Question

A sample of $0.50g$ of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in $50ml$ of $0.5M$ $H_{2}S{O}_4$. The residual acid required $60ml$ of $0.5M$ solution of $NaOH$ for neutralization. The percentage composition of nitrogen in the compound is:

• A. $14\%$
• B. $28\%$
• C. $56\%$
• D. $7\%$

Answer 1

Answer: (C)

$56\%$

60 mL of 0.5 M solution of NaOH will be used for 30 mL of 0.5 M $H_{2}S{O}_4$
The amount of $H_{2}S{O}_4$ used for the neutralization of ammonia is $50-30=20mL$
20 mL of 0.5 M $H_{2}S{O}_4$ has 20 meq, and due to which 20 meq of ammonia will be required.

Hence, 20 milimoles of ammonia will be required (monobasic) .
Now 1000 mL of 1 M ammonia contains 14 g of nitrogen.
Therefore, the ammonia contains $\frac{14\times 20}{1000}$ g of nitrogen.
0.50 g of organic compound contains $\frac{14\times 20}{1000}$ g of nitrogen.
Hence, the compound contains $\frac{14\times 20\times 100}{1000\times 0.5}=56\%$ of nitrogen.