Question

The ammonia $(NH_3)$ released on quantitative reaction of $0.6\, g$ urea $(NH_2CONH_2)$ with sodium hydroxide (NaOH) can be neutralized by :

• A. $200\, ml$ of $0.2\, N\, HCl$
• B. $200 \,ml$ of $0.4 \,N \,HCl$
• C. $100\, ml$ of $0.1\, N\, HCl$
• D. $100 \,ml$ of $0.2 \,N \,HCl$

Answer 1

Answer: (D)

$NH _{2} CONH _{2}+2 NaOH \rightarrow 2 NH _{3}+ Na _{2} CO _{3}$
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,
$NH _{3}+ HCl \rightarrow NH _{4} Cl$
$0.02$ mole $NH _{3}$ require $0.02$ mole $HCl$.
$n$-factor of $HCl$ is 1
$\therefore $ Molarity of $HCl =0.2\, M$
$0.2=\frac{n_{HCl}}{100\, ml } \times 1000$
$n _{ HCl }=\frac{0.2 \times 100}{1000}=0.02$ mol HCl
$\therefore 100\, ml$ of $0.2\, NHCl$