Question

During estimation of nitrogen present in an organic compound by Kjeldahl’s method, the ammonia evolved from 0.5 g of the compound in Kjeldahl’s estimation of nitrogen neutralized 10 mL of 1 M of ${{H}_{2}}S{{O}_{4}}$. The percentage of nitrogen in the compound is:
A. 56%
B. 45%
C. 50%
D. 40%

Answer 1

In Kjeldahl’s method, an organic compound having nitrogen is heated with concentrated sulphuric acid to form ammonium sulphate, which is then treated with NaOH to liberate ammonia gas.

Complete answer:
We have been given a nitrogen compound of 0.5 g, that neutralizes 10 mL of 1 M ${{H}_{2}}S{{O}_{4}}$. We have to calculate the percentage of ammonia in this organic compound.
Kjeldahl’s method uses conc. Sulphuric acid to liberate ammonia, which is titrated with alkali, and a neutralization reaction occurs. As, 10 mL of 1 M ${{H}_{2}}S{{O}_{4}}$is used, then 20 mL of 1 M ammonia neutralizes it.
As we know, 1000 mL of 1 M of ammonia has a mass of 14 g of nitrogen in it. So we will calculate, mass of nitrogen in 20 mL of ammonia through unitary method as,
N in 20mL $N{{H}_{3}}$ = $\dfrac{14}{1000}\times 20$
N in 20mL $N{{H}_{3}}$ = 0.28 g
So, 0.5 g of the organic compound contains 0.28 g of nitrogen. Thus, percentage of 0.28 g of N in 0.5 g of the compound is,
% of nitrogen = $\dfrac{0.28}{0.5}\times 100$
% of nitrogen = 56 %

Hence, the percentage of nitrogen in the compound is 56 %. So, option A is correct.

Note: 10 mL of 1 M ${{H}_{2}}S{{O}_{4}}$have a correspondence of 20 mL of ammonia as the basicity of conc. Sulphuric acid is 2. So, 2 ammonia molecules will neutralize 1 ${{H}_{2}}S{{O}_{4}}$ molecule, therefore, 20 mL of ammonia will neutralize 10 mL of ${{H}_{2}}S{{O}_{4}}$. 1 M of any gas contain the mass equal to to its molar mass, so, 1000 mL of 1 M ammonia has 14 g nitrogen (molar mass of N).