Vapour density of mixture of `NO_(2)` and `N_(2) O_(4)` is 34.5 , then percentage abundance of `NO_(2)` in mixture is

Options

A. `50%`
B. `25%`
C. `40%`
D. `60%`

Answer

To find the percentage abundance of \( NO_2 \) in a mixture of \( NO_2 \) and \( N_2O_4 \) given that the vapor density of the mixture is 34.5, we can follow these steps:

### Step 1: Understand the Reaction
The equilibrium reaction between \( N_2O_4 \) and \( NO_2 \) can be represented as:
\[ N_2O_4 \rightleftharpoons 2 NO_2 \]

### Step 2: Define Initial Moles
Assume initially we have 1 mole of \( N_2O_4 \) and 0 moles of \( NO_2 \):
- Initial moles of \( N_2O_4 = 1 \)
- Initial moles of \( NO_2 = 0 \)

### Step 3: Define Change in Moles
Let \( x \) be the amount of \( N_2O_4 \) that dissociates. At equilibrium, the moles will be:
- Moles of \( N_2O_4 = 1 - x \)
- Moles of \( NO_2 = 2x \)

### Step 4: Total Moles at Equilibrium
The total number of moles at equilibrium is:
\[ \text{Total moles} = (1 - x) + 2x = 1 + x \]

### Step 5: Calculate Molecular Weight of the Mixture
The vapor density (\( VD \)) is related to the molecular weight (\( MW \)) by the formula:
\[ MW = 2 \times VD \]
Given \( VD = 34.5 \):
\[ MW = 2 \times 34.5 = 69 \]

### Step 6: Initial Molecular Weight
The molecular weight of \( N_2O_4 \) is:
\[ MW_{N_2O_4} = 2 \times 14 + 4 \times 16 = 28 + 64 = 92 \]

### Step 7: Set Up the Equation
Using the relationship between initial moles, moles at equilibrium, and molecular weights:
\[ \frac{\text{Initial moles}}{\text{Moles at equilibrium}} = \frac{MW_{equilibrium}}{MW_{initial}} \]
Substituting the values:
\[ \frac{1}{1 + x} = \frac{69}{92} \]

### Step 8: Solve for \( x \)
Cross-multiplying gives:
\[ 92 = 69(1 + x) \]
Expanding and rearranging:
\[ 92 = 69 + 69x \\ 69x = 92 - 69 \\ 69x = 23 \\ x = \frac{23}{69} \approx 0.333 \]

### Step 9: Calculate Total Moles at Equilibrium
Substituting \( x \) back to find total moles:
\[ \text{Total moles} = 1 + x = 1 + 0.333 = 1.333 \]

### Step 10: Calculate Moles of \( NO_2 \)
Moles of \( NO_2 \) at equilibrium:
\[ \text{Moles of } NO_2 = 2x = 2 \times 0.333 \approx 0.666 \]

### Step 11: Calculate Percentage Abundance of \( NO_2 \)
The percentage abundance of \( NO_2 \) is given by:
\[ \text{Percentage of } NO_2 = \left( \frac{\text{Moles of } NO_2}{\text{Total moles}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage of } NO_2 = \left( \frac{0.666}{1.333} \right) \times 100 \approx 49.62\% \]

### Final Answer
Thus, the percentage abundance of \( NO_2 \) in the mixture is approximately:
\[ \boxed{50\%} \]