What is the partial pressure of oxygen at s.t.p. in a gaseous mixture containing 100 cm3 of oxygen and 900 cm3 of nitrogen gas?

Analyze the question: You have a gas mixture at s.t.p. (total pressure = 1 atm, or 760 mmHg). You need the partial pressure of oxygen.

Concept: Dalton's Law of Partial Pressures says that each gas in a mixture exerts a pressure equal to its mole fraction multiplied by the total pressure.

Partial pressure of a gas = (volume or mole fraction of that gas) × total pressure

At constant temperature and pressure, volume is proportional to moles (Avogadro's law), so the volume fraction can be used in place of the mole fraction.

Solution:

1. Total volume = 100 + 900 = 1000 cm³
2. Mole fraction of O₂ = 100 / 1000 = 0.1
3. Total pressure at s.t.p. = 1 atm
4. Partial pressure of O₂ = 0.1 × 1 atm = 0.1 atm

As a check, the nitrogen's partial pressure is 0.9 × 1 = 0.9 atm. Then 0.1 + 0.9 = 1.0 atm, which equals the total pressure.

Answer: The correct option is B. 0.1 atm.

Tip: At s.t.p. the total pressure is always 1 atm, so the partial pressure is simply the fraction of the gas in the mixture. Oxygen is 100 out of 1000, which is 1/10, or 0.1 atm. Watch for option D (0.9 atm), which is the nitrogen's partial pressure and a common trap.