1.5 gm sample of bleaching powder was suspended in water. It was treated with C H 3 C O O H followed by the addition of excess of KI. The liberated iodine required 150 mL of M 10 hypo solution for com

Question

1.5 gm sample of bleaching powder was suspended in water. It was treated with C H 3 C O O H followed by the addition of excess of KI. The liberated iodine required 150 mL of M 10 hypo solution for complete titration. The percentage of available chlorine in the sample is

TestHub · Accepted Answer

2 K I + C l 2 → 2 K C l + I 2 I 2 + 2 S 2 O 3 2 - → 2 I - + S 4 O 6 2 - Mol of available C l 2 ≡ mole of iodine = 1 × 150 × 1 0 - 3 10 × 2 Amount of available C l 2 = 150 × 1 0 - 3 20 × 71 g % of available C l 2 = 150 × 1 0 - 3 × 71 × 100 20 × 1.5 = 35.5

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