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5.7g of bleaching powder was suspended in 500ml of water. 25ml of this solution on treatment with KI in the presence of HCL liberated iodine which reacted with 24.35ml of N/10 na2s2o3 calculate the % of available chlorine in the bleaching powder...plz explain it in a simple manner..

5.7g of bleaching powder was suspended in 500ml of water. 25ml of this solution on treatment with KI in the presence of HCL liberated iodine which reacted with 24.35ml of N/10 na2s2o3 calculate the % of available chlorine in the bleaching powder...plz explain it in a simple manner..

Grade:10

1 Answers

vaibhav jon
11 Points
4 years ago
Mili eq of Na2 S2 O3 = 24.35*1/10=2.435This would be milli eq of I2 and therefore that of Cl2 (which liberates I2 from KI)Milli eq of Cl2 in 500 ml = 2.435*20=48.7Milli eq of Cl2 = meq of bleaching powder = meq of available Cl2 in bleaching powder% of Cl = 48.7/1000 * 35.5/5.7 * 100 =30.33%(here * means multiply)

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