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Question

The percent loss in weight after heating a pure sample of potassium chlorate (mol.wt.=122.5) will be?
($${ KClO }_{ 3 }\rightarrow { KCl+O }_{ 2 }$$)


A
12.25
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B
24.5
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C
39.18
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D
49
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Solution

The correct option is A 39.18
We have,

$$2KClO_3 \rightarrow 2KCl + 3O_2$$

molar mass of $$KClO_3$$ = 122.5 g

mass of KCl = 74.5 g

so, percentage loss = $$\dfrac{122.5-74.5}{122.5}*100 = 39.18 %$$

Chemistry

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