A mixture of `NO_2` and `N_2O_4` has a vapour density of 38.3 at 300K. What is the number of moles of `NO_2` in 130g of the mixture?

Options

A. 0.565
B. 0.658
C. 1.578
D. 1.75

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the molar mass of the mixture
Given the vapor density (VD) of the mixture is 38.3, we can calculate the molar mass (M) of the mixture using the formula:
\[ M = 2 \times \text{VD} \]
\[ M = 2 \times 38.3 = 76.6 \, \text{g/mol} \]

### Step 2: Calculate the number of moles of the mixture
We have a total mass of the mixture as 130 g. The number of moles (n) of the mixture can be calculated using the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} \]
\[ n = \frac{130 \, \text{g}}{76.6 \, \text{g/mol}} \]
\[ n \approx 1.697 \, \text{moles} \]

### Step 3: Set up the equations for the mixture
Let the mass of \( NO_2 \) be \( X \) grams. Then the mass of \( N_2O_4 \) will be \( 130 - X \) grams. The number of moles of \( NO_2 \) and \( N_2O_4 \) can be expressed as:
- Moles of \( NO_2 = \frac{X}{46} \)
- Moles of \( N_2O_4 = \frac{130 - X}{92} \)

### Step 4: Set up the equation for total moles
The total number of moles of the mixture is the sum of the moles of \( NO_2 \) and \( N_2O_4 \):
\[ \frac{X}{46} + \frac{130 - X}{92} = 1.697 \]

### Step 5: Solve the equation
To solve for \( X \), we first find a common denominator for the fractions, which is 92:
\[ \frac{2X}{92} + \frac{130 - X}{92} = 1.697 \]
Combining the fractions gives:
\[ \frac{2X + 130 - X}{92} = 1.697 \]
\[ \frac{X + 130}{92} = 1.697 \]
Now, multiply both sides by 92:
\[ X + 130 = 1.697 \times 92 \]
\[ X + 130 = 156.964 \]
Now, isolate \( X \):
\[ X = 156.964 - 130 \]
\[ X \approx 26.964 \, \text{g} \]

### Step 6: Calculate the number of moles of \( NO_2 \)
Now that we have the mass of \( NO_2 \), we can calculate the number of moles:
\[ \text{Moles of } NO_2 = \frac{X}{46} \]
\[ \text{Moles of } NO_2 = \frac{26.964}{46} \]
\[ \text{Moles of } NO_2 \approx 0.585 \, \text{moles} \]

### Final Answer
The number of moles of \( NO_2 \) in 130 g of the mixture is approximately **0.585 moles**.

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