JEE Main · 2004 · 24 Apr 2004 · Offline · Practical and Analytical Chemistry

Account for the acid. The sulphuric acid starts at $100\times 0.1=10\text{mmol}$. The sodium hydroxide used is $20\times 0.5=10\text{mmol}$, and since sulphuric acid is dibasic this neutralises $\frac{10}{2}=5\text{mmol}$ of leftover acid. So the ammonia took up $10-5=5\text{mmol}$ of acid, and because each mole of acid reacts with two of ammonia, $10\text{mmol}$ of $\mathrm{NH}{}_3$, that is $0.01\text{mol}$ of nitrogen, was evolved. That is $0.14\text{g}$ of nitrogen in the $0.30\text{g}$ sample, a nitrogen content of $46.67\%$, which matches urea, $\mathrm{NH}{}_2\mathrm{CONH}{}_2$, whose $\frac{28}{60}$ gives the same $46.67\%$.