JEE Main 2020 — Chemistry Physical Chemistry

$${\mathrm{NH}}_{2}{\mathrm{CONH}}_{2}+2\mathrm{NaOH}\rightarrow 2{\mathrm{NH}}_{3}+{\mathrm{Na}}_{2}{\mathrm{CO}}_{3}$$NH2​CONH2​+2NaOH→2NH3​+Na2​CO3​

By, stoichiometry, $1\mathrm{mol}$1mol of urea gives $2\mathrm{mol}$2mol ${\mathrm{NH}}_{3}$NH3​

Moles of urea =$=\frac{0.6\mathrm{gm}}{60\mathrm{gm}/\mathrm{mol}}=0.01\mathrm{mol}$=60gm/mol0.6gm​=0.01mol

So, $0.01$0.01 mole urea produce $0.02$0.02 mole ${\mathrm{NH}}_{3}$NH3​

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

$${\mathrm{NH}}_{3}+\mathrm{HCl}\rightarrow {\mathrm{NH}}_{4}\mathrm{Cl}$$NH3​+HCl→NH4​Cl

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

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

$$\therefore \mathrm{Mo}le of HCl=0.02 mole$$∴MoleofHCl=0.02mole

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

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

Hence, option $3$3 is correct.