Question

Ammonia evolved from the treatment of $0.30$g of an organic compound for the estimation of nitrogen was passed in $100$ mL of $0.1$M sulphuric acid. The excess of acid required $20$ mL of $0.5$M sodium hydroxide solution for complete neutralization. The organic compound is:

• A. thiourea
• B. benzamide
• C. urea
• D. acetamide

Answer 1

The correct option is C urea

Organic compound+ $H_{2}S{O}_4⟶$

Using Kjeldhal Process to calculate the percentage of nitrogen organic compound

Let unreacted $0.1M(=0.2N)H_{2}S{O}_4=Vml$

$\therefore ,20ml$ of $0.5M$ $NaOH=Vml$ of $0.2g$ $N{H}_{2}S{O}_4$

$\therefore 20\times 0.5=V\times 0.2$

or, $V=50ml$

used $H_{2}S{O}_4=100-50=50ml$

$\therefore ,$ % nitrogen=$14NV/w$

where $N$=normality of $H_{2}S{O}_4$

$V$=Volume of $H_{2}S{O}_4$ used

% nitrogen=$1.4\times 0.5\times 50\times 0.3=46.67$ %

Urea=$N{H}_{2}CON{H}_2=28\times 100/60$

$=46.67$ %

Thus, the organic compound is urea.