In the estimation of nitrogen by Kjeldahl's m | vedclass

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(A) The reaction for the neutralisation of $NH_3$ by $H_2SO_4$ is: $2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4$.Given,$20 \ mmol$ of $H_2SO_4$ is used.S

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$20$

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(A) The reaction for the neutralisation of $NH_3$ by $H_2SO_4$ is: $2NH_3 + H_2SO_4 ightarrow (NH_4)_2SO_4$.Given,$20 \ mmol$ of $H_2SO_4$ is used.Since $1 \ mol$ of $H_2SO_4$ reacts with $2 \ mol$ of $NH_3$,$20 \ mmol$ of $H_2SO_4$ reacts with $40 \ mmol$ of $NH_3$.$1 \ mmol$ of $NH_3$ contains $1 \ mmol$ of $N$ atoms,so $40 \ mmol$ of $NH_3$ contains $40 \ mmol$ of $N$ atoms.Mass of $N = 40 	imes 10^{-3} \ mol 	imes 14 \ g/mol = 0.56 \ g$.Percentage of $N = \frac{	ext{Mass of } N}{	ext{Mass of compound}} 	imes 100 = \frac{0.56}{2.8} 	imes 100 = 20 \%$.

In the estimation of nitrogen by Kjeldahl's method,$2.8 \ g$ of an organic compound required $20 \ mmol$ of $H_2SO_4$ for the complete neutralisation of $NH_3$ gas evolved. The percentage of nitrogen in the sample is?........$\%$

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