Vapour density of the equilibrium mixture of `NO_(2)` and `N_(2)O_(4)` is found to be `40` for the equilibrium
`N_(2)O_(4) hArr 2NO_(2)`
Calculate

Explanation

To solve the problem step by step, we will calculate the molecular weight, degree of dissociation, and the percentage of \( NO_2 \) in the equilibrium mixture of \( N_2O_4 \) and \( NO_2 \).

### Step 1: Calculate the Molecular Weight
Given that the vapor density of the equilibrium mixture is 40, we can use the relationship between vapor density and molecular weight:

\[ \text{Vapor Density} = \frac{\text{Molecular Weight}}{2} \]

From the problem, we have:

\[ 40 = \frac{\text{Molecular Weight}}{2} \]

To find the molecular weight, we rearrange the equation:

\[ \text{Molecular Weight} = 40 \times 2 = 80 \, \text{g/mol} \]

### Step 2: Calculate the Degree of Dissociation
The degree of dissociation \( x \) can be calculated using the formula:

\[ x = \frac{\text{Theoretical Molecular Weight} - \text{Observed Molecular Weight}}{\text{Observed Molecular Weight}} \]

First, we need to find the theoretical molecular weight of \( N_2O_4 \):

\[ \text{Molecular Weight of } N_2O_4 = (14 \times 2) + (16 \times 4) = 28 + 64 = 92 \, \text{g/mol} \]

Now we can substitute the values into the degree of dissociation formula:

\[ x = \frac{92 - 80}{80} = \frac{12}{80} = 0.15 \]

### Step 3: Calculate the Percentage of \( NO_2 \) in the Mixture
To find the percentage of \( NO_2 \) in the mixture, we need to determine the number of moles of \( NO_2 \) and the total number of moles at equilibrium.

- The number of moles of \( NO_2 \) produced is \( 2x = 2 \times 0.15 = 0.30 \).
- The number of moles of \( N_2O_4 \) remaining is \( 1 - x = 1 - 0.15 = 0.85 \).

Now, the total number of moles at equilibrium is:

\[ \text{Total Moles} = 2x + (1 - x) = 0.30 + 0.85 = 1.15 \]

Now we can calculate the percentage of \( NO_2 \):

\[ \text{Percentage of } NO_2 = \left( \frac{2x}{\text{Total Moles}} \right) \times 100 = \left( \frac{0.30}{1.15} \right) \times 100 \approx 26.08\% \]

### Final Results
1. **Molecular Weight**: 80 g/mol
2. **Degree of Dissociation**: 0.15
3. **Percentage of \( NO_2 \)**: 26.08%

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