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A sample of $\mathrm{NaClO}_3$ is converted by heat to $\mathrm{NaCl}$ with a loss of $0.16 \mathrm{~g}$ of oxygen. The residue is dissolved in water and precipitated as $\mathrm{AgCl}$. The mass of $\mathrm{AgCl}$ (in $\mathrm{g}$) obtained will be:
(Given: Molar mass of $\mathrm{AgCl}=143.5 \mathrm{~g} \mathrm{~mol}^{-1}$ )
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(Given: Molar mass of $\mathrm{AgCl}=143.5 \mathrm{~g} \mathrm{~mol}^{-1}$ )
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Verified Answer
The correct answer is:
$0.48$
$0.48$
No. of moles of oxygen in $0.16 \mathrm{~g}$ of oxygen molecule
$=\frac{0.16 \mathrm{~g}}{32 \mathrm{~g} / \mathrm{mol}}=0.005 \mathrm{~mole}$
$2 \mathrm{NaClO}_3 \stackrel{\Delta}{\longrightarrow} 2 \mathrm{NaCl}+3 \mathrm{O}_2$
According to the reaction,
$3$ moles of $\mathrm{O}_2=2$ moles of $\mathrm{NaCl}$
$=2$ moles of $\mathrm{AgCl}$
Molar mass of $\mathrm{AgCl}=143.5 \mathrm{~g} / \mathrm{mol}$
$0.005$ moles of $\mathrm{O}_2$ will ppt. $=0.005 \times \frac{2}{3}$ moles $\mathrm{AgCl}$
$=0.0033 \text { moles of } \mathrm{AgCl}$
$\therefore$ Mass of $\mathrm{AgCl}$ (in $\mathrm{g}$) obtained will be
$=143.5 \mathrm{~g} / \mathrm{mol} \times 0.0033$ moles $=0.48 \mathrm{~g}$.
$=\frac{0.16 \mathrm{~g}}{32 \mathrm{~g} / \mathrm{mol}}=0.005 \mathrm{~mole}$
$2 \mathrm{NaClO}_3 \stackrel{\Delta}{\longrightarrow} 2 \mathrm{NaCl}+3 \mathrm{O}_2$
According to the reaction,
$3$ moles of $\mathrm{O}_2=2$ moles of $\mathrm{NaCl}$
$=2$ moles of $\mathrm{AgCl}$
Molar mass of $\mathrm{AgCl}=143.5 \mathrm{~g} / \mathrm{mol}$
$0.005$ moles of $\mathrm{O}_2$ will ppt. $=0.005 \times \frac{2}{3}$ moles $\mathrm{AgCl}$
$=0.0033 \text { moles of } \mathrm{AgCl}$
$\therefore$ Mass of $\mathrm{AgCl}$ (in $\mathrm{g}$) obtained will be
$=143.5 \mathrm{~g} / \mathrm{mol} \times 0.0033$ moles $=0.48 \mathrm{~g}$.
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