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Percentage Composition

3.0 g of a sa...

Question

$3.0 g$ of a sample of impure ammonium chloride were boiled with an excess of caustic soda solution. Ammonia gas so evolved was passed into $120 m L$ of $N / 2 H_{2} S O_{4}$ . $28 m L$ of $N / 2 N a O H$ were required to neutralize residual acid. Calculate the percentage of purity of the given sample of ammonium chloride.

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Solution

$N H_{4} C l + N a O H \to N a C l + N H_{4} O H$
$N H_{3} + H_{2} S O_{4} \to (N H_{4})_{2} S O_{4}$

According to law of equivalence :

$(N_{1} V_{1})_{N a O H} = (N_{2} V_{2})_{H_{2} S O_{4}}$
By substitution
$\frac{1}{2} \times 28 = \frac{1}{2} \times V_{2}$

$V_{2} = 28 m L$

Volume of sulphuric acid consumed = 120 - 28 = 92mL
Volume of $N H_{3} = 2 \times 92 = 184 m L$

We know that 1 mole = 22.4 L
0.184 L has $\frac{1}{22.4} \times 0.184$ = 0.008 moles

Number of moles in 3g = $\frac{3}{53.5}$ =0.053

Precentage purity = $\frac{0.008}{0.0053} \times 100$ = 15%

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