A mixture contains `N_(2)O_(4)` and `NO_(2)` in the ratio 2:1 by volume. Calculate the vapour density of the mixture?

Options

A. `0.1`
B. `0.2`
C. `0.5`
D. `38.33`

Answer

To calculate the vapor density of a mixture containing \(N_2O_4\) and \(NO_2\) in the ratio of 2:1 by volume, we can follow these steps:

### Step 1: Determine the Molar Mass of Each Component

1. **Calculate the molar mass of \(N_2O_4\)**:
- The molar mass of nitrogen (N) = 14 g/mol
- The molar mass of oxygen (O) = 16 g/mol
- Molar mass of \(N_2O_4\) = \(2 \times 14 + 4 \times 16\)
- Molar mass of \(N_2O_4\) = \(28 + 64 = 92 \, \text{g/mol}\)

2. **Calculate the molar mass of \(NO_2\)**:
- Molar mass of \(NO_2\) = \(14 + 2 \times 16\)
- Molar mass of \(NO_2\) = \(14 + 32 = 46 \, \text{g/mol}\)

### Step 2: Calculate the Mass of Each Component in the Mixture

Given the volume ratio of \(N_2O_4\) to \(NO_2\) is 2:1, we can assume:
- Volume of \(N_2O_4\) = 2 units
- Volume of \(NO_2\) = 1 unit

3. **Calculate the mass of \(N_2O_4\)**:
- Mass of \(N_2O_4\) = Molar mass of \(N_2O_4\) × Volume of \(N_2O_4\)
- Mass of \(N_2O_4\) = \(92 \, \text{g/mol} \times 2 = 184 \, \text{g}\)

4. **Calculate the mass of \(NO_2\)**:
- Mass of \(NO_2\) = Molar mass of \(NO_2\) × Volume of \(NO_2\)
- Mass of \(NO_2\) = \(46 \, \text{g/mol} \times 1 = 46 \, \text{g}\)

### Step 3: Calculate the Total Mass and Total Volume of the Mixture

5. **Total mass of the mixture**:
- Total mass = Mass of \(N_2O_4\) + Mass of \(NO_2\)
- Total mass = \(184 \, \text{g} + 46 \, \text{g} = 230 \, \text{g}\)

6. **Total volume of the mixture**:
- Total volume = Volume of \(N_2O_4\) + Volume of \(NO_2\)
- Total volume = \(2 + 1 = 3 \, \text{units}\)

### Step 4: Calculate the Vapor Density of the Mixture

7. **Vapor density** is defined as:
\[ \text{Vapor Density} = \frac{\text{Total Mass}}{\text{Total Volume}} \times \frac{1}{2} \]
- Vapor Density = \(\frac{230 \, \text{g}}{3} \times \frac{1}{2} = \frac{230}{6} = 38.33 \, \text{g/L}\)

### Final Answer
The vapor density of the mixture is approximately **38.33 g/L**.

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