Na 2 S 2 O 3 . If N is normality of H 2 O 2 , then N x 25 = 0.3 x 20, N H 2 O 2 = 0 . 2 4 N M H 2 O 2 = 0.12 M N = 0.24 Volume strength = 0.12 × 11.2 =1.334 Vol Final answer = 1.334 × 1000 = 1334