Question

A sample of 0.50 g of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in 50 mL of 0.5 M $H_{2}S{O}_4$.The residue acid required 60 mL of 0.5 M solution of $NaOH$ for neutralisation. What would be the percentage composition of nitrogen in the compound?

• A. 50
• B. 60
• C. 56
• D. 44

Answer 1

Answer: (C)

Mass of organic compound = 0.50 g

60 mL of 0.5 M solution of $NaOH$ is required by residual acid for neutralisation

$\Rightarrow$ 60 mL of 0.5 M solution of $NaOH$ = $\frac{60}{2}$ mL of 0.5M $H_{2}S{O}_4$ = 30mL of 0.5M $H_{2}S{O}_4$

$\therefore$ Acid consumed in absorption of evolved ammonia is (50-30)mL = 20mL

Again, 20mL of 0.5M $H_{2}S{O}_4$ = 40mL of 0.5M $N{H}_3$

Since 1000mL of 1M $N{H}_3$ contains 14g of nitrogen,

$\therefore$ 40mL of 0.5M $N{H}_3$ will contain = $\frac{14\times 40}{1000}\times 0.5$ = 0.28g of $N$

Therefore, percentage of $N$ in 0.50g of organic compound = $\frac{0.28}{0.50}\times 100=56\%$

Hence, the correct answer is $56\%$.