Question

Calculate the percentage of available chlorine in a sample of $3.546g$ of bleaching powder which was dissolved in $100mL$ of water, $25mL$ of this solution, when treated with acetic acid and excess $KI$ , required $20mLof0.125N$ sodium thiosulphate solution.

• A. $15.01\%$
• B. $0.1\%$
• C. $10.01\%$
• D. $20.02\%$

Answer 1

The correct option is C $10.01\%$

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

$\text{Equivalents of}N{a}_2{S}_2{O}_3=\text{Equivalents of}{I}_2=\text{Equivalents of}C{l}_2$

$\therefore \text{Equivalents of}C{l}_2=\frac{0.125\times 20}{1000}=2.5\times {10}^{-3}$
$\text{Amount of}C{l}_2=\frac{2.5\times {10}^{-3}\times 71}{2}=0.08875g$

$\text{wt. of}C{l}_2\text{evolved from}100mL=\frac{0.08875\times 100}{25}=0.355$
$\therefore \%\text{of avaliable chlorine=}\frac{0.355}{3.546}\times 100=10.01\%$