Let Far = mole fraction of Argon and Fn2 = mole fraction of Nitrogen.
From the given information, the mass of 22.4 L of the gas mixture is 1.361 * 22.4 g.
According to the statement,
molmassargon * Far + molmassN2 * Fn2 = 1.361 * 22.4
And we know,
Far + Fn2 = 1
Now, we have two equations:
1) molmassargon * Far + molmassN2 * Fn2 = 1.361 * 22.4
2) Far + Fn2 = 1
The molar mass of Argon (Ar) is 39.948 g/mol, and the molar mass of Nitrogen (N2) is 28.014 g/mol.
So plug these values into equation 1:
39.948 * Far + 28.014 * Fn2 = 1.361 * 22.4
Now we need to solve these two equations for Far and Fn2:
First, we can solve equation 2 for Far:
Far = 1 - Fn2
Substitute this into equation 1:
39.948 * (1 - Fn2) + 28.014 * Fn2 = 1.361 * 22.4
Distribute and combine like terms:
39.948 - 39.948 * Fn2 + 28.014 * Fn2 = 30.4856
Combine the Fn2 terms:
11.934 * Fn2 = -9.4624
Now, solve for Fn2:
Fn2 = -9.4624 / 11.934
Fn2 ≈ 0.793
Now that we have the mole fraction of Nitrogen, we can find the mole fraction of Argon:
Far = 1 - Fn2
Far = 1 - 0.793
Far ≈ 0.207
So the mole fraction of Argon (Ar) is approximately 0.207, and the mole fraction of Nitrogen (N2) is approximately 0.793.
A mixture of Ar and N2 gases has a density of 1.361 g/L at STP. What is the mole fraction of each gas?
not sure on really where to start
Work this as the classic mixture:
the mass of 22.4 liters of the gas is 1.361*22.4 grams.
molmassargon*Far + molmassN2*Fn2=1.361*22.4
where Far + Fn2=1
You are solving for Far and Fn2.
What do u mean by Far and Fn2 where do i get those values?
That is the mole fractions: Far is the argon mole fraction, and fn2 is the nitrogen mole fraction. YOu get them by solving the two equations.
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