Density of dry air containing ony `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. `78%`
B. `85.5%`
C. `70.02%`
D. `62.75%`

Answer

To solve the problem of finding the percent composition of nitrogen (N₂) by mass in a dry air mixture containing only nitrogen and oxygen (O₂), we can follow these steps:

### Step 1: Convert Pressure to Atmospheres
Given the pressure in mm Hg, we need to convert it to atmospheres for use in the ideal gas law.

\[ P = \frac{740 \text{ mm Hg}}{760 \text{ mm Hg/atm}} = 0.974 \text{ atm} \]

**Hint:** Remember that 1 atm is equal to 760 mm Hg.

### Step 2: Use the Ideal Gas Law
The ideal gas law is given by the equation:

\[ PV = nRT \]

For one mole of gas, we can rearrange this to find the mass of the gas mixture:

\[ n = \frac{PV}{RT} \]

Where:
- \( R = 0.0821 \text{ atm L/(mol K)} \)
- \( T = 300 \text{ K} \)

### Step 3: Calculate the Mass of the Gas Mixture
We know the density (\( d \)) of the gas mixture is 1.15 g/L. We can express the mass of the mixture as follows:

\[ \text{Mass} = \text{Density} \times \text{Volume} \]

Using the ideal gas law, we can substitute for \( n \):

\[ \text{Mass} = d \times \frac{PV}{RT} \]

Substituting the known values:

\[ \text{Mass} = 1.15 \text{ g/L} \times \frac{(0.974 \text{ atm})(1 \text{ L})}{(0.0821 \text{ atm L/(mol K)})(300 \text{ K})} \]

Calculating this gives:

\[ \text{Mass} \approx 1.15 \times \frac{0.974}{24.63} \approx 0.045 \text{ g} \]

### Step 4: Calculate the Molar Mass of the Mixture
From the previous calculations, we can find the molar mass of the mixture:

\[ \text{Molar Mass} = \frac{\text{Mass}}{n} \]

Using the value of \( n \) from the ideal gas law:

\[ \text{Molar Mass} = \frac{1.15 \text{ g/L}}{0.974 \text{ atm}} \times \frac{0.0821 \text{ atm L/(mol K)} \times 300 \text{ K}}{1} \approx 29.09 \text{ g/mol} \]

### Step 5: Set Up the Equation for Molar Mass of the Mixture
Let \( x \) be the mole fraction of nitrogen (N₂) and \( (1 - x) \) be the mole fraction of oxygen (O₂). The molar masses are:

- Molar mass of N₂ = 28 g/mol
- Molar mass of O₂ = 32 g/mol

The equation for the average molar mass of the mixture is:

\[ 29.09 = 28x + 32(1 - x) \]

### Step 6: Solve for the Mole Fraction of Nitrogen
Rearranging the equation gives:

\[ 29.09 = 28x + 32 - 32x \]
\[ 29.09 = 32 - 4x \]
\[ 4x = 32 - 29.09 \]
\[ 4x = 2.91 \]
\[ x = \frac{2.91}{4} \approx 0.7275 \]

### Step 7: Calculate the Percent Composition of Nitrogen by Mass
The percent composition of nitrogen by mass is given by:

\[ \text{Percent N₂} = \left( \frac{x \times \text{Molar Mass of N₂}}{\text{Molar Mass of Mixture}} \right) \times 100 \]

Substituting the values:

\[ \text{Percent N₂} = \left( \frac{0.7275 \times 28}{29.09} \right) \times 100 \approx 70.27\% \]

### Final Answer
The percent composition of nitrogen by mass in the air mixture is approximately **70.27%**.

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