Density of dry air `(` only `N_(2)` and `O_(2))` is `1.24g` `litre^(-1)` at `760 m m` and `300 K`. Find the partial pressure of `N_(2)` gas in aire. ( Take `R=(1)/(12)` litre litre atm`//`mol K, mol. Wt. of `N_(2) =28)`

Options

A. 0.25
B. 0.365
C. 0.5
D. 0.75

Explanation

To find the partial pressure of nitrogen gas (N₂) in dry air, we will follow these steps:

### Step 1: Understand the given data
- Density of dry air (ρ) = 1.24 g/L
- Total pressure (P) = 760 mm Hg = 1 atm
- Temperature (T) = 300 K
- Gas constant (R) = 1/12 L atm/(mol K)
- Molar mass of nitrogen (N₂) = 28 g/mol

### Step 2: Use the density formula for gases
The formula for the density of a gas is given by:
\[ \rho = \frac{P \cdot M}{R \cdot T} \]
Where:
- \( \rho \) = density of the gas
- \( P \) = pressure of the gas
- \( M \) = molar mass of the gas mixture
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

### Step 3: Rearranging the formula to find the molar mass of the mixture
Rearranging the formula gives:
\[ M = \frac{\rho \cdot R \cdot T}{P} \]

### Step 4: Substitute the known values into the equation
Substituting the values into the equation:
\[ M = \frac{1.24 \, \text{g/L} \cdot \frac{1}{12} \, \text{L atm/(mol K)} \cdot 300 \, \text{K}}{1 \, \text{atm}} \]

### Step 5: Calculate the molar mass of the mixture
Calculating the above expression:
\[ M = \frac{1.24 \cdot \frac{1}{12} \cdot 300}{1} \]
\[ M = \frac{1.24 \cdot 25}{1} = 31 \, \text{g/mol} \]

### Step 6: Set up the equation for the molar mass of the mixture
The molar mass of the mixture can be expressed as:
\[ M = y \cdot M_{O_2} + (1 - y) \cdot M_{N_2} \]
Where:
- \( y \) = mole fraction of oxygen (O₂)
- \( M_{O_2} = 32 \, \text{g/mol} \)
- \( M_{N_2} = 28 \, \text{g/mol} \)

### Step 7: Substitute the known values into the equation
Substituting the values into the equation gives:
\[ 31 = y \cdot 32 + (1 - y) \cdot 28 \]

### Step 8: Simplify and solve for \( y \)
Expanding the equation:
\[ 31 = 32y + 28 - 28y \]
\[ 31 = 4y + 28 \]
\[ 3 = 4y \]
\[ y = \frac{3}{4} \]

### Step 9: Find the mole fraction of nitrogen
The mole fraction of nitrogen (N₂) is:
\[ x = 1 - y = 1 - \frac{3}{4} = \frac{1}{4} \]

### Step 10: Calculate the partial pressure of nitrogen
The partial pressure of nitrogen (Pₙ) is given by:
\[ P_n = x \cdot P \]
Substituting the values:
\[ P_n = \frac{1}{4} \cdot 1 \, \text{atm} = 0.25 \, \text{atm} \]

### Final Answer
The partial pressure of nitrogen gas in air is **0.25 atm**.

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