At `340 K` and `1` atm pressure, `N_(2)O_(4)` is `66%` into `NO_(2)`. What volume of `10 g N_(2)O_(4)` occupy under these conditions?

Explanation

To solve the problem, we need to determine the volume occupied by 10 g of \( N_2O_4 \) at 340 K and 1 atm pressure, given that 66% of \( N_2O_4 \) dissociates into \( NO_2 \). Here’s a step-by-step breakdown of the solution:

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

### Step 2: Determine the Degree of Dissociation
Given that \( N_2O_4 \) is 66% dissociated, we can express this as:
\[ \alpha = 0.66 \]
where \( \alpha \) is the degree of dissociation.

### Step 3: Calculate Moles at Equilibrium
Assume we start with 1 mole of \( N_2O_4 \):
- Initial moles of \( N_2O_4 = 1 \)
- Moles of \( N_2O_4 \) at equilibrium = \( 1 - \alpha = 1 - 0.66 = 0.34 \)
- Moles of \( NO_2 \) produced = \( 2\alpha = 2 \times 0.66 = 1.32 \)

Total moles at equilibrium:
\[ \text{Total moles} = \text{moles of } N_2O_4 + \text{moles of } NO_2 = 0.34 + 1.32 = 1.66 \]

### Step 4: Calculate Moles of \( N_2O_4 \) in 10 g
Next, we need to find out how many moles are in 10 g of \( N_2O_4 \). The molecular weight of \( N_2O_4 \) is calculated as follows:
- Atomic weight of Nitrogen (N) = 14 g/mol
- Atomic weight of Oxygen (O) = 16 g/mol
- Molecular weight of \( N_2O_4 = (2 \times 14) + (4 \times 16) = 28 + 64 = 92 \) g/mol

Now, calculate the number of moles in 10 g of \( N_2O_4 \):
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{10 \text{ g}}{92 \text{ g/mol}} \approx 0.1087 \text{ moles} \]

### Step 5: Calculate Total Moles at Equilibrium for 10 g
Using the ratio from Step 3, if 1 mole of \( N_2O_4 \) gives 1.66 moles at equilibrium, then:
\[ \text{Total moles at equilibrium} = 1.66 \times 0.1087 \approx 0.180 \text{ moles} \]

### Step 6: Use the Ideal Gas Law to Find Volume
We can now use the Ideal Gas Law \( PV = nRT \) to find the volume \( V \):
- \( P = 1 \text{ atm} \)
- \( n \approx 0.180 \text{ moles} \)
- \( R = 0.0821 \text{ L atm/(K mol)} \)
- \( T = 340 \text{ K} \)

Rearranging the Ideal Gas Law gives:
\[ V = \frac{nRT}{P} \]
Substituting the values:
\[ V = \frac{0.180 \text{ moles} \times 0.0821 \text{ L atm/(K mol)} \times 340 \text{ K}}{1 \text{ atm}} \approx 5.04 \text{ L} \]

### Final Answer
The volume occupied by 10 g of \( N_2O_4 \) under the given conditions is approximately **5.04 liters**.

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