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A 7.1 g sample of bleaching powder suspended in `H_(2)O` was treated with enough acetic acid and KI solution. Iodine thus liberated required 80 mL of 0.2 N hypo solution for titration. Calcutale the % of available chlorine :

Answer» Correct Answer - 8
molesof iodine = moles of chlorine `=(80xx0.2)/(2)xx10^(-3)=8xx10^(-3)`
so required % `=(8xx71xx10^(-3))/(7.1)xx100%=8%`


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