Question

$5$ g of a sample of bleaching powder is treated with excess acetic acid and $KI$ solution. The liberated $I_2$ required $50$ mL of $\frac{N}{10}$ hypo. The percent of available chlorine in the sample is :

• A. 3.55
• B. 7
• C. 35.5
• D. 28.2

Answer 1

The correct option is A 3.55

$CaOC{l}_2+2KI+2C{H}_{3}COOH\to I_2+CaC{l}_2+H_{2}O+2C{H}_{3}COOK$

$2N{a}_2{S}_2{O}_3+I_2\to N{a}_2{S}_4{O}_6+2NaI$

Moles of $N{a}_2{S}_2{O}_3=\frac{0.1\times 50}{1000}$.

So, moles of $I_2=CaOC{l}_2=2.5\times {10}^{-3}$,

mass of $CaOC{l}_2=\frac{2.5\times 127}{1000}=0.3175$ g
So, % of chlorine in sample $=\frac{0.3175\times 71\times 100}{127}=3.55$ %.