Question

The 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 neutralisation. The organic compound is:

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

Answer 1

Answer: (C)

urea

$2N{H}_3+H_{2}S{O}_4\left(aq\right)\to \left(N{H}_4{)}_{2}S{O}_4\right(aq)+H_{2}O$

$2NaOH\left(aq\right)+H_{2}S{O}_4\left(aq\right)\to N{a}_{2}S{O}_4\left(aq\right)+2H_{2}O\left(1\right)$

$2n_{H_{2}S{O}_4}=[n_{N{H}_3}+n_{NaOH}]$

$2\times \left(\frac{100}{1000}\right)\left(0.1\right)=[n_{N{H}_3}+\left(\frac{20}{1000}\right)(0.5\left)\right]$

$\frac{1}{50}=n_{N{H}_3}+\frac{1}{100}$

$n_{N{H}_3}=\frac{1}{100}$ (now check the answer one by one)

$\left[A\right]n_N$ in urea $=n_{N{H}_3}$

$2\times \frac{0.3}{60}=\left(\frac{1}{100}\right)$

$\frac{1}{100}=\frac{1}{100}$

Option C is correct.