Use Cl = 35.5. A 1.00 g sample of bleaching powder is treated with excess acidified KI (OCl- + 2I- + 2H+ -> Cl- + I2 + H2O). The I2 needs 30.0 mL of 0.200 M Na2S2O3 (I2 + 2S2O3^2- -> 2I- + S4O6^2-). Taking available chlorine as the mass of Cl2 equivalent to the I2 liberated (1 mol Cl2 per mol I2), the percentage of available chlorine is:

JEE Advanced levelSingle correct optionDifficulty 3 of 3rx148

Use Cl = 35.5. A $1.00 g$ sample of bleaching powder is treated with excess acidified $KI$ ( $OCl X^{-} + 2 I X^{-} + 2 H X^{+} Cl X^{-} + I X_{2} + H X_{2} O$ ). The $I X_{2}$ needs $30.0 mL$ of $0.200 M$ $Na X_{2} S X_{2} O X_{3}$ ( $I X_{2} + 2 S X_{2} O X_{3} X^{2 -} 2 I X^{-} + S X_{4} O X_{6} X^{2 -}$ ). Taking available chlorine as the mass of $Cl X_{2}$ equivalent to the $I X_{2}$ liberated (1 mol $Cl X_{2}$ per mol $I X_{2}$ ), the percentage of available chlorine is:

Thiosulfate

= 30.0 \times 0.200 = 6.0

mmol, so

I X_{2}

= 3.0

mmol, equivalent to

3.0

mmol

Cl X_{2}

= 3.0 \times 1 0^{- 3} \times 71 = 0.213 g

. Percentage

= 21.3%