Question

$0.50 \,g$ of an organic compound was Kjeldahlised and the $NH _{3}$ evolved was absorbed in $50 \,mL$ of $0.5\, M H _{2} SO _{4}$ The residual acid required $60\, cm ^{3}$ of $0.5\, M\, NaOH$. The percentage of nitrogen in the organic compound is

• A. 14
• B. 28
• C. 56
• D. 42

Answer 1

Answer: (C)

Vol. of $1 M H _{2} SO _{4}$ taken $=50 \times 0.5=25 mL$

Vol. of $1 M NaOH$ used for neutralisation of the residual acid

$=60 \times 0.5=30 mL$

Since one mole of $H _{2} SO _{4}$ neutralises 2 moles of $NaOH$.

$\therefore $ Vol. of $1 M H _{2} SO _{4}$ left unused $=30 / 2=15 mL$

Vol. of $1 M H _{2} SO _{4}$ used $=(25-15) 1=10 mL$

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

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