Question

$1.25g$ of bleaching powder when treated with acetic acid and excess of $KI$, liberated iodine which required $37.5mL$ of $0.2N$ sodium thiosulphate solution. Calculate the percentage of available chlorine in the sample of bleaching powder.

• A. $30.0\%$
• B. $21.3\%$
• C. $12.5\%$
• D. $9.7\%$

Answer 1

The correct option is B $21.3\%$

$CaOC{l}_2\left(aq\right)+2C{H}_{3}COOH\left(aq\right)\to Ca(C{H}_{3}COO{)}_2+C{l}_2(g)+H_{2}O(l)$
$C{l}_2\left(g\right)+2KI\left(aq\right)\to 2KCl\left(s\right)+I_2\left(g\right)$
$I_2\left(aq\right)+2N{a}_2{S}_2{O}_3\left(aq\right)\to N{a}_2{S}_4{O}_6\left(aq\right)+2NaI\left(aq\right)$

$\text{Equivalents of}N{a}_2{S}_2{O}_3=\text{Equivalents of}{I}_2=\text{Equivalents of}C{l}_2$
$\therefore \text{Equivalents of}C{l}_2=\frac{0.2\times 37.5}{1000}=0.0075$
$\text{Amount of}C{l}_2=\frac{0.0075\times 71}{2}=0.266g$
$\%$ of $C{l}_2=\frac{0.266}{1.25}\times 100\%=21.3\%$