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 180g of the mixture?

Options

A. 0.587
B. 0.782
C. 1.45
D. 3.45

Explanation

To solve the problem step-by-step, we will follow these calculations:

### 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 have 180 grams of the mixture. The number of moles (n) can be calculated using the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} \]
Substituting the values:
\[ n = \frac{180 \, \text{g}}{76.6 \, \text{g/mol}} \approx 2.349 \, \text{moles} \]

### Step 3: Set up the equation for the mixture
Let \( x \) be the mass of \( \text{NO}_2 \) in the mixture. The molar mass of \( \text{NO}_2 \) is 46 g/mol, and the molar mass of \( \text{N}_2\text{O}_4 \) is 92 g/mol. The mass of \( \text{N}_2\text{O}_4 \) can be expressed as \( 180 - x \).

The number of moles of \( \text{NO}_2 \) and \( \text{N}_2\text{O}_4 \) can be expressed as:
\[ \text{Moles of } \text{NO}_2 = \frac{x}{46} \]
\[ \text{Moles of } \text{N}_2\text{O}_4 = \frac{180 - x}{92} \]

### Step 4: Set up the equation for total moles
The total number of moles of the mixture is given as:
\[ \frac{x}{46} + \frac{180 - x}{92} = 2.349 \]

### Step 5: Solve the equation
To solve this equation, we first find a common denominator, which is 92:
\[ \frac{2x}{92} + \frac{180 - x}{92} = 2.349 \]
Combining the fractions gives:
\[ \frac{2x + 180 - x}{92} = 2.349 \]
This simplifies to:
\[ \frac{x + 180}{92} = 2.349 \]
Multiplying both sides by 92:
\[ x + 180 = 2.349 \times 92 \]
Calculating the right side:
\[ x + 180 = 216.058 \]
Now, solving for \( x \):
\[ x = 216.058 - 180 = 36.058 \, \text{g} \]

### Step 6: Calculate the number of moles of \( \text{NO}_2 \)
Now, we can find the number of moles of \( \text{NO}_2 \):
\[ \text{Moles of } \text{NO}_2 = \frac{x}{46} = \frac{36.058}{46} \approx 0.782 \, \text{moles} \]

### Final Answer:
The number of moles of \( \text{NO}_2 \) in 180 grams of the mixture is approximately **0.782 moles**.

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