Question

A mixture of hydrogen and argon gases, at a total pressure of 980 mm Hg, contains 0.291 grams of hydrogen and 5.62 grams of

argon. What is the partial pressure of each gas in the mixture?
PH2
Par
mm Hg
mm Hg

Answer 1

Partial pressure of H₂ = 499 mmHg

Partial pressure of Ar = 481 mmHg

Step-by-step explanation:

We'll begin by calculating the number of mole of each gas. This can be obtained as follow:

For Hydrogen:

Molar mass of H₂ = 2 × 1 = 2 g/mol

Mass of H₂ = 0.291 g

Mole of H₂ =?

Mole = mass /molar mass

Mole of H₂ = 0.291/ 2

Mole of H₂ = 0.1455 mole

For Argon:

Molar mass of Ar = 40 g/mol

Mass of Ar = 5.62 g

Mole of Ar =?

Mole = mass /molar mass

Mole of Ar = 5.62 / 40

Mole of Ar = 0.1405 mole

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

Mole of H₂ = 0.1455 mole

Mole of Ar = 0.1405 mole

Total mole = 0.1455 + 0.1405

Total mole = 0.286 mole

Mole fraction of H₂ (nₕ₂) = mole of H₂ / total mole

Mole fraction of H₂ (nₕ₂) = 0.1455/0.286

Mole fraction of H₂ (nₕ₂) = 0.509

Mole fraction of Ar (nₐᵣ) = mole of Ar / total mole

Mole fraction of Ar (nₐᵣ) = 0.1405/0.286

Mole fraction of Ar (nₐᵣ) = 0.491

Finally, we shall determine the partial pressure of each gas. This can be obtained as follow:

For Hydrogen:

Mole fraction of H₂ (nₕ₂) = 0.509

Total pressure (Pₜ) = 980 mmHg

Partial pressure of H₂ (Pₕ₂) =?

Pₕ₂ = nₕ₂ × Pₜ

Pₕ₂ = 0.509 × 980

Partial pressure of H₂ (Pₕ₂) = 499 mmHg

For Argon:

Partial pressure of H₂ (Pₕ₂) = 499 mmHg

Total pressure (Pₜ) = 980 mmHg

Partial pressure of Ar (Pₐᵣ) =?

Pₜ = Pₕ₂ + Pₐᵣ

980 = 499 + Pₐᵣ

Collect like terms

980 – 499 = Pₐᵣ

481 = Pₐᵣ

Partial pressure of Ar (Pₐᵣ) = 481 mmHg

SUMMARY:

Partial pressure of H₂ (Pₕ₂) = 499 mmHg

Partial pressure of Ar (Pₐᵣ) = 481 mmHg