A mixture of argon (Ar) and nitrogen `(N_(2))` has a density of 1.40 g `L^(-1)` at STP. Mole fraction of `N_(2)` in the mixture is (Given atomic mass of Ar=40)

Options

A. 0.3
B. 0.4
C. 0.5
D. 0.7

Explanation

To find the mole fraction of nitrogen (N₂) in a mixture of argon (Ar) and nitrogen with a given density of 1.40 g/L at STP, we can follow these steps:

### Step 1: Use the formula for molar mass from density
The formula relating density (D), molar mass (M), temperature (T), and pressure (P) is given by:
\[ P_m = DRT \]
Where:
- \( P_m \) = molar mass of the gas mixture
- \( D \) = density of the gas mixture
- \( R \) = ideal gas constant (0.0831 L·bar/K·mol)
- \( T \) = temperature in Kelvin (STP = 273 K)

### Step 2: Calculate the molar mass of the gas mixture
At STP, the pressure \( P \) is 1 bar, and the density \( D \) is given as 1.40 g/L. Plugging these values into the equation:
\[ P_m = DRT \]
\[ P_m = (1.40 \, \text{g/L}) \times (0.0831 \, \text{L·bar/K·mol}) \times (273 \, \text{K}) \]

Calculating this gives:
\[ P_m = 31.37 \, \text{g/mol} \]

### Step 3: Set up the equation for effective molar mass
The effective molar mass \( M \) of the mixture can be expressed as:
\[ M = M_{N_2} \cdot X_{N_2} + M_{Ar} \cdot X_{Ar} \]
Where:
- \( M_{N_2} = 28 \, \text{g/mol} \) (molar mass of nitrogen)
- \( M_{Ar} = 40 \, \text{g/mol} \) (molar mass of argon)
- \( X_{N_2} \) = mole fraction of nitrogen
- \( X_{Ar} = 1 - X_{N_2} \) (mole fraction of argon)

### Step 4: Substitute values into the equation
Substituting the known values into the effective molar mass equation:
\[ 31.37 = 28X_{N_2} + 40(1 - X_{N_2}) \]

### Step 5: Solve for the mole fraction of nitrogen
Expanding and rearranging the equation:
\[ 31.37 = 28X_{N_2} + 40 - 40X_{N_2} \]
\[ 31.37 - 40 = -12X_{N_2} \]
\[ -8.63 = -12X_{N_2} \]
\[ X_{N_2} = \frac{8.63}{12} \]
\[ X_{N_2} = 0.719 \]

### Conclusion
The mole fraction of nitrogen (N₂) in the mixture is approximately **0.719**.

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