Question

The vapour density of undecomposed ${N_2}{O_4}$ is $46$ .when heated, vapour density decreases to $24.5$ due to its dissociation to $N{O_2}$ . The $\%$dissociation of ${N_2}{O_4}$ is
A) $40$
B) $57$
C) $67$
D) $87$

Answer 1

We need to know that the vapour density is used to measure the molecular weight of the molecule. On applying heating, the vapour density of the molecule decreases. ${N_2}{O_4}$ is nitrogen pentoxide. It is one of the oxide compounds of nitrogen. The oxidation state of nitrogen in ${N_2}{O_4}$ is $+ 5$.
Formula used : Molecular weight is equal to twice of the vapour density of the molecule.
${\text{Molecular weight}} = 2 \times {\text{Vapour density}}$
${\text{Molecular weight}} = {\text{Sum of the atomic weight of the atoms in the molecules}}$
Average molecular weight after dissociation $= \dfrac{{{\text{Number of moles}} \times {\text{Molecular weight}}}}{{{\text{Total number of the product formed}}}}$
The approximate % of dissociation $= \dfrac{{{\text{Total number of moles after dissociation}}}}{{{\text{Total number of moles before dissociation}}}} \times 100$

Complete step by step answer:
The complete equation of the decomposition of ${N_2}{O_4}$
$2{N_2}{O_4} \to 4N{O_2}$
The vapour density of ${N_2}{O_4}$ is 46
The vapour density of undissociated of ${N_2}{O_4}$ is 24.5
The molecular weight before dissociation of ${N_2}{O_4}$ is equal to twice of the vapour density before dissociation of ${N_2}{O_4}$
The molecular weight before dissociation of ${N_2}{O_4}$ is calculated,
${\text{Molecular weight}} = 2 \times {\text{Vapour density}}$
$= 2 \times 46$
$= 92$
The molecular weight of ${N_2}{O_4}$ is $92g$
The molecular weight after dissociation of ${N_2}{O_4}$ is equal to twice of the vapour density after dissociation of ${N_2}{O_4}$
The molecular weight after dissociation of ${N_2}{O_4}$ is calculated,
The vapour density of undissociated of ${N_2}{O_4}$ is 24.5
${\text{Molecular weight}} = 2 \times {\text{Vapour density}}$
Now we can substitute the known values we get,
$= 2 \times 24.5$
On simplification we get,
$= 49$
The molecular weight of after undissociated ${N_2}{O_4}$ is $49g$
The molecular weight of $N{O_2}$ ,
The atomic weight of nitrogen is $14g$
The atomic weight of oxygen is $16g$
${\text{Molecular weight}} = {\text{Sum of the atomic weight of the atoms in the molecules}}$
One nitrogen and two oxygen in $N{O_2}$
= atomic weight of nitrogen + 2(atomic weight of oxygen)
$= 14 + \left( {2 \times 16} \right)$
On simplification we get,
$= 46g$
The molecular weight of $N{O_2}$ is $46g$
The initial total number of $100$ moles of before dissociated in ${N_2}{O_4}$.
$X$ Moles of ${N_2}{O_4}$ is dissociated in ${N_2}{O_4}$.
$100 - X$ Moles after dissociated in ${N_2}{O_4}$ .
$2X$ Moles of $N{O_2}$ is produced after dissociation.
The total number of moles produced after dissociation is $(100 - X) + 2X = 100 + X$
The average molecular weight of the molecule after dissociation is equal to the product of the moles of the product and molecular weight of the product divided by the total number of the moles after dissociation.
Average molecular weight after dissociation $= \dfrac{{{\text{Number of moles}} \times {\text{Molecular weight}}}}{{{\text{Total number of the product formed}}}}$
$\dfrac{{92(100 - X) + 46(2X)}}{{(100 + X)}} = 49$
On simplification we get,
$X = 87.7551$
The number of moles dissociated is $88$ .
The percentage of the dissociation is equal to the total number of moles dissociation is divided by the total number of moles multiplied by $100$.
The approximate % of dissociation $= \dfrac{{{\text{Total number of moles after dissociation}}}}{{{\text{Total number of moles before dissociation}}}} \times 100$
$= \dfrac{{88}}{{100}} \times 100$
The approximately $\%$ of dissociation is $88$
Option D is correct, because value is $87$ nearest value of calculation in dissociation.

Note:
We need to know the molecular weight of the molecules calculated by the sum of the atomic weight of the atom in molecules. The molecular weight is calculated by using the vapour density of the molecule. The association and dissociation of the molecules is dependent on the stability of the product. Percentage of dissociation is used to write the partial pressure and concentration. By using vapour density, various physical properties of the molecules are determined.