Question

$6.35$ g of impure bleaching powder paste is mixed with $KI$. $I_3^{-}$ formed requires $50$ mL of $1M$ hypo in neutral medium. Thus, $\%$ purity of bleaching powder is :

Answer 1

The reactions are as follows:
$CaOC{l}_2+2KI+H_{2}S{O}_4\to I_2+CaC{l}_2+H_{2}O+K_{2}S{O}_4$
$2$ $1$
$I_2+2S_2{O}_3^{-2}\to S_4{O}_6^{-2}+2I^{-}$
$1$ $2$
So, the number of moles of hypo solution $=50\times \frac{1}{1000}=50\times {10}^{-3}$ moles
So, mass of bleaching powder $50\times {10}^{-3}\times \frac{127}{2}=3.175$ g
So, $\%$ purity of sample $=3.175\times \frac{100}{6.35}=50\%$