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$0.05\;g$ of a commercial sample $KCl_3$ on decomposition liberated just sufficient oxygen for complete oxidation of $20\;ml\; CO$ at $27^{\large\circ}C$ and $750\;mm$ pressure.Calculate % of $KClO_3$ in sample.


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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.
answered Nov 6, 2013 by sreemathi.v
edited Mar 18, 2014 by mosymeow_1

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