Ammoina carbonate when heated to `200^(@)C` gives a mixture of `NH_(3)` and `CO_(2)` vapour with a density of `13.0` What is the degree of dissociation of ammonium carbonate ?

Options

A. `3//2`
B. `1//2`
C. `2`
D. `1`

Answer

D

Explanation

`{:((NH_(4))CO_(3)(g),overset(200^(@)C)(hArr),2NH_(3)(g)+,CO_(2)(g)+,H_(2)O(g)),("Initial moles",1,0,0,0),("Equilibrium moles",1-alpha,2alpha,alpha,alpha):}`
Total moles `= (1-alpha)+(2alpha)+(alpha)+(alpha)`
`=1+3alpha`
During dissociation, the number of moles increases while the vapor density decreases, i.e., vapor density and number of moles are inversely related:
`(d_(i))/(d_(f))=(n_(f))/(n_(i))`
`(d_(i))/(d_(f))=(1+3alpha)/(1)`
`3alpha=(d_(i))/(d_(f))-1`
or `alpha=(1)/(3)((d_(i)-d_(f))/(d_(f)))`
The molecular mass of ammonium carbonate is 96. Since the vapor density is half of the molecular mass, we have
`alpha=(1)/(3)((48-13)/(13))`
`=(35)/(39)=0.9~~1`