Question

For the estimation of nitrogen, $1.4g$ of an organic compound was digested by Kjeldahl's method and the evolved ammonia was absorbed in $60mL$ of $M/10$ sulphuric acid. The unreacted acid required $20mL$ of $M/10$ sodium hydroxide for complete neutralisation. The percentage of nitrogen in the compound is:

• A. 6%
• B. 10%
• C. 3%
• D. 5%

Answer 1

The correct option is B 10%

Mass of organic compound $=1.4g$
Let it contain $xmmol$ of $N$ atom.

$\text{Organic compound}\to \underset{xmmol}{N{H}_3}2N{H}_3+\underset{\text{6 mmol initialy taken}}{H_{2}S{O}_4}\to (N{H}_4{)}_{2}S{O}_4...(i)H_{2}S{O}_4+NaOH\to N{a}_{2}S{O}_4+2H_{2}O...(ii)$

$2mmolNaOH$ reacted.

$\text{Hence,}mmol\text{of}{H}_{2}S{O}_4\text{reacted in Eq. (ii)}=1$

$\therefore mmol\text{of}{H}_{2}S{O}_4\text{reacted in Eq. (i)}=6-1=5mmol$

$\Rightarrow mmol\text{of}N{H}_3\text{reacted in Eq. (i)}=2\times 5=10mmol$

$\therefore mmol\text{of}N\text{atom in the organic compound}=10mmol$

$\Rightarrow \text{Mass of N}=10\times {10}^{-3}\times 14=0.14g$
$\%\text{of}N=\frac{\text{Mass of}N\text{in organic compound}}{\text{Mass of organic compound}}\times 100$

$\therefore \%\text{of}N=\frac{0.14}{1.4}\times 100=10\%$