The vapour density of a mixture containing `N_(2) (g) and O_(2) (g)` is 14.4. The percentage of `N_(2)` in the mixture is :

Options

A. 20
B. 80
C. 60
D. 50

Explanation

To solve the problem, we need to find the percentage of nitrogen (N₂) in a mixture of nitrogen and oxygen (O₂) given that the vapor density of the mixture is 14.4.

### Step-by-Step Solution:

1. **Understand the relationship between vapor density and molecular mass:**
The vapor density (VD) of a gas mixture is given by the formula:
\[ \text{VD} = \frac{\text{Molecular mass of the mixture}}{2} \]
Given that the vapor density of the mixture is 14.4, we can find the effective molecular mass (M) of the mixture:
\[ 14.4 = \frac{M}{2} \implies M = 14.4 \times 2 = 28.8 \]

2. **Set up the equation for the effective molecular mass:**
The effective molecular mass of the mixture can be expressed in terms of the mole fractions of N₂ and O₂:
\[ M = (y \cdot M_{N2}) + ((1 - y) \cdot M_{O2}) \]
where:
- \( y \) = mole fraction of N₂
- \( M_{N2} = 28 \) g/mol (molecular mass of nitrogen)
- \( M_{O2} = 32 \) g/mol (molecular mass of oxygen)

Substituting the known values:
\[ 28.8 = (y \cdot 28) + ((1 - y) \cdot 32) \]

3. **Simplify the equation:**
Expanding the equation:
\[ 28.8 = 28y + 32 - 32y \]
Combining like terms:
\[ 28.8 = 32 - 4y \]

4. **Solve for the mole fraction of N₂ (y):**
Rearranging the equation:
\[ 4y = 32 - 28.8 \]
\[ 4y = 3.2 \implies y = \frac{3.2}{4} = 0.8 \]

5. **Calculate the percentage of N₂ in the mixture:**
To find the percentage of N₂, we multiply the mole fraction by 100:
\[ \text{Percentage of N₂} = y \times 100 = 0.8 \times 100 = 80\% \]

### Final Answer:
The percentage of N₂ in the mixture is **80%**.