20% `N_(2)O_(4)` molecules are dissociated in a sample of gas at `27^(@)C` and 1 atm. Calculate the density of the equilibrium mixture.

Answer

To solve the problem of calculating the density of the equilibrium mixture of \( N_2O_4 \) and \( NO_2 \) after 20% of \( N_2O_4 \) has dissociated, we can follow these steps:

### Step 1: Determine the initial moles of \( N_2O_4 \)
Assume we start with 1 mole of \( N_2O_4 \).

### Step 2: Calculate the amount of \( N_2O_4 \) that dissociates
Since 20% of \( N_2O_4 \) is dissociated:
\[ \text{Dissociated } N_2O_4 = 0.2 \text{ moles} \]
Remaining \( N_2O_4 \):
\[ \text{Remaining } N_2O_4 = 1 - 0.2 = 0.8 \text{ moles} \]

### Step 3: Calculate the moles of \( NO_2 \) produced
From the dissociation reaction:
\[ N_2O_4 \rightleftharpoons 2 NO_2 \]
For every mole of \( N_2O_4 \) that dissociates, 2 moles of \( NO_2 \) are produced. Therefore, from 0.2 moles of \( N_2O_4 \) dissociating:
\[ \text{Moles of } NO_2 = 2 \times 0.2 = 0.4 \text{ moles} \]

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

### Step 5: Calculate the average molecular weight of the mixture
The molecular weight of \( N_2O_4 \) is 92 g/mol and for \( NO_2 \) it is 46 g/mol. Now, we calculate the total mass of the mixture:
\[ \text{Mass of } N_2O_4 = 0.8 \text{ moles} \times 92 \text{ g/mol} = 73.6 \text{ g} \]
\[ \text{Mass of } NO_2 = 0.4 \text{ moles} \times 46 \text{ g/mol} = 18.4 \text{ g} \]
\[ \text{Total mass} = 73.6 \text{ g} + 18.4 \text{ g} = 92 \text{ g} \]

### Step 6: Calculate the average molecular weight of the mixture
\[ \text{Molecular weight of mixture} = \frac{\text{Total mass}}{\text{Total moles}} = \frac{92 \text{ g}}{1.2 \text{ moles}} = 76.67 \text{ g/mol} \]

### Step 7: Calculate the density of the equilibrium mixture
Using the ideal gas law, the density \( d \) can be calculated using:
\[ d = \frac{PM}{RT} \]
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:
\[ d = \frac{1 \text{ atm} \times 76.67 \text{ g/mol}}{0.0821 \text{ L atm/(K mol)} \times 300 \text{ K}} \]
\[ d = \frac{76.67}{24.63} \approx 3.11 \text{ g/L} \]

### Final Answer
The density of the equilibrium mixture is approximately **3.11 g/L**.