2.00 g of bleaching powder is made up to 250 mL. A 25.0 mL portion is treated with excess KI and acid (Cl2 + 2I- -> 2Cl- + I2) and the iodine needs 22.6 mL of 0.100 M Na2S2O3. Taking the molar mass of Cl2 as 71, the percentage of available chlorine is

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$2.00 g$ of bleaching powder is made up to $250 mL$ . A $25.0 mL$ portion is treated with excess $KI$ and acid ( $Cl X_{2} + 2 I X^{-} 2 Cl X^{-} + I X_{2}$ ) and the iodine needs $22.6 mL$ of $0.100 M$ $Na X_{2} S X_{2} O X_{3}$ . Taking the molar mass of $Cl X_{2}$ as $71$ , the percentage of available chlorine is

A4.0%
B40.1%
C80.2%
D20.1%

Answer: B

mol thiosulphate = 2.26e-3; I2 = 1.13e-3 = mol Cl2 in 25.0 mL; in 250 mL = 1.13e-2 mol. Mass Cl2 = 1.13e-2 x 71 = 0.8023 g. Percent = 0.8023/2.00 x 100 = 40.11%.

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2.00 g of bleaching powder is made up to 250 mL. A 25.0 mL portion is treated with excess KI and acid (ClX2​+2IX−​2ClX−+IX2​) and the iodine needs 22.6 mL of 0.100 M NaX2​SX2​OX3​. Taking the molar mass of ClX2​ as 71, the percentage of available chlorine is

Answer: B

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