Question:
A sample of $NaClO_3$ is converted by heat to NaCl with a loss of $0.16\, g$ of oxygen. The residue is dissolved in water and precipitated as $AgCl$. The mass of $AgCl$ (in g) obtained will be : (Given : Molar mass of $AgCl = 143.5\, g\, mol^{-1}$)
A sample of $NaClO_3$ is converted by heat to NaCl with a loss of $0.16\, g$ of oxygen. The residue is dissolved in water and precipitated as $AgCl$. The mass of $AgCl$ (in g) obtained will be : (Given : Molar mass of $AgCl = 143.5\, g\, mol^{-1}$)
Updated On: Sep 30, 2024
- 0.35
- 0.41
- 0.48
- 0.54
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The Correct Option is C
Solution and Explanation
The reaction is $2 NaClO _{3} \stackrel{\Delta}{\longrightarrow} 2 NaCl +3 O _{2}$
$3\, mol$ of $O _{2}=2\, mol$ of $NaCl =2\, mol$ of $AgCl$
Number of moles of $O _{2}=\frac{0.16}{32}=0.005 mol$
$0.005\, mol$ of $O _{2}=0.005 \times \frac{2}{3}=0.0033\, mol$ of $AgCl$
The mass of $AgCl$ obtained will $be =143.5 \times 0.0033=0.48\, g$
$3\, mol$ of $O _{2}=2\, mol$ of $NaCl =2\, mol$ of $AgCl$
Number of moles of $O _{2}=\frac{0.16}{32}=0.005 mol$
$0.005\, mol$ of $O _{2}=0.005 \times \frac{2}{3}=0.0033\, mol$ of $AgCl$
The mass of $AgCl$ obtained will $be =143.5 \times 0.0033=0.48\, g$
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Chemistry is a vast subject and for understanding its significance we can take help of following points:
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