Question

In the estimation of nitrogen by Kjeldahl's method, the ammonia evolved from 0.30 g of an organic compound (X) was passed into 100 mL of 0.1 M $\text{H}_2\text{SO}_4$. The unreacted acid required 20 mL of 0.5 M NaOH for complete neutralization. What is X?

• A. $\text{CH}_3\text{CONH}_2$
• B. $\text{C}_6\text{H}_5\text{CONH}_2$
• C. $(\text{NH}_2)_2\text{CO}$
• D. $\text{C}_6\text{H}_5\text{NH}_2$

Hint:

Remember that Urea contains two nitrogen atoms per molecule, making its nitrogen content exceptionally high (46.7%).
This value is a standard constant in organic biochemistry questions.

Answer 1

The Correct Option is C

Step 1: Find the total acid taken and the leftover acid.
Total $\text{H}_2\text{SO}_4$ taken is $100 \times 0.1 \times 2 = 20$ meq, using the factor 2 since it is dibasic. The unreacted acid, found from the NaOH used, is $20 \times 0.5 \times 1 = 10$ meq.
Step 2: Find the acid that actually neutralised the ammonia.
\[ \text{meq reacted with NH}_3 = 20 - 10 = 10 \text{ meq} \]
Step 3: Convert to percentage nitrogen.
Using $\%\text{N} = \dfrac{1.4 \times \text{meq of acid reacted}}{W}$, we get $\%\text{N} = \dfrac{1.4 \times 10}{0.30} \approx 46.7\%$.
Step 4: Match this against the molar masses of the options.
Only urea, $(\text{NH}_2)_2\text{CO}$, with molar mass 60 and two nitrogen atoms, gives $\%\text{N} = 28/60 \approx 46.7\%$, matching our calculated value, while acetamide, benzamide and aniline all give much lower percentages.
Final answer: Option 3, X is urea, $(\text{NH}_2)_2\text{CO}$.