Density of dry air containing only `N_2 and O_2` is `1.15 g/L` at `740 mm ` of `Hg and 300 K`. What is % composition of `N_2` by mass in the air ?

Options

A. `82.5% N_2 & 17.5 %O_2`
B. `62.5%N_2 & 37.5% O_2`
C. `42.5%N_2 & 57.5% O_2`
D. `22.5%N_2 & 77.5% O_2`

Answer

To find the percentage composition of nitrogen (N₂) by mass in dry air containing only nitrogen and oxygen (O₂), we can follow these steps:

### Step 1: Convert Pressure to Atmospheres
Given:
- Pressure (P) = 740 mm Hg

We know that:
- 1 atm = 760 mm Hg

To convert the pressure to atmospheres:
\[ P = \frac{740 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} = 0.974 \, \text{atm} \]

### Step 2: Use the Ideal Gas Law
The ideal gas equation is given by:
\[ PV = nRT \]
Where:
- \( P \) = pressure in atm
- \( V \) = volume in liters
- \( n \) = number of moles
- \( R \) = ideal gas constant = 0.0821 L·atm/(K·mol)
- \( T \) = temperature in Kelvin = 300 K

For one mole of gas, we can rearrange the equation to find the mass of the gas:
\[ PV = \frac{m}{M}RT \]
Where \( m \) is the mass and \( M \) is the molar mass.

### Step 3: Calculate the Mass of the Gas Mixture
We can express the volume \( V \) in terms of density \( d \):
\[ V = \frac{m}{d} \]
Substituting this into the ideal gas equation gives:
\[ P \left(\frac{m}{d}\right) = \frac{m}{M}RT \]
Rearranging gives:
\[ m = \frac{dRT}{P} \]

Given:
- Density \( d = 1.15 \, \text{g/L} \)
- \( R = 0.0821 \, \text{L·atm/(K·mol)} \)
- \( T = 300 \, \text{K} \)
- \( P = 0.974 \, \text{atm} \)

Substituting the values:
\[ m = \frac{1.15 \, \text{g/L} \times 0.0821 \, \text{L·atm/(K·mol)} \times 300 \, \text{K}}{0.974 \, \text{atm}} \]
Calculating this gives:
\[ m \approx 29.09 \, \text{g} \]

### Step 4: Set Up the Equation for Molar Mass
Let \( x \) be the mole fraction of N₂ and \( (1-x) \) be the mole fraction of O₂. The molar masses are:
- Molar mass of N₂ = 28 g/mol
- Molar mass of O₂ = 32 g/mol

Now, we can express the average molar mass of the mixture:
\[ 29.09 = 28x + 32(1-x) \]
Expanding this gives:
\[ 29.09 = 28x + 32 - 32x \]
Combining like terms:
\[ 29.09 = 32 - 4x \]
Solving for \( x \):
\[ 4x = 32 - 29.09 \]
\[ 4x = 2.91 \quad \Rightarrow \quad x = 0.7275 \]

### Step 5: Calculate the Percentage Composition of N₂
The percentage composition of N₂ by mass is given by:
\[ \text{Percentage of N₂} = \left( \frac{x \cdot M_{N_2}}{m} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage of N₂} = \left( \frac{0.7275 \cdot 28}{29.09} \right) \times 100 \]
Calculating this gives:
\[ \text{Percentage of N₂} \approx 70.02\% \]

### Final Answer
The percentage composition of N₂ by mass in the air is approximately **70.02%**.