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 190g of the mixture?

Options

A. 1.829
B. 0.829
C. 3
D. 1.15

Explanation

To solve the problem step by step, we will follow the outlined approach:

### Step 1: Calculate the molar mass of the mixture
The vapor density (VD) of the mixture is given as 38.3. The molar mass (M) can be calculated using the formula:
\[ M = \text{Vapor Density} \times 2 \]
Substituting the value:
\[ M = 38.3 \times 2 = 76.6 \, \text{g/mol} \]

### Step 2: Calculate the number of moles of the mixture
We know the mass of the mixture is 190 g. The number of moles (n) of the mixture can be calculated using the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} \]
Substituting the values:
\[ n = \frac{190 \, \text{g}}{76.6 \, \text{g/mol}} \approx 2.48 \, \text{moles} \]

### Step 3: Set up the equation for the masses of NO2 and N2O4
Let the mass of \( NO_2 \) be \( x \) grams. Therefore, the mass of \( N_2O_4 \) will be \( 190 - x \) grams.

### Step 4: Calculate the number of moles of NO2 and N2O4
The number of moles of \( NO_2 \) can be calculated as:
\[ \text{moles of } NO_2 = \frac{x}{46} \]
The number of moles of \( N_2O_4 \) can be calculated as:
\[ \text{moles of } N_2O_4 = \frac{190 - x}{92} \]

### Step 5: 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 \( N_2O_4 \):
\[ \frac{x}{46} + \frac{190 - x}{92} = 2.48 \]

### Step 6: Solve the equation
To solve the equation, we first find a common denominator, which is 92:
\[ \frac{2x}{92} + \frac{190 - x}{92} = 2.48 \]
Combining the fractions:
\[ \frac{2x + 190 - x}{92} = 2.48 \]
This simplifies to:
\[ \frac{x + 190}{92} = 2.48 \]
Multiplying both sides by 92:
\[ x + 190 = 2.48 \times 92 \]
Calculating the right side:
\[ x + 190 = 228.16 \]
Now, solving for \( x \):
\[ x = 228.16 - 190 = 38.16 \, \text{g} \]

### Step 7: Calculate the number of moles of NO2
Now, we can find the number of moles of \( NO_2 \):
\[ \text{moles of } NO_2 = \frac{x}{46} = \frac{38.16}{46} \approx 0.829 \, \text{moles} \]

### Final Answer
The number of moles of \( NO_2 \) in 190 g of the mixture is approximately **0.829 moles**.

---