Question involving Gas Laws: A 6.0L flask

[Chandasouk]

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

P_total=1.75atm

mole fraction: X_He=0.25

X_Ar=0.35

X_CH4=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]

Looks OK. Watch significant digits.

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methods

 

[Blueshift5]

Based on the given information, there are approximately 9.69 x 10^22 molecules of methane present in the 6.0L flask. This calculation was done using the ideal gas law, which relates the pressure, volume, temperature, and number of moles of a gas. The mole fractions of helium and argon were used to determine the partial pressure of methane, which was then used to calculate the number of moles and molecules of methane present in the flask. This calculation can be used to better understand the composition and behavior of gases in a given system.