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Question

The % loss in weight after heating a pure sample of potassium chlorate (Molecular weight =122.5 g/mol) will be:

A
12.25
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B
24.5
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C
39.2
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D
19.6
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Solution

The correct option is D 39.2
The balanced chemical equation for the thermal decomposition of potassium chlorate is,
2KClO3Δ2KCl+3O2
2 moles of potassium chlorate corresponds to 3 moles of oxygen.
Thus, 2×122.5=245g of potassium chlorate corresponds to 3×32=96g of oxygen.
The % loss in weight after heating a pure sample of potassium chlorate (Mol. wt. =122.5 g/mol) will be 96245×100=39.2%.

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