The vapour density of undecomposed `N_2O,` is 46. When heated, the vapour density decreases to 24.5 due to its dissociation to `NO_2.` The percentage dissociation of `N_2O_4` at the final temperature is

Options

A. 87
B. 60
C. 40
D. 70

Answer

To solve the problem, we need to determine the percentage dissociation of \( N_2O_4 \) into \( NO_2 \) based on the given vapor densities. Here’s a step-by-step solution:

### Step 1: Understand the relationship between vapor density and molecular weight
The vapor density (VD) is related to the molecular weight (MW) by the formula:
\[ \text{VD} = \frac{\text{MW}}{2} \]
This means that if we know the vapor density, we can find the molecular weight of the substance.

### Step 2: Calculate the molecular weight of undecomposed \( N_2O_4 \)
Given that the vapor density of undecomposed \( N_2O_4 \) is 46:
\[ \text{MW of } N_2O_4 = 2 \times 46 = 92 \, \text{g/mol} \]

### Step 3: Calculate the molecular weight of the dissociated mixture
When heated, the vapor density decreases to 24.5:
\[ \text{MW of the mixture} = 2 \times 24.5 = 49 \, \text{g/mol} \]

### Step 4: Set up the dissociation equation
Assume we start with 100 moles of \( N_2O_4 \). Let \( x \) be the number of moles that dissociate into \( NO_2 \):
\[ N_2O_4 \rightleftharpoons 2 NO_2 \]
If \( x \) moles of \( N_2O_4 \) dissociate, then:
- Moles of \( N_2O_4 \) remaining = \( 100 - x \)
- Moles of \( NO_2 \) formed = \( 2x \)

### Step 5: Calculate the total number of moles after dissociation
The total number of moles after dissociation is:
\[ \text{Total moles} = (100 - x) + 2x = 100 + x \]

### Step 6: Calculate the average molecular weight of the mixture
The average molecular weight of the mixture can be calculated using the formula:
\[ \text{Average MW} = \frac{\text{Total mass}}{\text{Total moles}} \]
The total mass of the mixture is:
\[ \text{Total mass} = (100 - x) \times 92 + 2x \times 46 \]
Thus,
\[ \text{Total mass} = 9200 - 92x + 92x = 9200 + 92x \]
Now substituting into the average molecular weight formula:
\[ 49 = \frac{9200 + 92x}{100 + x} \]

### Step 7: Solve for \( x \)
Cross-multiplying gives:
\[ 49(100 + x) = 9200 + 92x \]
Expanding and rearranging:
\[ 4900 + 49x = 9200 + 92x \]
\[ 4900 - 9200 = 92x - 49x \]
\[ -4300 = 43x \]
\[ x = \frac{-4300}{43} = 100 \]

### Step 8: Calculate percentage dissociation
The percentage dissociation is given by:
\[ \text{Percentage dissociation} = \left( \frac{x}{100} \right) \times 100 = x\% \]
Substituting \( x = 100 \):
\[ \text{Percentage dissociation} = 100\% \]

### Final Answer
The percentage dissociation of \( N_2O_4 \) at the final temperature is **100%**.