Which of the following gases will have a density of 2.104 g/L at 303 K and 1.31 atm? A. He
B. Ne
C. Ar
D. Kr
E. Xe

The gas with the density of 2.104 g/L has been Argon. Thus, option C is correct.

The gas has been assumed to be the ideal gas. According to the ideal gas equation:

Pressure Volume = Moles × Rydberg constant × Temperature

Density can be defined as:

Density =

Moles can be defined as:

Moles =

Substituting moles in the ideal gas equation:

Pressure Volume = × Rydberg constant × Temperature

Pressure = × Rydberg constant × Temperature

Pressure = Density  × Rydberg constant × Temperature

Substituting the values from the question:

1.31 atm = 2.104 g/L    × 0.08206 L.atm/K.mol × 303 K

1.31 atm = 52.314

Molecular mass = 39.934 g/mol

From the following gases, the gas with the molecular mass of 39.934 g/mol has been Ar. Thus, the gas with the density of 2.104 g/L has been Argon, Option C is correct.

For more information about the density refer to the link:

link

C. Ar

Explanation:

From the question, we are given the flowing;

T = 303 K

P = 1.31 atm

V = 1L

Density = Mass / volume;

Mass = Density * Volume = 2.104 g/L  * 1L

Mass = 2.104 g

The only way we can identify the element, is by obtaining he molar mass. Since we have the mass, we just need to obtain the number of moles.  

Using the general gas equation;

PV = nRT

n = PV / RT; where R = gas constant = 8.314472  L kPa K−1 mol−1

1.31 atm = 132.7358 kpa

n = (132.7358 * 1) / (8.314472 * 303)

n = 132.7358 / 2519.285 = 0.0527 mol

Molar mass = Mass / Number of moles = 2.104 / 0.0527 = 39.92 g/mol

Looking up the periodic table, the only element that has that molar mass is Argon (Ar)

the answer is d. matter from the air combined with the steel wool.