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0.5 g of bleaching powder liberates iodine from ki that reacts with 20 ml of 0.1n sodiumthiosulphate solution. what is the active oxygen content in bleaching powder?
CaOCl2 + 2KI ------ 2KCl + I2 + oxygen + CaI2 + 2Na2S2O3 --------- Na2S4O6 + 2NaIThe above reaction should be remembered Moles of Na2S2O3 = NV/1000 = 20×0.1/1000 = 0.002 moles1 mole of I2 react with 2 mole of Na2S2O3Hence moles of Na2S2O3 = 0.002÷2 = 0.001Number of moles of I2 = number of moles of O = 0.001 moles Mass of O = 0.001×16 = 0.016 g%of oxygen in given sample = 0.016 × 100/ 0.5 = 3.2 %
Dear student 60mL of 0.5N Na2S2O3 = 60mL 0.5N I2 Amt of Cl = 35.5*0.5*60/1000 = 1.065 g % of available cl = 1.065 /3.55 * 100 = 30%
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