A gaseous mixture of O2 and N2 contains 37.8 % nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 485 mmHg ?
To find the partial pressure of oxygen in the mixture, we need to determine the mole fractions of both oxygen and nitrogen.
Let's start by assuming we have 100 grams of the gaseous mixture. Since we know that the mixture contains 37.8% nitrogen by mass, we can calculate the mass of nitrogen by multiplying 100 grams by 0.378:
Mass of nitrogen = 100 g × 0.378 = 37.8 g
Similarly, the mass of oxygen in the mixture can be calculated as:
Mass of oxygen = 100 g - 37.8 g = 62.2 g
Next, we need to convert these masses into moles. We can do this by dividing the masses by their respective molar masses. The molar mass of nitrogen (N₂) is approximately 28 g/mol, and the molar mass of oxygen (O₂) is approximately 32 g/mol.
Moles of nitrogen = 37.8 g / 28 g/mol ≈ 1.35 mol
Moles of oxygen = 62.2 g / 32 g/mol ≈ 1.94 mol
Now that we have the mole fractions, we can calculate the partial pressure of oxygen. The partial pressure of a gas in a mixture is proportional to its mole fraction. Since the total pressure is given as 485 mmHg, we can use the mole fraction of oxygen to find its partial pressure.
Mole fraction of oxygen = Moles of oxygen / Total moles
Mole fraction of oxygen = 1.94 mol / (1.35 mol + 1.94 mol) ≈ 0.59
Partial pressure of oxygen = Mole fraction of oxygen × Total pressure
Partial pressure of oxygen = 0.59 × 485 mmHg ≈ 286 mmHg
Therefore, the partial pressure of oxygen in the mixture is approximately 286 mmHg.