Question

In the analysis of $0.5$ g sample of feldspar, a mixture of chlorides of $Na$ and $K$ are obtained which weighs $0.1180$ g. Subsequent treatment of mixed chlorides with $AgN{O}_3$ gives $0.2451$ g $AgCl$. The $\%$ of $K_{2}O$ in feldspar is (as the nearest integer) :

Answer 1

Let feldspar contains $a$ g of${Na}_{2}O$ and $b$ g of $K_{2}O$
${Na}_{2}O\to 2NaCl\stackrel{AgN{O}_3}{\to }2AgCl$
$K_{2}O\to 2KCl\stackrel{AgN{O}_3}{\to }2AgCl$
$\because 62g{Na}_{2}O=2\times 58.5gNaCl$
$\therefore ag{Na}_{2}O=\frac{2\times 58.5\times a}{62}gNaCl$ $=1.8871\times ag$
Also, $\because 94g{K}_{2}O=2\times 74.5gKCl$
$\therefore bg{K}_{2}O=\frac{2\times 74.5\times b}{94}gKCl$ $=1.5851\times bgKCl$
$\therefore 1.8871a+1.5851b=0.1180...\left(i\right)$
Similarly,
$\because 62g{Na}_{2}O=2\times 143.5gAgCl$
$\therefore ag{Na}_{2}O=\frac{2\times 143.5\times a}{62}gAgCl$ $=4.6290\times agAgCl$
Also, $\because 94g{K}_{2}O=2\times 143.5gAgCl$
$\therefore bg{K}_{2}O=\frac{2\times 143.5\times b}{94}gAgCl$ $=3.0532\times bgAgCl$
$\therefore 4.6290a+3.0532b=0.2451$ $...\left(ii\right)$
Solving Eqs. (i) and (ii), $a$ $=$ $0.0179$ g, $b$ $=$ $0.0531$ g
$\%of{K}_{2}O=\frac{0.0531\times 100}{0.5}=10.62$
So, nearest integer is $11$.