Question

0.6g of organic compound on Kjeldahl's analysis produced Ammonia which is absorbed in 90 mL of N/9 $H_{2}S{O}_4$. The excess acid left requires 20 mL of N/10 of NaOH for complete neutrilisation. Calculate the percentage of Nitrogen in the organic.

Answer 1

$20ml$ of $\frac{N}{10}NaOH$ was required by residual acid for neutralisation.

$20ml$ of $\frac{N}{10}NaOH$ solution= $20ml$ of $\frac{N}{10}{H}_{2}S{O}_4$

$\therefore$ Acid consumed in absorption of evolved ammonia is $(90-20)ml=70ml$

$70ml$ of $\frac{N}{10}{H}_{2}S{O}_4=70ml$ of $\frac{N}{10}N{H}_3$

Since, $1000ml$ of $1N$ $N{H}_3$ contains $1g$ of $N_2$

$\therefore 70ml$ of $\frac{N}{10}N{H}_3$ will contain= $\frac{14\times 70}{1000}\times \frac{1}{10}=0.098gm$ of Nitrogen

$\therefore$ Percentage of nitrogen in $0.60gm$ of organic compound= $\frac{0.098}{0.6}\times 100$%=$16.3$%