Question

$5.0$ g of bleaching powder was suspended in water and volume made up to half a litre. $20$ mL of this suspension when acidified with acetic acid and treated with excess of potassium iodide solution liberated iodine, which required $20$ mL of a deci-normal hypo solution for titration. Calculate the value of $10\times x$, where $x$ is the percentage of available chlorine in bleaching powder?

Answer 1

The chemical reactions are as follows:

$CaOC{l}_2+H_{2}2O\to Ca\left(OH\right)+2+C{l}_2$

$C{l}_2+2KI\to 2KCl+I_2$

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

$1$ mole $CaOC{l}_2=1$ mole $C{l}_2=1$ mole $I_2=2$ moles $N{a}_2{S}_2{O}_3$

The equivalent mass of $C{l}_2=\frac{71}{2}=35.5$

$N_1{V}_1(C{l}_2$ in bleaching powder) $=N_2{V}_2$ (hypo)
$N_1\times 20=0.1\times 20$
$N_1=0.1N$

The normality of $CaOC{l}_2$ solution is $0.1N$.

The amount of chlorine is $0.1\times 35.5g{L}^{-1}=3.55g{L}^{-1}$ in pure sample.

The impure sample contains $\frac{5.0g}{0.5L}=10g{L}^{-1}$chlorine.

The percentage of pure chlorine $\left(x\right)$ available in the bleaching powder is $\frac{3.55}{10}\times 100=35.5\%$.

So, $10x=355$.