Question

The vapor density of undecomposed ${{{N}}_2}{{{O}}_4}$ is $46$. When heated, vapor density decreases to $24.5$ due to its dissociation to ${{N}}{{{O}}_2}$. The percentage dissociation of ${{{N}}_2}{{{O}}_4}$ at the final temperature is:
A. $88$
B. $60$
C. $40$
D. $70$

Answer 1

Hint: We know that the vapor density is an example of a colligative property. Thus it is obvious that it is dependent on the amount of solute. Van’t Hoff factor is used to determine how the amount of solute influences the vapor density.

Complete step by step answer:
The chemical reaction showing the decomposition of ${{{N}}_2}{{{O}}_4}$ to ${{N}}{{{O}}_2}$ is given below:
${{{N}}_2}{{{O}}_4} \rightleftharpoons {{N}}{{{O}}_2}$
First we have to calculate the molecular mass of undecomposed ${{{N}}_2}{{{O}}_4}$ and decomposed ${{{N}}_2}{{{O}}_4}$. It is given that the vapor density of undecomposed ${{{N}}_2}{{{O}}_4}$, ${\rho _{{{un}}}} = 46$ and that of decomposed ${{{N}}_2}{{{O}}_4}$, ${\rho _{{{dec}}}} = 24.5$.
Thus the molecular weight can be calculated from the vapor density values since molecular weight is the double of vapor density.
i.e. molecular weight of undecomposed ${{{N}}_2}{{{O}}_4}$, ${{{M}}_{{{un}}}} = 2 \times {\rho _{{{un}}}} = 2 \times 46 = 92$
Molecular weight of decomposed ${{{N}}_2}{{{O}}_4}$, ${{{M}}_{{{dec}}}} = 2 \times {\rho _{{{dec}}}} = 2 \times 24.5 = 49$
Now let’s calculate the van’t Hoff factor, ${{i}}$. It is the ratio of observed value of weight of solute to the theoretical value of weight of solute. It can be calculated by dividing ${{{M}}_{{{un}}}}$ by ${{{M}}_{{{dec}}}}$.
i.e. ${{i = }}\dfrac{{{{{M}}_{{{un}}}}}}{{{{{M}}_{{{dec}}}}}} = \dfrac{{92}}{{49}} = 1.88$
Next, we have to calculate the degree of dissociation, $\alpha $ using the formula given below:
$\alpha = \dfrac{{{{i}} - 1}}{{{{n}} - 1}}$, where ${{n}}$ is the number of moles formed after decomposition, i.e. ${{n = 2}}$.
On substitution, we get
$\alpha = \dfrac{{{{i}} - 1}}{{{{n}} - 1}} = \dfrac{{1.88 - 1}}{{2 - 1}} = 0.88$
Percentage dissociation can be calculated by just multiplying the degree of dissociation by $100$.
i.e. $\% = \alpha \times 100 = 0.88 \times 100 = 88\% $
Thus the percentage dissociation is $88$.

Hence, option A is the correct answer.

Note: The van’t Hoff factor value may vary depending on the decomposition or combination of the atoms or molecules. If the van’t Hoff factor value is less than one, then the atoms or molecules are combined in the solution. When it is greater than one, the solute gets decomposed in the solution. When it is equal to one, it neither gets decomposed nor combined in the solution.