$1.500 g$ of a fertiliser containing $(NH X_{4}) X_{2} SO X_{4}$ (molar mass $132$ ) is boiled with excess $NaOH$ ; the ammonia is absorbed in $50.0 mL$ of $0.400 M$ $HCl$ . The unreacted $HCl$ needs $8.6 mL$ of $0.500 M$ $NaOH$ . Assuming no other source of ammonia, find the percentage purity of $(NH X_{4}) X_{2} SO X_{4}$ (two decimals).

Numerical answer type, answer in %, tolerance 0.05.

Answer: 69.08 %

HCl taken

= 20.0

mmol; excess

= 8.6 \times 0.500 = 4.30

mmol;

NH X_{3}

absorbed

= 15.70

mmol. Each

(NH X_{4}) X_{2} SO X_{4}

gives 2

NH X_{3}

, so salt

= 7.85

mmol

= 1036.2

mg. Percentage

= 1036.2/1500 \times 100 = 69.08%