Question

The vapour density of undecomposed \[{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}\] is 46. When heated, vapour density decreases to 24.5 due to its dissociation \[\text{ N}{{\text{O}}_{\text{2}}}\] . The ${\scriptstyle{}^{0}/{}_{0}}$ dissociation of \[{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}\] is:
A. 40
B. 57
C. 67
D. 87

Answer 1

Hint: The vapour density is a colligative property, so it depends on the amount of solute. The measure of the effect of solute on the vapour density is determined by Van't Hoff’s factor ‘i’ .It is the ratio of observed value to the theoretical or actual value of the solute. Since some of the solutes may undergo dissociation or association in the solution. The ‘i’ is related to the dissociation of solute.

Complete step by step answer:
We know that,
Vapour density of \[{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}\] is 46 (decomposed)
Vapour density of \[{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}\] is 24.5 (after having \[{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}\])

We know that the molecular weight ${{\text{M}}_{\text{w}}}$ can be expressed in terms of vapour density,
$\text{ }{{\text{M}}_{\text{w}}}\text{ = 2 }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\times \text{Vapour Density}$

Now we have to find the molecular weight before the dissociation. So, the molecular weight before the dissociation is,
$\text{ }{{\text{M}}_{\text{w}}}\text{ = 2 }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\times \text{Vapour Density = 2 }\times \text{ 46 = 92}$

So, now the molecular weight after the dissociation will be
 $\text{ }{{\text{M}}_{\text{w}}}\text{ = 2 }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\times \text{Vapour Density = 2 }\times \text{ 24}\text{.5 = 49}$

Now we have to find the Van't Hoff factor, which can also be denoted as i. So, the Van’t Hoff faction is,
$\text{Van't Hoff }\!\!'\!\!\text{ s factor = }\dfrac{\text{Observed value}}{\text{Actual value}}\text{ = }\dfrac{92}{49}\text{ = 1}\text{.877}\simeq \text{ 1}\text{.88}$

We know that the degree of dissociation α is given in terms of Van’t Hoff’s factor ‘i’ as follow:
  \[\text{ }\!\!\alpha\!\!\text{ =}\dfrac{i-1}{n-1}\].
In the above-mentioned formula,
$\text{ }\!\!\alpha\!\!\text{ }$ is the degree of dissociation, n is the number of particles after the dissociation.

So, we are given that nitrogen trioxide dissociates into nitrogen dioxide. The reaction is as shown below:
${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}\to \text{ 2N}{{\text{O}}_{\text{2}}}$
In the above-mentioned reaction, the value of n is 2.

So, putting the values in the formula of degree of dissociation.
We know that, \[\text{ i = 1}\text{.88}\] and \[\text{n =2}\] . So, the dissociation constant is obtained as follows:
\[\text{ }\!\!\alpha\!\!\text{ =}\dfrac{i-1}{n-1}\text{ = }\dfrac{1.88-1}{2-1}\text{ = 0}\text{.88}\]
There the percentage dissociation of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ into the $\text{N}{{\text{O}}_{\text{2}}}$ equals to the,
$\alpha \text{ }{\scriptstyle{}^{0}/{}_{0}}\text{ = 0}\text{.88 }\times \text{100}{\scriptstyle{}^{0}/{}_{0}}\text{ = 88}{\scriptstyle{}^{0}/{}_{0}}$
So the closest answer is 87 %.
So, the correct answer is “Option D”.

Note: When particles associates in the solution, the values of ‘i’ is:
$\text{i }\langle \text{ 1}$ , ‘i’ is less than unity.
Example, benzoic acid in benzene forms dimer
When the solute dissociates in the solution, then the value of ‘i’ is:
$\text{i }\rangle \text{ 1}$ , ‘i’ is greater than the unity.
For example, sodium chloride is a solution to form sodium and chloride ions.
When solute does not dissociate or associate in the solution, the value of ‘i’ is:
$\text{i = 1}$ , ‘i’ is equal to the unity ,For example, Glucose in water.