Calculate the percentage of available chlorine in a sample of 3.546 g of bleaching powder $\left(CaOC{l}_2\right)$ which is dissolved in water to have 100 mL of solution, if 25 mL of this solution, when treated with acetic acid and excess KI , requires 20 mL of 0.125 N sodium thiosulphate solution.

The correct option is C 10.01 %
 ${\text{CaOCl}}_2\left(\text{aq}\right)+2{\text{CH}}_3\text{COOH}\left(\text{aq}\right)\to \text{Ca}({\text{CH}}_3{\text{COO}}_2+{\text{Cl}}_2( \text{g})+{\text{H}}_2\text{O}\left(\text{l}\right)$
 ${\text{Cl}}_2( \text{g})+2\text{KI}\left(\text{aq}\right)\to 2\text{KCl}\left(\text{s}\right)+{\text{I}}_2( \text{g})$
 ${\text{I}}_2\left(\text{aq}\right)+2{\text{Na}}_2 {\text{S}}_2{\text{O}}_3\left(\text{aq}\right)\to {\text{Na}}_2 {\text{S}}_4{\text{O}}_6\left(\text{aq}\right)+2\text{NaI}\left(\text{aq}\right)$
Equivalents of ${\text{Na}}_2 {\text{S}}_2{\text{O}}_3=$  Equivalents of ${\text{I}}_2=$  Equivalents of  ${\text{Cl}}_2$
 $\therefore$ Equivalents of  ${\text{Cl}}_2=\frac{0.125\times 2\text{O}}{1000}=2.5\times {10}^{-3}$
Amount of  ${\text{Cl}}_2=\frac{2.5\times {10}^{-3}\times 71}{2}=0.08875 \text{g}$
wt. of  ${\text{Cl}}_2$  evolved from $100 \text{mL}=\frac{0.08875\times 100}{25}=0.355$ 
$\therefore \%$  of avaliable chlorine  $=\frac{0.355}{3546}\times 100=10.01\%$