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  4. The % loss in weight after heating a pure sample of potassium chlorate (M. wt. 122.5) will be- 2 KClO 3 stackrelΔ longrightarrow 2 KCl +3 O 2

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Q. The $\%$ loss in weight after heating a pure sample of potassium chlorate (M. wt. 122.5) will be-
$2 KClO _{3} \stackrel{\Delta}{\longrightarrow } 2 KCl +3 O _{2}$

Some Basic Concepts of Chemistry

Solution:

moles of $KCl =2$

$Wt =150$

Molar mass of $KClO _{3}=122.5 g$

Mass of $KCl =74.5 \,g$

$\%$ loss

$\% $ loss $=\frac{122.5-74.5}{122.5} \times 100$

$=\frac{48}{122.5} \times 100=39.18$