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Lunna [17]
3 years ago
10

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

Chemistry
1 answer:
xenn [34]3 years ago
6 0

Answer:

Partial pressure of H₂ = 499 mmHg

Partial pressure of Ar = 481 mmHg

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

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