Question

The ammonia ${\mathrm{NH}}_3$ released on the quantitative reaction of $0.6\mathrm{g}$ urea (${\mathrm{NH}}_2{\mathrm{CONH}}_2$) with Sodium hydroxide ($\mathrm{NaOH}$)can be neutralized by

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

Answer 1

The correct option is D

$100\mathrm{ml}$ of $0.2\mathrm{N}\mathrm{HCl}$

Step 1: Given data

Mass of urea ${\mathrm{NH}}_2{\mathrm{CONH}}_2$$=0.6\mathrm{g}$

Step 2: Calculating normality

$\underset{\mathrm{urea}}{{\mathrm{NH}}_2{\mathrm{CONH}}_2\left(\mathrm{s}\right)}+\underset{\mathrm{Sodium}\mathrm{hydroxide}}{2\mathrm{NaOH}\left(\mathrm{l}\right)}\to \underset{\mathrm{Ammonia}}{2{\mathrm{NH}}_3\left(\mathrm{g}\right)}+\underset{\mathrm{Sodium}\mathrm{carbonate}}{{\mathrm{Na}}_2{\mathrm{CO}}_3\left(\mathrm{s}\right)}$

From the reaction, we can say that one mole of urea gives two moles of ammonia.

Moles of urea$=\frac{\mathrm{Given}\mathrm{mass}}{\mathrm{Molecular}\mathrm{mass}}$

$=\frac{0.6}{60}\Rightarrow 0.01$moles

Moles of Ammonia$=2\times$moles of urea

$=2\times 0.01\Rightarrow 0.02$moles

Now, by reacting Ammonia with Hydrochloric acid we get $\underset{\mathrm{Ammonia}}{{\mathrm{NH}}_3\left(\mathrm{l}\right)}+\underset{\mathrm{Hydrochloric}\mathrm{acid}}{\mathrm{HCl}\left(\mathrm{l}\right)}\to \underset{\mathrm{Ammonium}\mathrm{chloride}}{{\mathrm{NH}}_4\mathrm{Cl}\left(\mathrm{s}\right)}$

From the reaction, we can say that $0.02$ moles required $0.02$ moles of Hydrochloric acid

n factor of $\mathrm{HCl}=1$ as it contains one Hydrogen ion

Therefore, the Molarity of $\mathrm{HCl}=\frac{\mathrm{moles}}{\mathrm{Volume}\mathrm{in}\mathrm{litres}}$

$=\frac{0.02}{100}\times 1000\Rightarrow 0.2\mathrm{mol}$

Calculating moles from molarity

Molarity of$\mathrm{HCl}=\frac{\mathrm{Moles}}{\mathrm{Volume}\mathrm{in}\mathrm{litre}}$

$0.02=\frac{\mathrm{moles}\mathrm{HCl}}{100}\times 1000$

Moles of $\mathrm{HCl}=0.02$

Normality$=\frac{\mathrm{Molarity}}{\mathrm{Volume}\mathrm{in}\mathrm{Litres}}$

$=\frac{0.02}{0.1}\Rightarrow 0.2\mathrm{N}$

Therefore, $100\mathrm{ml}$ of $0.2\mathrm{N}\mathrm{HCl}$ is required for neutralizing ammonia.