Final answer:
option c,The gas with a density of 1.06 g/L at 300 K and 0.93 atm is nitrogen, as the calculated molar mass using the Ideal Gas Law is approximately 29 g/mol, closely aligning with the molar mass of nitrogen.
Step-by-step explanation:
To determine which gas will have a density of 1.06 g/L at 300 K and 0.93 atm, we can use the Ideal Gas Law rearranged to solve for molar mass (M), as density (d) is related to molar mass and molar volume (Vm) by the equation d = M/Vm. This can be combined with the Ideal Gas Law PV = nRT, where n (number of moles) can be expressed as mass (m) divided by molar mass (M), thus giving us a new formula for density: d = PM/RT.
To find the molar mass M we rearrange the equation to M = dRT/P. Given the values R = 0.0821 L*atm/(mol*K), T = 300 K, P = 0.93 atm, and d = 1.06 g/L, we can plug in these values to solve for M. Substituting into the equation gives us M = (1.06 g/L)(0.0821 L*atm/(mol*K))(300 K) / (0.93 atm), which simplifies to M ≈ 29 g/mol.
Now we compare this result with the molar mass of the gases listed:
The molar mass closest to our calculated 29 g/mol is that of nitrogen, so the correct answer is (c) Nitrogen.