$(a)\;70.2\%\qquad(b)\;61.3\%\qquad(c)\;65.4\%\qquad(d)\;62.5\%$

Step 1:

$2KClO_3\rightarrow 2KCl+3O_2$

$CO+\large\frac{1}{2}$$O_2\rightarrow CO_2$

For, $CO$ $PV=\large\frac{w}{m}$$RT$

$\therefore$ Mole of CO = $\large\frac{PV}{RT}$

$= \large\frac{750\times 20}{760\times 0.081\times 300\times 1000}$

$= 8.01\times 10^{-4}$

Step 2:

Mole ratio for combination $CO : O_2$ $= 2 : 1$

$\therefore$ Mole of $O_2$ required for CO $=\large\frac{1}{2}$$\times 8.01\times 10^{-4}$

3 mole of $O_2$ are given by 2 mole of $KClO_3$

$\therefore\large\frac{8.01\times 10^{-4}}{2}$ of $O_2$ are given by $\large\frac{2\times 8.01\times 10^{-4}}{2\times 3}$

Mole of $KClO_3=2.67\times 10^{-4}$ Mole of $KClO_3$

$\therefore$ weight of $KClO_3=2.67\times 10^{-7}\times 122.5g$ $= 3.27\times 10^{-2}g$

$\%$ of $KClO_3$ in sample =$\large\frac{3.27\times 10^{-2}}{0.05}$$\times 100 =65.4$%

Hence (c) is the correct answer.

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