The vapoour density of a mixture containing `NO_(2)` and `N_(2)O_(4)` is 38.3 at 300 K. the number of moles of `NO_(2)` in 100 g of the mixture is approximately

Options

A. 0.44
B. 4.4
C. 33.4
D. 3.34

Answer

To solve the problem, we need to find the number of moles of \( NO_2 \) in a 100 g mixture of \( NO_2 \) and \( N_2O_4 \) given that the vapor density of the mixture is 38.3 at 300 K.

### Step 1: Calculate the average molar mass of the mixture
The vapor density (VD) is related to the average molar mass (M) of the gas mixture by the formula:
\[ \text{Vapor Density} = \frac{\text{Average Molar Mass}}{2} \]
Given that the vapor density is 38.3, we can calculate the average molar mass:
\[ M = 2 \times \text{Vapor Density} = 2 \times 38.3 = 76.6 \, \text{g/mol} \]

### Step 2: Set up the equations for the mixture
Let \( x \) be the mass of \( NO_2 \) in grams in the 100 g mixture. Then, the mass of \( N_2O_4 \) will be \( 100 - x \) grams.

The molar masses are:
- Molar mass of \( NO_2 \) = 46 g/mol
- Molar mass of \( N_2O_4 \) = 92 g/mol

The number of moles of \( NO_2 \) and \( N_2O_4 \) can be expressed as:
\[ \text{Moles of } NO_2 = \frac{x}{46} \]
\[ \text{Moles of } N_2O_4 = \frac{100 - x}{92} \]

### Step 3: Write the equation for total moles
The total number of moles in the mixture can also be expressed as:
\[ \text{Total moles} = \frac{100}{76.6} \]

Setting the two expressions for total moles equal gives us:
\[ \frac{x}{46} + \frac{100 - x}{92} = \frac{100}{76.6} \]

### Step 4: Solve the equation
To solve for \( x \), we first find a common denominator for the left-hand side:
\[ \frac{2x}{92} + \frac{100 - x}{92} = \frac{100}{76.6} \]
This simplifies to:
\[ \frac{2x + 100 - x}{92} = \frac{100}{76.6} \]
\[ \frac{x + 100}{92} = \frac{100}{76.6} \]

Cross-multiplying gives:
\[ 76.6(x + 100) = 9200 \]
\[ 76.6x + 7660 = 9200 \]
\[ 76.6x = 9200 - 7660 \]
\[ 76.6x = 1540 \]
\[ x = \frac{1540}{76.6} \approx 20.1 \, \text{g} \]

### Step 5: Calculate the number of moles of \( NO_2 \)
Now that we have \( x \), we can find the number of moles of \( NO_2 \):
\[ \text{Moles of } NO_2 = \frac{x}{46} = \frac{20.1}{46} \approx 0.44 \, \text{moles} \]

### Final Answer
The number of moles of \( NO_2 \) in 100 g of the mixture is approximately **0.44 moles**.

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