Question

What is the partial pressure of O2 gas in a gas mixture containing 40.0 g of N2 and 12.0 g of O2, if the gas mixture exerts a total pressure of 1.86 atm at 25oC?
Group of answer choices

0.429 atm

0.698 atm

1.47 atm

0.387 atm

1.20 atm

Answer 1

The partial pressure of O₂ gas in the mixture containing 40.0 g of N₂ and 12.0 g of O₂, is 0.387 atm

We'll begin by calculating the number of mole of each gas present in the mixture.

For N₂:

Molar mass of N₂ = 2 × 14

= 28 g/mol

Mass of N₂ = 40 g

Mole of N₂ =?

Mole = mass / molar mass

Mole of N₂ = 40 / 28

Mole of N₂ = 1.429 mole

For O₂:

Molar mass of O₂ = 2 × 16

= 32 g/mol

Mass of O₂ = 12 g

Mole of O₂ =?

Mole = mass / molar mass

Mole of O₂ = 12 / 32

Mole of O₂ = 0.375 mole

Next, we shall determine the mole fraction of O₂. This can be obtained as follow:

Mole of O₂ = 0.375 mole

Mole of N₂ = 1.429 mole

Total mole = 1.429 + 0.375

Total mole = 1.804 mole

Mole fraction of O₂ =?

Mole fraction = mole of gas / total mole

Mole fraction of O₂ = 0.375 / 1.804

Mole fraction of O₂ = 0.208

Finally, we shall determine the partial pressure of O₂ gas in the mixture.

Mole fraction of O₂ = 0.208

Total pressure = 1.86 atm

Partial pressure of O₂ =?

Partial pressure = mole fraction × Total pressure

Partial pressure of O₂ = 0.208 × 1.86

Partial pressure of O₂ = 0.387 atm

Therefore, the partial pressure of O₂ gas in the mixture is 0.387 atm

Learn more: https://brainly.com/question/10737617