3.6 g mixture of sodium chloride and potassium chloride is dissolved in water. The solution is treated with excess of silver nitrate solution. 7.74 g of silver chloride is obtained. Find the percentage of sodium chloride and potassium chloride in the mixture.

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Solution

The balanced equation between $N a C l$ and $A g N O_{3}$ is:

$N a C l 1 m o l 58.5 g + A g N O_{3} 1 m o l 143.5 g \to A g C l + N a N O_{3}$

Let x g of $N a C l$ be present in the mixture.
$58.5 g N a C l$ produce $= 143.5 A g C l$

$x g N a C l$ will produce $= \frac{143.5}{58.5} \times x g A g C l$

The balanced equation between $K C l$ and $A g N O_{3}$ is:
$K C l 1 m o l 74.5 g + A g N O_{3} \to A g C l_{3} 1 m o l 143.5 g + K N O_{3}$

$K C l$ present in the mixture $= (3.60 - x) g$
$74.5 g o f C l$ produce $= 143.5 g$ of $A g C l$

$(3.6 - x) g$ of $K C l$ will produce $= \frac{14.3}{74.5} \times (3.6 - x) g o f A g C l$

Thus, $\frac{14.3}{74.5} x + \frac{14.3}{74.5} (3.6 - x) = 7.74$
$x = 1.54$
% of $N a C l =$ $= \frac{1.54}{3.60} \times 100 = 42.7$

% of $K C l =$ $= \frac{(3.60 - 1.54)}{3.60} \times 100 = 57.3$

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