Answer:
the partial pressure of methane CH₄ is 239.4 torr.
Explanation:
Given;
molar mass of methane, CH₄ = 90.0 g
molar mass of argon, Ar = 10.0 g
total pressure of the gases, Pt = 250 torr
number of moles of methane CH₄;
[tex]n_{CH_4} = \frac{90}{16} = 5.625 \ moles[/tex]
number of moles of Ar;
[tex]n_{Ar} = \frac{10}{39.95} = 0.25 \ mole[/tex]
The mole fraction of methane XCH₄ is calculated as;
[tex]X_{CH_4} = \frac{n_{CH_4} }{n_{CH_4} \ + \ n_{Ar}} \\\\X_{CH_4} = \frac{5.625}{5.625 \ + \ 0.25} \\\\X_{CH_4} = 0.9575[/tex]
Apply Dalton's law of partial pressure, to determine the partial pressure of methane CH₄;
[tex]P_{CH_4} = X_{CH_4} P_t\\\\P_{CH_4} = 0.9575 \ \times \ 250 \ torr\\\\P_{CH_4} = 239.4 \ torr[/tex]
Therefore, the partial pressure of methane CH₄ is 239.4 torr.
The partial pressure of CH₄ in the mixture containing 90 g of CH₄ and 10 g of Ar having a total pressure of 250 torr is 239.25 torr
We'll begin by calculating the number of mole of CH₄ and Ar. This can be obtained as follow:
Mass of CH₄ = 90 g
Molar mass of CH₄ = 12 + (4×1) = 16 g/mol
Mole = mass / molar mass
Mole of CH₄ = 90 / 16
Mass of Ar = 10 g
Molar mass of Ar = 40 g/mol
Mole = mass / molar mass
Mole of Ar = 10 / 40
Mole of CH₄ = 5.625 mole
Mole of Ar = 0.25 mole
Total mole = 5.625 + 0.25
Total mole = 5.875 mole
Mole fraction = mole of substance / total mole
Mole fraction of CH₄ = 5.625 / 5.875
Mole fraction of CH₄ = 0.957
Total pressure = 250 torr
Partial pressure = mole fraction × total pressure
Partial pressure of CH₄ = 0.957 × 250
Therefore, the partial pressure of CH₄ in the mixture is 239.25 torr
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