Question

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

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

Answer 1

Answer: (C)

urea

By Kjeldahl process, % $N=\frac{1.4\times N\times V}{W}$

$N⟶$ Normality of $H_{2}S{O}_4,V⟶$ Volume of $H_{2}S{O}_4$ used, $W⟶$ Weight of sample

Eq. in $20ml$ of $0.5M$ $NaOH$= Eq, in $V^{\prime }$ ml of $0.2N$ $H_{2}S{O}_3$

$\Rightarrow 20\times 5=V^{\prime }\times 0.2$

$\Rightarrow V^{\prime }=50ml$

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

%$N=\frac{1.4\times 0.2\times 50}{0.3}=46.67$%

Out of given options, urea has $46.67$% $N$. Hence, urea is the correct answer.