Question

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.

Answer 1

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.