6.35g of impure bleaching powder paste is mixed with $K I$ , and $I_{2}$ formed requires 50 ml of 1M hypo in neutral medium. Thus, the percentage purity of the bleaching powder is__________.

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Solution

The reactions involved are:
$C a O C l_{2} + 2 K I + H_{2} S O_{4} \to I_{2} + C a C l_{2} + H 2 O + K_{2} S O_{4}$
2 1
$I_{2} + 2 S_{2} O_{3}^{- 2} \to S_{4} O_{6}^{- 2} + 2 I^{-}$
The number of moles of hypo solution $= \frac{50 \times 1}{1000} = 50 \times 10^{- 3}$ mole.
The number of moles of $I_{2} = \frac{50 \times 10^{- 3}}{2} = 25 \times 10^{- 3}$ mole.
The number of moles of $C a O C l_{2} = 25 \times 10^{- 3}$ mole.
The mass of bleaching powder $25 \times 10^{- 3} \times 127.2 = 3.18$ g.
So, the percentage purity of the sample is $= \frac{3.18 \times 100}{6.35} = 50$ %..

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