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 90g of the mixture?

Options

A. 0.38
B. 0.98
C. 1.58
D. 2

Explanation

To solve the problem, we need to find the number of moles of \( NO_2 \) in a 90 g mixture of \( NO_2 \) and \( N_2O_4 \) with a given vapor density of 38.3 at 300 K. Here’s a step-by-step breakdown of the solution:

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

**Hint**: Remember that vapor density is half the molar mass for gases.

### Step 2: Calculate the Number of Moles of the Mixture
Now, we can calculate the total number of moles of the mixture using the formula:
\[ \text{Moles of mixture} = \frac{\text{mass of mixture}}{\text{molar mass of mixture}} \]
Substituting the values:
\[ \text{Moles of mixture} = \frac{90 \, \text{g}}{76.6 \, \text{g/mol}} \approx 1.17 \, \text{moles} \]

**Hint**: Make sure to use the total mass of the mixture for this calculation.

### Step 3: Set Up the Equation for Masses
Let \( x \) be the mass of \( NO_2 \) in the mixture. Therefore, the mass of \( N_2O_4 \) will be \( 90 - x \).

### Step 4: Calculate Moles of Each Component
The moles of \( NO_2 \) can be expressed as:
\[ \text{Moles of } NO_2 = \frac{x}{46} \]
And the moles of \( N_2O_4 \) can be expressed as:
\[ \text{Moles of } N_2O_4 = \frac{90 - x}{92} \]

### Step 5: Set Up the Total Moles Equation
According to the problem, the total moles of the mixture is equal to the sum of the moles of \( NO_2 \) and \( N_2O_4 \):
\[ \frac{x}{46} + \frac{90 - x}{92} = 1.17 \]

### Step 6: Solve for \( x \)
To solve this equation, we can find a common denominator, which is 92:
\[ \frac{92x + 46(90 - x)}{46 \times 92} = 1.17 \]
Simplifying the numerator:
\[ 92x + 4140 - 46x = 1.17 \times (46 \times 92) \]
Calculating \( 1.17 \times 46 \times 92 \):
\[ 1.17 \times 4232 = 4963.84 \]
So we have:
\[ 46x + 4140 = 4963.84 \]
Solving for \( x \):
\[ 46x = 4963.84 - 4140 \]
\[ 46x = 823.84 \]
\[ x \approx 17.64 \, \text{g} \]

### Step 7: Calculate Moles of \( NO_2 \)
Now, we can find the number of moles of \( NO_2 \):
\[ \text{Moles of } NO_2 = \frac{x}{46} = \frac{17.64}{46} \approx 0.383 \, \text{moles} \]

### Final Answer
The number of moles of \( NO_2 \) in 90 g of the mixture is approximately:
\[ \boxed{0.38} \]