5.00 g of bleaching powder is shaken with water and made up to 500 mL. A 25.0 mL portion is treated with excess KI and acid, and the iodine liberated needs 20.0 mL of 0.100 M $Na X_{2} S X_{2} O X_{3}$ . The percentage of available chlorine (as $Cl X_{2}$ , Cl = 35.5) in the sample is:

Answer: C

Thiosulphate = 2.00 mmol, so

I X_{2}

= 1.00 mmol =

Cl X_{2}

available in 25.0 mL. In 500 mL: 20.0 mmol. Mass =

20.0 \times 71 = 1420

mg. Percentage = 1420/5000 x 100 = 14.2%.