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Oxyacids Of Sulphur

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To determine the $N a C l$ content in commercial $N a O H$ , $2 g$ of the latter was dissolved in water and an excess amount of an $A g N O_{3}$ solution was added to this solution. The precipitate formed was washed and dried. Its mass was $0.287 g$ . Find the mass of $N a C l$ in the initial sample.

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Solution

Reaction as follows:
$N a C l + A g N O_{3} ⟶ N a N O_{3} + A g C l ↓$
$0.287 g m$ of $p p t$ $A g C l = \frac{0.287}{143.32} = 0.002$ moles of $A g C l$
Therefore $0.002$ moles of $N a C l$ forms $0.002$ moles of $A g C l$
$∴$ Mass of $N a C l = 0.002 \times 58.5 = 0.117 g$

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