The ammonia
Question:

The ammonia $\left(\mathrm{NH}_{3}\right)$ released on quantitative reaction of $0.6 \mathrm{~g}$ urea $\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)$ with sodium hydroxide $(\mathrm{NaOH})$ can be neutralized by:

1. $200 \mathrm{~mL}$ of $0.4 \mathrm{~N} \mathrm{HCl}$

2. $200 \mathrm{~mL}$ of $0.2 \mathrm{NHCl}$

3. $100 \mathrm{~mL}$ of $0.2 \mathrm{~N} \mathrm{HCl}$

4. $100 \mathrm{~mL}$ of $0.1 \mathrm{~N} \mathrm{HCl}$

Correct Option: , 3

Solution:

$1 \mathrm{~mol}$ of urea $=2 \mathrm{~mol}$ of $\mathrm{NH}_{3}$

$60 \mathrm{~g}$ of urea $=2 \mathrm{~mol}$ of $\mathrm{NH}_{3}$

$0.6 \mathrm{~g}$ of urea $=\frac{2}{60} \times 0.6 \mathrm{~mol}=0.02 \mathrm{~mol}$ of $\mathrm{NH}_{3}$

$\mathrm{mol}$ of $\mathrm{NH}_{3}=\mathrm{mol}$ of $\mathrm{HCl}$

$\therefore \mathrm{mol}$ of $\mathrm{HCl}=0.02 \mathrm{~mol}$

$\Rightarrow$ Normality of $\mathrm{HCl}=0.2 \mathrm{~N}$

Volume of $\mathrm{HCl}=100 \mathrm{~mL}$