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Molality

Chloride samp...

Question

Chloride samples are prepared for analysis by using $N a C l, K C l$ and $N H_{4} C l$ separately or as mixture. What minimum volume of $5$ % by weight $A g N O_{3}$ solution (specific gravity $= 1.02 g m L^{- 1})$ must be added to a sample of $0.321$ g, in order to ensure complete precipitation of chloride in every possible case?

A

20

mL

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B

30

mL

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C

40

mL

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D

50

mL

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Solution

The correct option is A $20$ mL
$0.321$ g of a sample of $N H_{4} C l$ contains $\frac{0.321}{53.5} = 0.006$ mol, so it will provide $0.006$ moles of chloride ion.
$A g N O_{3} + N H_{4} C l \to A g C l$
For complete precipitation, moles of $A g N O_{3}$ required is $0.006$ mol.
Let, $V$ mL of solution is required, so, $V \times 1.02 \times \frac{0.05}{160} = 0.006 ⟹ V = 20$ mL.

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