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The impure $6 g o f N a C l$ is dissolved in water and then treated with excess of silver nitrate solution. The mass of precipitate of silver chloride is found to be $14 g .$ The $%$ purity of $N a C l$ solution would be approximately.

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answer is 95.

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Detailed Solution

The reaction that takes place is $N a C l + A g N O_{3} \to A g C l ↓ + N a N O_{3}$

$∵ 143.5 g o f A g C l$ is produced from $58.5 g N a C l$

$∴ 14 g o f A g C l$ will be produced from $\frac{58.5 \times 14}{143.5} = 5.70 g N a C l$

This is the amount of $N a C l$ in common salt; $% p u r i t y = \frac{5.70}{6} \times 100 = 95 %$

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