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Question involving Gas Laws: A 6.0L flask

  1. Dec 12, 2009 #1
    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?
     
  2. jcsd
  3. Dec 13, 2009 #2

    Borek

    User Avatar

    Staff: Mentor

    Looks OK. Watch significant digits.

    --
    methods
     
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