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The partial pressures of CH4, N2, and O2 in a sample of gas were found to be 143 mmHg, 469 mmHg, and 563 mmHg, respectively. Calculate the mole fraction of oxygen. 20.1 0.399 0.741 0.479 0.359

Respuesta :

Answer: The mole fraction of oxygen gas is 0.479

Explanation:

We are given:

Partial pressure of methane = 143 mmHg

Partial pressure of nitrogen gas = 469 mmHg

Partial pressure of oxygen gas = 563 mmHg

Total pressure = (143 + 469 + 563) = 1175 mmHg

To calculate the mole fraction of oxygen gas, we use the equation given by Raoult's law, which is:

[tex]p_{O_2}=p_T\times \chi_{O_2}[/tex]

where,

[tex]p_{O_2}[/tex] = partial pressure of oxygen gas = 563 mmHg

[tex]p_T[/tex] = total pressure = 1175 mmHg

[tex]\chi_{O_2}[/tex] = mole fraction of oxygen gas = ?

Putting values in above equation, we get:

[tex]563mmHg=1175mmHg\times \chi_{O_2}\\\\\chi_{O_2}=\frac{563}{1175}=0.479[/tex]

Hence, the mole fraction of oxygen gas is 0.479

Eduard22sly

The mole fraction of oxygen gas, O₂ in the sample of the gas is 0.479

  • We'll begin by calculating the total pressure. This can be obtained as follow:

Partial pressure of CH₄ = 143 mmHg

Partial pressure of N₂ = 469 mmHg

Partial pressure of O₂ = 563 mmHg

Total pressure =?

Total pressure = 143 + 469 + 563

Total pressure = 1175 mmHg

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

Total pressure = 1175 mmHg

Partial pressure of O₂ = 563 mmHg

Mole fraction of O₂ =?

Mole fraction = Partial pressure / total pressure

Mole fraction of O₂ = 563 / 1175

Mole fraction of O₂ = 0.479

Therefore, the mole fraction of O₂ is 0.479

Learn more: https://brainly.com/question/2060778

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