The percentage loss in weight after heating a pure sample of potassium chlorate will be:
[Molar mass of potassium chlorate $= 122.5$ ]

12.25

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24.50

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39.18

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49.0

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Solution

The correct option is C $39.18$
On heating, $K C l O_{3}$ decomposes into $K C l$ and $O_{2}$ . The loss in weight is due to the oxygen gas evolved. The decomposition reaction is shown below

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

$% W e i g h t l o s t = \frac{3 \times 32}{2 \times 122.5} \times 100 = 39.18$

Hence, the correct option is $C$

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Similar questions

w t . = 122.5

%

= 122.5

{K C l O}_{3} \to {K C l + O}_{2}

K C l O_{3} (s) \to K C l (s) + O_{2} (g)

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