1.50 g of bleaching powder is suspended in water, acidified and treated with excess KI. The liberated iodine requires 42.0 mL of 0.200 M Na2S2O3. Taking available chlorine as Cl2 (71 g/mol; Cl2 + 2I- -> 2Cl- + I2), find the percentage available chlorine by mass (one decimal place).

JEE Main levelNumerical answerDifficulty 2 of 3eq140

1.50 g of bleaching powder is suspended in water, acidified and treated with excess KI. The liberated iodine requires 42.0 mL of 0.200 M $Na X_{2} S X_{2} O X_{3}$ . Taking available chlorine as $Cl X_{2}$ (71 g/mol; $Cl X_{2} + 2 I X^{-} 2 Cl X^{-} + I X_{2}$ ), find the percentage available chlorine by mass (one decimal place).

Numerical answer type, answer in %, tolerance 0.06.

Thiosulfate = 0.200

\times

42.0 = 8.40 mmol, which equals 4.20 mmol

I X_{2}

, and hence 4.20 mmol

Cl X_{2}

. Mass of

Cl X_{2}

= 4.20

\times

71 = 298.2 mg. Percentage = 0.2982/1.50

\times

100 = 19.9 %.