Question

If the density of a mixture of nitrogen and oxygen gases at 400 K and 1 atm pressure is 0.920 g L\(^{-1}\), what is the mole fraction of nitrogen in the mixture?

• A. \( 0.456 \)
• B. \( 0.554 \)
• C. \( 0.432 \)
• D. \( 0.568 \)

Hint:

Use the ideal gas law to find molar mass when density is given.
- Mole fraction \( x \) can be determined using weighted averages of molar masses.

Answer 1

The Correct Option is A

We use the ideal gas equation: \[ PV = nRT \] Rearranging for molar mass: \[ M = \frac{dRT}{P} \] Given: \[ d = 0.920 \text{ g/L}, \quad R = 0.0821 \text{ L atm mol}^{-1} \text{K}^{-1}, \quad T = 400 K, \quad P = 1 \text{ atm} \] Substituting values: \[ M = \frac{0.920 \times 0.0821 \times 400}{1} = 30.05 \text{ g/mol} \] For a mixture of \( N_2 \) (M = 28 g/mol) and \( O_2 \) (M = 32 g/mol), the molar mass is given by: \[ M = x \times 28 + (1-x) \times 32 \] Solving for \( x \): \[ 30.05 = 28x + 32(1 - x) \] \[ 30.05 = 28x + 32 - 32x \] \[ 30.05 - 32 = -4x \] \[ x = \frac{1.95}{4} = 0.456 \] Thus, the mole fraction of nitrogen is \(\boxed{0.456}\).