During thermal dissociation, the observed vapour density of `N_(2)O_(4)(g)` is 26.0. The extent of dissociation of `N_(2)O_(4)` is:

Options

A. `50%`
B. `87%`
C. `77%`
D. `23%`

Answer

To solve the problem of finding the extent of dissociation of \( N_2O_4(g) \) given its observed vapor density, we can follow these steps:

### Step 1: Understand the Reaction
The dissociation of \( N_2O_4 \) can be represented as:
\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]

### Step 2: Calculate the Molecular Mass of \( N_2O_4 \)
The molecular mass of \( N_2O_4 \) can be calculated as follows:
- Nitrogen (N) has a molar mass of approximately 14 g/mol. Since there are 2 nitrogen atoms:
\[ 2 \times 14 = 28 \text{ g/mol} \]
- Oxygen (O) has a molar mass of approximately 16 g/mol. Since there are 4 oxygen atoms:
\[ 4 \times 16 = 64 \text{ g/mol} \]
- Therefore, the total molecular mass of \( N_2O_4 \) is:
\[ 28 + 64 = 92 \text{ g/mol} \]

### Step 3: Calculate the Initial Vapor Density
The vapor density (VD) is given by the formula:
\[ VD = \frac{\text{Molecular Mass}}{2} \]
For \( N_2O_4 \):
\[ VD = \frac{92}{2} = 46 \]

### Step 4: Use the Given Observed Vapor Density
The observed vapor density is given as 26.0.

### Step 5: Set Up the Equation for Degree of Dissociation
Let \( x \) be the degree of dissociation of \( N_2O_4 \). The relationship between the initial vapor density (D), observed vapor density (d), and degree of dissociation can be expressed as:
\[ D - d = (n - 1) \cdot d \]
Where \( n \) is the number of moles of products formed from one mole of reactant. In this case, \( n = 2 \) (since 1 mole of \( N_2O_4 \) produces 2 moles of \( NO_2 \)):
\[ D - d = (2 - 1) \cdot d \]
Substituting the values:
\[ 46 - 26 = 1 \cdot 26 \]
This simplifies to:
\[ 20 = 26x \]

### Step 6: Solve for \( x \)
Rearranging gives:
\[ x = \frac{20}{26} = \frac{10}{13} \approx 0.769 \]

### Step 7: Calculate the Percentage of Degree of Dissociation
To find the percentage of degree of dissociation:
\[ \text{Percentage of } x = x \times 100 = 0.769 \times 100 \approx 77\% \]

### Final Answer
The extent of dissociation of \( N_2O_4 \) is approximately **77%**.

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