Question

3.0 g sample of bleaching powder is dissolved in 100 mL water. 20 mL of this suspension is acidified with HCl and treated with
excess Kl. The la liberated exactly requires 15.21 mL of 0.1 Na2S2O3 for complete reduction. Percentage available chlorine in the
sample of bleaching powder is​

Answer 1

To find:- Percentage available chlorine in the  sample of bleaching powder.

Solution:-

                             KL + H C L          Na₂S₂O₃

Bleaching powder-----------------> I₂--------------> I⁻ + Na₂S₂O₆

In following way redox changes,

I₂ +2 e------>2 I⁻

2(S²⁺)------->(S^5/2⁺)₄+2 e

Milliequivalent of bleaching powder = Milliequivalent of chlorine=  Milliequivalent of Na₂S₂O₃ used= 15.21  * 0.1

Available chlorine in the  sample of bleaching powder=15.21 × 0.1× 100/20

                                                                                        =7.605

$or,\frac{w}{(71/2)} *1000 =7.605$

$or, w_{Cl 2} = 0.269 g$

Percentage available chlorine in the  sample of bleaching powder is ,(0.629/3.0)×100=20.96 %.

Hence, percentage available chlorine in the  sample of bleaching powder is 20.96 %.(Ans)