The vapour density of mixture consisting of `NO_2` and `N_2O_4` is 38.3 at `26.7^@C`. Calculate the number of moles of `NO_2` I `100g` of the mixture.

Options

A. `0.2`
B. `0.4`
C. `0.8`
D. `1.6`

Explanation

To solve the problem of finding the number of moles of \( NO_2 \) in a 100g mixture of \( NO_2 \) and \( N_2O_4 \) with a vapor density of 38.3 at 26.7°C, we can follow these steps:

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

### Step 2: Calculate the Total Number of Moles in the Mixture
The total number of moles (\( N_T \)) in the 100g mixture can be calculated using the formula:
\[ N_T = \frac{\text{mass of mixture}}{\text{molar mass of mixture}} = \frac{100 \, \text{g}}{76.6 \, \text{g/mol}} \approx 1.306 \, \text{moles} \]

### 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. The mass of \( N_2O_4 \) will then be \( 100 - A \) grams. The molar masses of \( NO_2 \) and \( N_2O_4 \) are 46 g/mol and 92 g/mol, respectively.

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

### Step 4: Write the Total Moles Equation
The total number of moles in the mixture can be expressed as:
\[ n_{NO_2} + n_{N_2O_4} = N_T \]
Substituting the expressions for \( n_{NO_2} \) and \( n_{N_2O_4} \):
\[ \frac{A}{46} + \frac{100 - A}{92} = 1.306 \]

### Step 5: Solve the Equation
To solve for \( A \), we can find a common denominator (which is 92):
\[ \frac{2A}{92} + \frac{100 - A}{92} = 1.306 \]
Combining the fractions:
\[ \frac{2A + 100 - A}{92} = 1.306 \]
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 6: Calculate the Number of Moles of \( NO_2 \)
Now we can calculate the number of moles of \( NO_2 \):
\[ n_{NO_2} = \frac{A}{46} = \frac{20.072}{46} \approx 0.436 \, \text{moles} \]

### Conclusion
The number of moles of \( NO_2 \) in 100g of the mixture is approximately \( 0.436 \, \text{moles} \).

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