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A mixture of neon and methane gases, in a 8.53 L flask at 72 °C, contains 1.56 grams of neon and 4.85 grams of methane. The partial pressure of methane in the flask is atm and the total pressure in the flask is atm.

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Answer:

The partial pressure of methane in the flask is 1.00 atm and the total pressure in the flask is 1.26 atm.

Step-by-step explanation:

You must apply the Ideal Gases Law to solve this;

Pressure . volume = n° moles . Gases constant (R) . T° in K

T° in K = T°C + 273

72 + 273 = 345K

Gases constant (R) = 0.082 L.atm /mol.K

n° moles = in mixture, the total moles are the sum of each gas.

Moles = mass / molar mass

Moles Ne = 1.56 g / 20.17 g/m = 0.077 moles

Moles CH₄ = 4.85 g/ 16.04 g/m = 0.302 moles

0.077 + 0.302 = 0.379 moles

Pressure . 8.53L = 0.379 moles . 0.082 L.atm /mol.K . 345K

Pressure = ( 0.379 moles . 0.082 L.atm /mol.K . 345K ) / 8.53L

Pressure = 1.26 atm (The total pressure in system)

We can use the molar fraction to know, the partial pressure of CH₄.

Moles CH₄ / Total moles = Partial pressure CH₄ / Total pressure

0.302 / 0.379 = Partial pressure CH₄ / 1.26 atm

(0.302 / 0.379 ) . 1.26 atm = Partial pressure CH₄

1.00 atm = Partial pressure CH₄

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