Question

$0.50g$ of an organic compound was Kjeldahlised and the $N{H}_3$ evolved was absorbed in $50mL$ of $0.5M{H}_{2}S{O}_4$ The residual acid required $60c{m}^3$ of $0.5MNaOH$. The percentage of nitrogen in the organic compound is

• A. 14
• B. 28
• C. 56
• D. 42

Answer 1

Answer: (C)

Vol. of $1M{H}_{2}S{O}_4$ taken $=50\times 0.5=25mL$

Vol. of $1MNaOH$ used for neutralisation of the residual acid

$=60\times 0.5=30mL$

Since one mole of $H_{2}S{O}_4$ neutralises 2 moles of $NaOH$.

$\therefore$ Vol. of $1M{H}_{2}S{O}_4$ left unused $=30/2=15mL$

Vol. of $1M{H}_{2}S{O}_4$ used $=(25-15)1=10mL$

Normality $=n\times M=2\times 1=2N$

$So,\%N=\frac{1.4\times 10\times 2}{0.5}=56$