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4.9 gram $K C l O_{3}$ on calcinations suffers a loss in weight of 0.384 grams. The percentage of $K C l O_{3}$ that got decomposed

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Detailed Solution

$2 k C l O_{3} \to 2 K C l + 3 O_{2}$

2 x 122.5g of $k C l O_{3} \to$ 3 x 32g of O₂

? $\leftarrow$ 0.384g

= $\frac{2 \times 22.5 \times 0.384}{3 \times 32} = 0.98 g$

Mass of 100% pure KClO₃ undergoes calcination=0.98g

Percentage purity of 4.9 g. $k C l O_{3} = \frac{0.98}{4.9} \times 100 = 20 %$

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