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

Options

A. `3.11 g//` litre
B. `2.11g//` litre
C. `4.5g//` litre
D. None of these

Explanation

To solve the problem of finding 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: Understand the dissociation reaction
The dissociation of \( N_2O_4 \) into \( NO_2 \) can be represented as:
\[ N_2O_4 \rightleftharpoons 2NO_2 \]

### Step 2: Determine the degree of dissociation
Given that \( N_2O_4 \) is 20% dissociated, we can express this as:
\[ \alpha = 0.2 \]

### Step 3: Calculate the moles of reactants and products at equilibrium
Assuming we start with 1 mole of \( N_2O_4 \):
- Moles of \( N_2O_4 \) that dissociate = \( \alpha \times 1 = 0.2 \) moles
- Moles of \( NO_2 \) produced = \( 2 \times 0.2 = 0.4 \) moles

At equilibrium:
- Moles of \( N_2O_4 \) remaining = \( 1 - 0.2 = 0.8 \) moles
- Moles of \( NO_2 \) = \( 0.4 \) moles

### Step 4: Calculate the total moles at equilibrium
Total moles at equilibrium:
\[ \text{Total moles} = \text{Moles of } N_2O_4 + \text{Moles of } NO_2 = 0.8 + 0.4 = 1.2 \text{ moles} \]

### Step 5: Calculate the molar mass of the mixture
The molar mass of the components is:
- Molar mass of \( N_2O_4 \) = 92 g/mol
- Molar mass of \( NO_2 \) = 46 g/mol

Using the formula for the average molar mass of the mixture:
\[ M_{mixture} = \frac{(M_{N_2O_4} \times \text{moles of } N_2O_4) + (M_{NO_2} \times \text{moles of } NO_2)}{\text{Total moles}} \]
Substituting the values:
\[ M_{mixture} = \frac{(92 \times 0.8) + (46 \times 0.4)}{1.2} \]
Calculating:
\[ M_{mixture} = \frac{73.6 + 18.4}{1.2} = \frac{92}{1.2} = 76.67 \text{ g/mol} \]

### Step 6: Calculate the density of the mixture
Using the ideal gas law to find the density:
\[ \text{Density} = \frac{PM}{RT} \]
Where:
- \( P = 1 \text{ atm} \)
- \( M = 76.67 \text{ g/mol} \)
- \( R = 0.0821 \text{ L atm/(K mol)} \)
- \( T = 27^{\circ}C = 300 \text{ K} \)

Substituting the values:
\[ \text{Density} = \frac{1 \times 76.67}{0.0821 \times 300} \]
Calculating:
\[ \text{Density} = \frac{76.67}{24.63} \approx 3.11 \text{ g/L} \]

### Final Answer
The density of the equilibrium mixture of \( N_2O_4 \) and \( NO_2 \) at \( 27^{\circ}C \) and 1 atm is approximately \( 3.11 \text{ g/L} \).

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