Question

$1.4\, g$ of an organic compound was digested according to Kjeldahl's method and the ammonia evolved was absorbed in $60\, mL$ of $M/10\, H_2SO_4$ solution. The excess sulphuric acid required $20\, mL$ of $M/10\, NaOH$ solution for neutralization. The percentage of nitrogen in the compound is :

• A. 3
• B. 5
• C. 10
• D. 24

Answer 1

Answer: (C)

Out of $60 mL$ of $\frac{ M }{10} H _{2} SO _{4}, 10 mL$ will be neutralized with $20 mL$ of $\frac{ M }{10} NaOH$.
Hence, $60-10=50 mL$ of $\frac{ M }{10} H _{2} SO _{4}$ will be neutralized with ammonia. This will be equal to $100 mL$ ammonia.
$1000 mL$ of $1 M$ ammonia corresponds to $14 g$.
$100 mL$ of $\frac{ M }{10}$ ammonia will correspond to $\frac{100}{1000} \times \frac{1}{10} \times 14=0.14 g$ nitrogen
Thus, $1.4 g$ of organic compound contains $0.14 g$ nitrogen.
Hence, $100 g$ of organic compound will contain $\frac{0.14}{1.4} \times 100=10 g$ nitrogen
Hence, the per cent of nitrogen in the organic compound is $10 \%$.