|Dec12-09, 07:51 PM||#1|
Question involving Gas Laws: A 6.0L flask...
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
mole fraction: XHe=0.25
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=(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?
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