If air contains `N_(2)` and `O_(2)` in volume ratio `4 : 1`, the average vapour density of air is:

Options

A. 14.4
B. 15.5
C. 16.5
D. 29

Explanation

To find the average vapor density of air containing nitrogen (N₂) and oxygen (O₂) in a volume ratio of 4:1, we can follow these steps:

### Step 1: Identify the molecular masses of the gases
- The molecular mass of nitrogen (N₂) is calculated as:
\[ \text{Molecular mass of } N_2 = 14 \times 2 = 28 \text{ g/mol} \]
- The molecular mass of oxygen (O₂) is calculated as:
\[ \text{Molecular mass of } O_2 = 16 \times 2 = 32 \text{ g/mol} \]

### Step 2: Determine the contribution of each gas to the average vapor density
- The volume ratio of N₂ to O₂ is given as 4:1. This means:
- For N₂:
\[ \text{Contribution of } N_2 = \text{Molecular mass of } N_2 \times \text{Volume ratio} = 28 \times 4 = 112 \]
- For O₂:
\[ \text{Contribution of } O_2 = \text{Molecular mass of } O_2 \times \text{Volume ratio} = 32 \times 1 = 32 \]

### Step 3: Calculate the total contribution
- Add the contributions of both gases:
\[ \text{Total contribution} = 112 + 32 = 144 \]

### Step 4: Calculate the average vapor density
- To find the average vapor density, divide the total contribution by the total volume ratio:
\[ \text{Total volume ratio} = 4 + 1 = 5 \]
- Therefore, the average vapor density (D) is:
\[ D = \frac{\text{Total contribution}}{\text{Total volume ratio}} = \frac{144}{5} = 28.8 \]

### Final Answer:
The average vapor density of air containing N₂ and O₂ in a volume ratio of 4:1 is **28.8 g/mol**.

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