Question

a volume of 18.0 l contains a mixture of 0.250 mole n₂, 0.250 mole o₂, and an unknown quantity of he. the temperature of the mixture is 0 °c, and the total pressure is 1.00 atm. how many grams of helium are present in the gas mixture? express your answer to three significant figures and include the appropriate units. view available hint(s) mass of he = value units

Explanation:

Step1: Find total moles using ideal gas law

The ideal gas law is \( PV = nRT \). At \( 0^\circ\text{C} \) (273.15 K), \( P = 1.00 \) atm, \( V = 18.0 \) L, \( R = 0.0821 \frac{\text{L·atm}}{\text{mol·K}} \).
Rearranging for \( n \): \( n = \frac{PV}{RT} \).
Substitute values: \( n = \frac{1.00 \times 18.0}{0.0821 \times 273.15} \approx 0.803 \) mol (total moles).

Step2: Calculate moles of He

Moles of \( \text{N}_2 = 0.250 \) mol, \( \text{O}_2 = 0.250 \) mol.
Moles of He: \( n_{\text{He}} = 0.803 - 0.250 - 0.250 = 0.303 \) mol.

Step3: Find mass of He

Molar mass of He is \( 4.00 \frac{\text{g}}{\text{mol}} \).
Mass = moles × molar mass: \( 0.303 \times 4.00 = 1.21 \) g.

Answer:

1.21 g