Question

Ammonium carbamate when heated to ${200}^{0}C$ gives a mixture of $N{H}_3$ and $C{O}_2$ with vapour density of 13.
$N{H}_2-\stackrel{\stackrel{O}{||}}{C}-ON{H}_4\left(s\right)\rightleftharpoons 2N{H}_3\left(g\right)+C{O}_2\left(g\right)$
The degree of dissociation of ammonium carbonate is:

• A. 1
• B. 0.5
• C. 0.75
• D. 0.9

Answer 1

The correct option is A 1

According to question-
$N{H}_2-\stackrel{\stackrel{O}{||}}{C}-ON{H}_4\left(s\right)\rightleftharpoons 2N{H}_3\left(g\right)+C{O}_2\left(g\right)$

Density initial (d) = $13$
Here, n = 3
Vapour density (D):
$D=\frac{MolecularmassofAmmoniumcarbamate}{2}$
$=\frac{78}{2}=39$
we know that the relation between density, vapor density, degree of dissociation and mole is given by
$\alpha =\frac{D-d}{(n-1)d}=\frac{39-13}{(3-1)\times 13}=\frac{26}{26}=1$
Hence degree of dissociation of ammonium carbonate is 1