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Concentration Terms - An Introduction (w/w etc)

A quantity of...

Question

A quantity of ammonium chloride was heated with $100 m L$ of $0.8 N N a O H$ solution till the reaction was complete. The excess of $N a O H$ was neutralised with $12.5 m L$ of $0.75 N H_{2} S O_{4}$ . Calculate the quantity of ammonium chloride.

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Solution

$12.5 m L$ of $0.75 N H_{2} S O_{4} \equiv 12.5 m L$ of $0.75 N N a O H$
$12.5 m L$ of $0.75 N N a O H \equiv 11.72 m L$ of $0.8 N N a O H$
$N a O H$ solution used by $N H_{4} C l$
$= (100 - 11.72) m L$ of $0.8 N N a O H$
$= 88.28 m L$ of $0.8 N N a O H$
$\equiv 88.28 m L$ of $0.8 N N H_{4} C l$
Mass of $N H_{4} C l$ present in $88.28 m L$ of $0.8 N N H_{4} C l$ solution
$= \frac{N \times E \times V}{1000} = \frac{0.8 \times 53.5 \times 88.28}{1000} = 3.7793 g$
[Eq. mass of $N H_{4} C l = 53.5$ ].

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