Question involving Gas Laws: A 6.0L flask (1 Viewer)

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A 6.0L flask contains a mixture of methane, argon, and helium at 45C and 1.75atm. If the mole fractions of helium and Argon are 0.25 and 0.35, respectively, how many molecules of methane are present?

V = 6.0L

T=318.15K

Ptotal=1.75atm

mole fraction: XHe=0.25

XAr=0.35

XCH4=0.40

Because mole fractions always add up to 1

Then I used Partial Pressure formula

Partial Pressure of CH4 = 0.40 X 1.75atm = 0.7atm

Then Ideal gas law

n=PV/RT

n=(0.7atm)(6L)/(.o8206)(318.15K) = 0.16087 moles of CH4

0.16087 moles of CH4 X 6.022 x 1023 molecules = 9.69 X 10^22 molecules of methane?
 

Borek

Mentor
27,852
2,424
Looks OK. Watch significant digits.

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methods
 

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