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2.480 g of ...

Question

$2.480 g$ of $K C I O_{3}$ are dissolved in conc. $H C I$ and the solution was boiled. Chlorine gas evolved in the reactions was then passed through a solution of $K I$ and liberated iodine was titrated with $100 m l$ of hypo. $12.3 m l$ of same hypo solutions required $24.6 m l$ of $0.5$ iodine for complete neutralization. Calculate $%$ purity of $K C I O_{3}$ sample.

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Solution

$2 K C I O_{3} + 12 H C l ⟶ 2 K C l + 6 H_{2} O + 6 C l_{2} C l_{2} + 2 K I ⟶ 2 K C l + I_{2}$
Also Meq. of $I_{2}$ = Meq.of Hypo= 100x 1
Also mM of $C l_{2}$ =mM of $I_{2}$ = $\frac{100}{2}$ =50
Also mM of $K C I O_{3}$ $= \frac{2 \times m M o f {C l}_{2}}{6} = \frac{2 \times 50}{6} = \frac{50}{3}$
Therefore $= \frac{w}{122.5} \times 1000 = \frac{50}{3}$
$∴ w ({K C I O}_{3}) = 2.042$
$P e r c e n t a g e$ $p u r i t y o f {K C I O}_{3} = \frac{2.042}{2.48} \times 100 = 82.32$ %.

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