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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..

Himanshu Yadav , 12 Years ago
Grade 10
anser 1 Answers
vaibhav jon

Last Activity: 8 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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