Question

Ammonium carbamate when heated to ${200}^{o}C$ gives a mixture of ${NH}_3$ and ${CO}_2$ vapour with a density of $13$. What is the degree of dissociation of ammonium carbamate?

• A. $\frac{3}{2}$
• B. $\frac{3}{4}$
• C. $2$
• D. $1$

Answer 1

Answer: (D)

$1$

$Explanation:$

Ammonium carbamate formula is $N{H}_{2}COON{H}_4$

when it is heated ;

$N{H}_{2}COON{H}_{4\left(s\right)}\leftrightharpoons 2N{H}_{3\left(g\right)}+C{O}_{2\left(g\right)}$

Let $\alpha =$ degree of dissociation of Ammonium carbamate

Let $D=$ density of ammonium carbamate

$\because D=\frac{mol.wt}{2}$

where molecular wt. of $N{H}_{2}COON{H}_4=14+2+12+32+14+4=78$

Then $D=\frac{78}{2}=39$

Give vapour density $d=13$

$\alpha =\frac{D-d}{(n-1)d}$

where $n=$ no of products $=3$

$\alpha =\frac{39-13}{(3-1)\times 13}=\frac{26}{\left(2\right)\times 13}=1$

$\therefore \alpha =1$

Hence the correct option is option $D.$