Question

To a 25 mL H2O2 solution, excess of acidified solution of KI was added. The iodine liberated required 20 mL

Answer 1

M of H2O2 = M of Na2S2O3.If N = normality of H2O2; Then, N× 25 = 0.3 × 20N=0.24Now , volume strength=N × 5.6=0.24× 5.6 =》 1.344g/L