Some Basic Concepts of Chemistry - Result Question 59
58. In the analysis of $0.5 g$ sample of feldspar, a mixture of chlorides of sodium and potassium is obtained, which weighs $0.1180 g$. Subsequent treatment of the mixed chlorides with silver nitrate gives $0.2451 g$ of silver chloride. What is the percentage of sodium oxide and potassium oxide in the sample?
(1979,5 M)
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Answer:
Correct Answer: 58. (i) 0.0179 g, (ii) 10.6 %
Solution:
- Oxides of sodium and potassium are converted into chlorides according to following reactions :
$\begin{aligned} Na _2 O+2 HCl & \longrightarrow 2 NaCl+H _2 O \\ K _2 O+2 HCl & \longrightarrow 2 KCl+H _2 O \end{aligned}$
Finally all the chlorides of $NaCl$ and $KCl$ are converted into $AgCl$, hence
$\text { moles of }(NaCl+KCl)=\text { moles of } AgCl$
(one mole of either $NaCl$ or $KCl$ gives one mole of $AgCl$ )
Now, let the chloride mixture contain $x g NaCl$.
$\Rightarrow \quad \frac{x}{58.5}+\frac{0.118-x}{74.5}=\frac{0.2451}{143.5}$
Solving for $x$ gives $x=0.0338 g$ (mass of $NaCl$ )
$\Rightarrow \quad$ Mass of $KCl=0.118-0.0338$
$=0.0842 g$
Also, moles of $Na _2 O=\frac{1}{2} \times$ moles of $NaCl$
$\Rightarrow \quad$ Mass of $Na _2 O=\frac{1}{2} \times \frac{0.0338}{58.5} \times 62=0.0179 g$
Similarly, mass of $K _2 O=\frac{1}{2} \times \frac{0.0842}{74.5} \times 94=0.053 g$
$\Rightarrow \quad$ Mass $%$ of $Na _2 O=\frac{0.0179}{0.5} \times 100=3.58 %$
Mass $%$ of $K _2 O=\frac{0.053}{0.5} \times 100=10.6 %$
BETA
