Question

The ammonia $(N{H}_3)$ released on quantitative reaction of $0.6g$ urea $(N{H}_{2}CON{H}_2)$ with sodium hydroxide (NaOH) can be neutralized by :

• A. $200ml$ of $0.2NHCl$
• B. $200ml$ of $0.4NHCl$
• C. $100ml$ of $0.1NHCl$
• D. $100ml$ of $0.2NHCl$

Answer 1

Answer: (D)

$N{H}_{2}CON{H}_2+2NaOH\to 2N{H}_3+N{a}_{2}C{O}_3$
1 mole urea gives 2 moles ammonia as per the balance reaction.
$n_{urea}=\frac{0.6}{60}=0.01$ mole
$\therefore n_{\text{ammonia }}=2\times 0.01=0.02$ mole
Now,
$N{H}_3+HCl\to N{H}_{4}Cl$
$0.02$ mole $N{H}_3$ require $0.02$ mole $HCl$.
$n$-factor of $HCl$ is 1
$\therefore$ Molarity of $HCl=0.2M$
$0.2=\frac{n_{HCl}}{100ml}\times 1000$
$n_{HCl}=\frac{0.2\times 100}{1000}=0.02$ mol HCl
$\therefore 100ml$ of $0.2NHCl$