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 $0.5M$ sodium hydroxide solution for complete neutralization. The organic compound is

• A. Urea $\left(N{H}_{2}CON{H}_2\right)$
• B. Benzamide $\left(C_6{H}_{5}CON{H}_2\right)$
• C. Acetamide $\left(C{H}_{3}CON{H}_2\right)$
• D. Thiourea $\left(N{H}_{2}CSN{H}_2\right)$

Answer 1

The correct option is A Urea $\left(N{H}_{2}CON{H}_2\right)$

$H_{2}S{O}_4$ is dibasic.
$0.1M{H}_{2}S{O}_4=0.2N{H}_{2}S{O}_4$ $[\because M=2\times N]$
$M_{eq}of{H}_{2}S{O}_{4}taken=100\times 0.2=20$
$M_{eq}of{H}_{2}S{O}_{4}neutralisedbyNaOH=20\times 0.5=10$
$M_{eq}of{H}_{2}S{O}_{4}neutralisedbyN{H}_3=20-10=10$
$\%of{N}_2=\frac{1.4\times {\text{M}}_{eq}\text{of acid neutralised by}N{H}_3}{\text{weight of organic compound}}=\frac{1.4\times 10}{0.3}=46.6\%\text{of nitrogen in urea}=\frac{14\times 2\times 100}{60}=46.6\left[\text{Molecular weight of urea = 60}\right]$
$\text{Similarly \% of Nitrogen in Benzamide}=\frac{14\times 10}{121}=11.5\%[M.W.of{C}_6{H}_{5}CON{H}_2=121]\text{\% of nitrogen in Acetamide}=\frac{14\times 1\times 100}{59}=23.4[M.W.ofC{H}_{3}CON{H}_2=59]\text{\% of nitrogen in Thiourea}=\frac{14\times 2\times 100}{76}=36.8\%[M.W.ofN{H}_{2}CSN{H}_2=76]$
Hence, the compound must be urea.