Some Basic Concepts of Chemistry 1 Question 38
58. In the analysis of $0.5 \mathrm{~g}$ sample of feldspar, a mixture of chlorides of sodium and potassium is obtained, which weighs $0.1180 \mathrm{~g}$. Subsequent treatment of the mixed chlorides with silver nitrate gives $0.2451 \mathrm{~g}$ of silver chloride. What is the percentage of sodium oxide and potassium oxide in the sample?
$(1979,5 \mathrm{M})$
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Solution:
- Oxides of sodium and potassium are converted into chlorides according to following reactions :
$$ \begin{aligned} \mathrm{Na}{2} \mathrm{O}+2 \mathrm{HCl} & \longrightarrow 2 \mathrm{NaCl}+\mathrm{H}{2} \mathrm{O} \ \mathrm{K}{2} \mathrm{O}+2 \mathrm{HCl} & \longrightarrow 2 \mathrm{KCl}+\mathrm{H}{2} \mathrm{O} \end{aligned} $$
Finally all the chlorides of $\mathrm{NaCl}$ and $\mathrm{KCl}$ are converted into $\mathrm{AgCl}$, hence
$$ \text { moles of }(\mathrm{NaCl}+\mathrm{KCl})=\text { moles of } \mathrm{AgCl} $$
(one mole of either $\mathrm{NaCl}$ or $\mathrm{KCl}$ gives one mole of $\mathrm{AgCl}$ )
Now, let the chloride mixture contain $x \mathrm{~g} \mathrm{NaCl}$.
$$ \Rightarrow \quad \frac{x}{58.5}+\frac{0.118-x}{74.5}=\frac{0.2451}{143.5} $$
Solving for $x$ gives $x=0.0338 \mathrm{~g}$ (mass of $\mathrm{NaCl}$ )
$\Rightarrow \quad$ Mass of $\mathrm{KCl}=0.118-0.0338$
$$ =0.0842 \mathrm{~g} $$
Also, moles of $\mathrm{Na}_{2} \mathrm{O}=\frac{1}{2} \times$ moles of $\mathrm{NaCl}$
$\Rightarrow \quad$ Mass of $\mathrm{Na}_{2} \mathrm{O}=\frac{1}{2} \times \frac{0.0338}{58.5} \times 62=0.0179 \mathrm{~g}$
Similarly, mass of $\mathrm{K}_{2} \mathrm{O}=\frac{1}{2} \times \frac{0.0842}{74.5} \times 94=0.053 \mathrm{~g}$
$\Rightarrow \quad$ Mass $%$ of $\mathrm{Na}_{2} \mathrm{O}=\frac{0.0179}{0.5} \times 100=3.58 %$
Mass $%$ of $\mathrm{K}_{2} \mathrm{O}=\frac{0.053}{0.5} \times 100=10.6 %$
