At `27^(@)C` and 1 atm pressure ,`N_(2)O_(4)` is 20% dissociation into `NO_(@)` .What is the density of equilibrium mixture of `N_(2)O_(4) and NO_(2)` at `27^(@)C` and 1 atm?

Options

A. (a) `3.11 g//litre`
B. (b) `2.11 g//litre`
C. (c) `4.5 g//litre`
D. (d) none of these

Answer

To find the density of the equilibrium mixture of \( N_2O_4 \) and \( NO_2 \) at \( 27^\circ C \) and 1 atm, we can follow these steps:

### Step 1: Determine the degree of dissociation
Given that \( N_2O_4 \) is 20% dissociated, we can express this as:
\[ \alpha = \frac{20}{100} = 0.2 \]

### Step 2: Calculate the molecular mass of \( N_2O_4 \)
The molecular mass of \( N_2O_4 \) is:
\[ M_{N_2O_4} = 92 \text{ g/mol} \]

### Step 3: Calculate the molecular mass of the mixture
Using the formula for the molecular mass of the mixture:
\[ M_{mixture} = \frac{M_{N_2O_4} - \alpha \cdot M_{N_2O_4}}{N - 1} \]
Where \( N \) is the number of gaseous moles of products. For the dissociation:
\[ N = 2 \quad (\text{since } N_2O_4 \rightarrow 2 NO_2) \]
Thus, substituting in the values:
\[ M_{mixture} = \frac{92 - 0.2 \cdot 92}{2 - 1} = \frac{92 - 18.4}{1} = 73.6 \text{ g/mol} \]

### Step 4: Calculate the density of the mixture
Using the ideal gas equation \( PV = nRT \), we can express density \( \rho \) as:
\[ \rho = \frac{P \cdot M_{mixture}}{R \cdot T} \]
Where:
- \( P = 1 \text{ atm} \)
- \( R = 0.0821 \text{ L atm/(mol K)} \)
- \( T = 27^\circ C = 300 \text{ K} \)

Substituting the values:
\[ \rho = \frac{1 \cdot 73.6}{0.0821 \cdot 300} \]

### Step 5: Calculate the final density
Calculating the above expression:
\[ \rho = \frac{73.6}{24.63} \approx 2.99 \text{ g/L} \]

### Conclusion
The density of the equilibrium mixture of \( N_2O_4 \) and \( NO_2 \) at \( 27^\circ C \) and 1 atm is approximately \( 2.99 \text{ g/L} \).

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