Question

The ammonia evolved from the treatment of $1.50$ g of an organic compound for the estimation of nitrogen was passed in $100$ 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

For neutralization, millimoles of $NaOH$ used $=100\times 0.5=50$

So, the millimoles of ammonia evolved are $50$.

As each ammonia molecule contains one atom of $N$, so millimoles of $N$ in an organic compound is $50$.

Mass of $N=\frac{50}{1000}\times 14=0.7$gm

% of $N$ in organic compound $=\frac{100\times 0.7}{1.5}=46.7$%

The molar mass of urea is $60$ g/mol.

Percent of N in urea is $\frac{100\times 28}{60}=46.7$%

Hence, the given compound is urea.