In the problem number `21`, the number of mole of `N_(2)O_(4)` in `100 g` of the mixture is:

Options

A. `0.43`
B. `0.86`
C. `0.57`
D. `0.2`

Explanation

To solve the problem of finding the number of moles of \( N_2O_4 \) in a 100 g mixture of \( NO_2 \) and \( N_2O_4 \) given that the vapor density of the mixture is 38.3 g/L, we can follow these steps:

### Step 1: Calculate the Molar Mass of the Mixture
The molar mass of a gas mixture can be calculated using the formula:
\[ \text{Molar Mass} = 2 \times \text{Vapor Density} \]
Given that the vapor density of the mixture is 38.3 g/L, we can calculate:
\[ \text{Molar Mass} = 2 \times 38.3 = 76.6 \, \text{g/mol} \]

### Step 2: Calculate the Total Number of Moles of the Mixture
Using the mass of the mixture (100 g) and the molar mass calculated in Step 1, we can find the total number of moles (\( N_T \)):
\[ N_T = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{100 \, \text{g}}{76.6 \, \text{g/mol}} \approx 1.306 \, \text{moles} \]

### Step 3: Set Up the Equation for Moles of \( NO_2 \) and \( N_2O_4 \)
Let \( A \) be the mass of \( NO_2 \) in grams. Then the mass of \( N_2O_4 \) will be \( 100 - A \) grams. The number of moles of each gas can be expressed as:
- Moles of \( NO_2 = \frac{A}{46} \)
- Moles of \( N_2O_4 = \frac{100 - A}{92} \)

According to the problem, the total moles of the mixture can be expressed as:
\[ \frac{A}{46} + \frac{100 - A}{92} = N_T \]
Substituting \( N_T \) from Step 2:
\[ \frac{A}{46} + \frac{100 - A}{92} = 1.306 \]

### Step 4: Solve for \( A \)
To solve for \( A \), we can find a common denominator (which is 92) and rewrite the equation:
\[ \frac{2A}{92} + \frac{100 - A}{92} = 1.306 \]
Combining the fractions gives:
\[ \frac{2A + 100 - A}{92} = 1.306 \]
This simplifies to:
\[ \frac{A + 100}{92} = 1.306 \]
Multiplying both sides by 92:
\[ A + 100 = 120.072 \]
Thus:
\[ A = 120.072 - 100 = 20.072 \, \text{g} \]

### Step 5: Calculate the Mass of \( N_2O_4 \)
Now we can find the mass of \( N_2O_4 \):
\[ \text{Mass of } N_2O_4 = 100 - A = 100 - 20.072 = 79.928 \, \text{g} \]

### Step 6: Calculate the Number of Moles of \( N_2O_4 \)
Finally, we can calculate the number of moles of \( N_2O_4 \):
\[ \text{Moles of } N_2O_4 = \frac{79.928}{92} \approx 0.868 \, \text{moles} \]

### Final Answer
The number of moles of \( N_2O_4 \) in 100 g of the mixture is approximately **0.868 moles**.

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