The vapour density of undecomposed `N_(2)O_(4)` is 46. When heated, the vapour density decreases to 24.5 due to its dissociation to `NO_(2)`. The per cent dissociation of `N_(2)O_(4)` at the final temperature is

Options

A. 88
B. 60
C. 40
D. 70

Answer

To solve the problem, we need to determine the percentage dissociation of \( N_2O_4 \) when it dissociates into \( NO_2 \). We will use the given vapor densities to find the percentage dissociation.

### Step-by-Step Solution:

1. **Understand the relationship between vapor density and molar mass**:
- The vapor density (VD) is related to the molar mass (M) of a gas by the formula:
\[ \text{VD} = \frac{M}{2} \]
- Therefore, the molar mass of \( N_2O_4 \) can be calculated from its vapor density of 46:
\[ M_{N_2O_4} = 2 \times 46 = 92 \, \text{g/mol} \]

2. **Calculate the molar mass of \( NO_2 \)**:
- The molar mass of \( NO_2 \) is:
\[ M_{NO_2} = 14 + 2 \times 16 = 46 \, \text{g/mol} \]

3. **Set up the dissociation reaction**:
- The dissociation of \( N_2O_4 \) can be represented as:
\[ N_2O_4 \rightleftharpoons 2 NO_2 \]

4. **Determine the initial and final conditions**:
- Let the initial number of moles of \( N_2O_4 \) be 1.
- If \( x \) is the amount that dissociates, then at equilibrium:
- Moles of \( N_2O_4 \) = \( 1 - x \)
- Moles of \( NO_2 \) = \( 2x \)

5. **Calculate the total number of moles at equilibrium**:
- Total moles at equilibrium = \( (1 - x) + 2x = 1 + x \)

6. **Use the vapor density after dissociation**:
- The vapor density after dissociation is given as 24.5, leading to:
\[ M = 2 \times 24.5 = 49 \, \text{g/mol} \]

7. **Set up the equation for average molar mass**:
- The average molar mass at equilibrium can be calculated as:
\[ \text{Average Molar Mass} = \frac{(1 - x) \cdot 92 + (2x) \cdot 46}{1 + x} \]
- Setting this equal to 49:
\[ \frac{(1 - x) \cdot 92 + (2x) \cdot 46}{1 + x} = 49 \]

8. **Cross-multiply and simplify**:
- Multiply both sides by \( (1 + x) \):
\[ (1 - x) \cdot 92 + (2x) \cdot 46 = 49(1 + x) \]
- Expanding both sides:
\[ 92 - 92x + 92x = 49 + 49x \]
- This simplifies to:
\[ 92 = 49 + 49x \]
- Rearranging gives:
\[ 43 = 49x \quad \Rightarrow \quad x = \frac{43}{49} \approx 0.8776 \]

9. **Calculate the percentage dissociation**:
- The percentage dissociation \( \alpha \) is given by:
\[ \alpha = \frac{x}{\text{initial moles}} \times 100 = \frac{0.8776}{1} \times 100 \approx 87.76\% \]

10. **Final answer**:
- The percentage dissociation of \( N_2O_4 \) is approximately 88%.