JEE Main 2020 — Some Basic Concepts of Chemistry Question with Solution

${\mathrm{NH}}_2{\mathrm{CONH}}_2 + 2\mathrm{NaOH} \to 2{\mathrm{NH}}_3 +{\mathrm{Na}}_2{\mathrm{CO}}_3$

By, stoichiometry, $1 \mathrm{mol}$ of urea gives $2 \mathrm{mol}$ ${\mathrm{NH}}_3$

Moles of urea =$=\frac{0.6 \mathrm{gm}}{60 \mathrm{gm}/\mathrm{mol}}=0.01 \mathrm{mol}$

So, $0.01$ mole urea produce $0.02$ mole ${\mathrm{NH}}_3$

To neutralize ammonia $\mathrm{HCl}$ is required

${\mathrm{NH}}_3 +\mathrm{HCl}\to {\mathrm{NH}}_4\mathrm{Cl}$

By equation, it is clear that $1 \mathrm{mol}$ of ammonia is neutralized by 1mol of $\mathrm{HCl}$.

So, $0.02 \mathrm{mol}$ of ammonia neutralized by $0.02 \mathrm{mol}$ of $\mathrm{HCl}$

$\therefore \mathrm{Mo}\mathrm{l}\mathrm{e }\mathrm{o}\mathrm{f }\mathrm{H}\mathrm{C}\mathrm{l}=\mathrm{0.02 }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}$

$\mathrm{Equivalent} \mathrm{of} \mathrm{HCl}=\mathrm{normality}\times \mathrm{volume}=0.2 \mathrm{N}\times 100 \mathrm{mL}=0.02 \mathrm{equivalent}$

As $\mathrm{n}$-factor of $\mathrm{HCl}$ is $1$, so $\mathrm{number} \mathrm{of} \mathrm{equivalents}=\mathrm{number} \mathrm{of} \mathrm{moles}=0.02 \mathrm{moles}$

Hence, option $3$ is correct.