Vapour density of `N_(2)O_(4)` of x `60^(@)` C is found to be 30.6 The degree of dissociation of `N_(2) O_(4)` is :

Options

A. 0.1
B. 0.2
C. 0.4
D. 0.5

Answer

To find the degree of dissociation of \( N_2O_4 \) at \( 60^\circ C \) given that its vapor density is 30.6, we can follow these steps:

### Step 1: Understand the dissociation reaction
The dissociation of \( N_2O_4 \) can be represented as:
\[ N_2O_4 \rightleftharpoons 2 NO_2 \]
This means that one mole of \( N_2O_4 \) produces two moles of \( NO_2 \).

### Step 2: Calculate the molecular mass of \( N_2O_4 \)
The molecular mass of \( N_2O_4 \) is calculated as follows:
- Nitrogen (N) has an atomic mass of approximately 14 g/mol.
- Oxygen (O) has an atomic mass of approximately 16 g/mol.

Thus, the molecular mass of \( N_2O_4 \) is:
\[ M_{N_2O_4} = 2 \times 14 + 4 \times 16 = 28 + 64 = 92 \text{ g/mol} \]

### Step 3: Determine the initial vapor density
The initial vapor density (\( D \)) can be calculated using the formula:
\[ D = \frac{M}{2} \]
where \( M \) is the molecular mass of the gas. Therefore,
\[ D = \frac{92}{2} = 46 \text{ g/L} \]

### Step 4: Use the given vapor density to find the degree of dissociation
Let \( \alpha \) be the degree of dissociation. At equilibrium, the vapor density (\( d \)) is given as 30.6 g/L. The relationship for degree of dissociation is:
\[ \alpha = \frac{D - d}{d \cdot (n - 1)} \]
Here, \( n \) is the number of moles of products formed from one mole of reactants. Since one mole of \( N_2O_4 \) produces two moles of \( NO_2 \), \( n = 2 \).

Substituting the values:
\[ \alpha = \frac{46 - 30.6}{30.6 \cdot (2 - 1)} \]
\[ \alpha = \frac{15.4}{30.6 \cdot 1} \]
\[ \alpha = \frac{15.4}{30.6} \]
Calculating this gives:
\[ \alpha \approx 0.5032 \]

### Step 5: Conclusion
The degree of dissociation of \( N_2O_4 \) at \( 60^\circ C \) is approximately \( 0.5032 \), which can be rounded to \( 0.5 \).

### Final Answer
The degree of dissociation of \( N_2O_4 \) is \( 0.5 \).

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