Question

In the analysis of $0.50$ g sample of feldspar, a mixture of the 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 feldspar?

• A. $N{a}_{2}O=4.5\%$, $K_{2}O=6.6\%$
• B. $N{a}_{2}O=3.6\%$, $K_{2}O=10.6\%$
• C. $N{a}_{2}O=1.8\%$, $K_{2}O=5.3\%$
• D. None of the above

Answer 1

Answer: (B)

$N{a}_{2}O=3.6\%$, $K_{2}O=10.6\%$

Let the amount of sodium chloride be $a$ g.
The amount of potassium chloride is $(0.1180-a)$ g.
$NaCl+AgN{O}_3\to AgCl+NaN{O}_3$
$NaCl\equiv AgN{O}_3$

The mass of $AgN{O}_3$ formed $=\frac{a}{58.5}\times 143.5$

$KCl+AgN{O}_3\to AgCl+KN{O}_3$
$KCl\equiv AgN{O}_3$

The mass of $AgN{O}_3$ formed $=\frac{(0.1180-a)}{58.5}\times 143.5$

$\therefore \left[\frac{a}{58.5}\times 143.5\right]+\left[\frac{(0.1180-a)}{74.5}\times 143.5\right]=0.2451$

On solving, we get $a=0.0338$ (mass of $NaCl$)
$N{a}_{2}O\equiv 2NaCl$

$\therefore \frac{W_{NaO}}{62}=\frac{0.0038}{2\times 58.5}$ $⟹W_{N{a}_{2}O}=0.0179g$

Percentage of $N{a}_{2}O=\frac{0.079\times 100}{0.500}=3.58\%$

Mass of $KCl$ $=(0.1180-a)$ $=(0.1180-0.0338)=0.0842$ g

$K_{2}O\equiv 2KCl$

$\frac{W_{K_{2}O}}{94}=\frac{0.0038}{2\times 74.5}$ or $W_{K_{2}O}=0.031$ g

Percentage of $K_{2}O=\frac{0.0531\times 100}{0.500}=10.62\%$