A 1.00 g sample of bleaching powder is treated with excess KI and dilute acid. The liberated iodine requires 98.4 mL of 0.100 M Na2S2O3. Take Cl = 35.5, so the molar mass of Cl2 is 71 g mol^-1. Calculate the percentage of available chlorine by mass, correct to one decimal place.

JEE Advanced levelNumerical answerDifficulty 3 of 3ph189

A $1.00 g$ sample of bleaching powder is treated with excess KI and dilute acid. The liberated iodine requires $98.4 mL$ of $0.100 M$ $Na X_{2} S X_{2} O X_{3}$ . Take Cl = 35.5, so the molar mass of $Cl X_{2}$ is $71 g mo l^{- 1}$ . Calculate the percentage of available chlorine by mass, correct to one decimal place.

Numerical answer type, answer in %, tolerance 0.05.

Available chlorine is the

Cl X_{2}

liberated by acid, and each

Cl X_{2}

forms one

I X_{2}

, which needs 2

S X_{2} O X_{3} X^{2 -}

n (S X_{2} O X_{3} X^{2 -}) = 0.100 \times 0.0984 = 9.84 \times 1 0^{- 3} mol

, so

n (Cl X_{2}) = 4.92 \times 1 0^{- 3} mol

. Mass of

Cl X_{2} = 4.92 \times 1 0^{- 3} \times 71 = 0.3493 g

. Percentage

= 0.3493/1.00 \times 100 = 34.9%