Given that a sample of air is made up of nitrogen, oxygen, and argon in the mole fractions \(0.78 \mathrm{~N}_{2}, 0.21 \mathrm{O}_{2}\), and \(0.010 \mathrm{Ar}\), what is the density of air at standard temperature and pressure?

To calculate the average molar mass of air, we need to take into account the mole fractions and molar masses of each component. The molar masses of N2, O2, and Ar are approximately 28, 32, and 40 g/mol, respectively. The average molar mass of air, M, is the sum of the products of the mole factions and their respective molar masses: \(M = x_{N_2}M_{N_2} + x_{O_2}M_{O_2} + x_{Ar}M_{Ar}\) Plugging in the given values: \(M = (0.78)(28) + (0.21)(32) + (0.01)(40)\)

Now, we calculate M: \(M \approx (21.84) + (6.72) + (0.4)\) \(M \approx 28.96~g/mol\) The average molar mass of air is approximately 28.96 g/mol.

The ideal gas law is given by the formula: \(PV = nRT\) Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. We can rearrange this formula to solve for the density of the gas, ρ: \(\rho = \frac{m}{V} = \frac{nM}{V}\) Since \(n = \frac{PV}{RT}\), we can substitute this into the density formula: \(\rho = \frac{(\frac{PV}{RT})M}{V}\) \(ρ = \frac{PM}{RT}\) At standard temperature and pressure (STP), the pressure is 1 atm, and the temperature is 273.15 K. The ideal gas constant for these units is 0.0821 L atm/mol K. We have all the necessary variables to find the density at STP: \(\rho = \frac{(1~atm)(28.96~g/mol)}{(0.0821~L~atm/mol~K)(273.15~K)}\)

Now, we calculate the density: \(\rho \approx \frac{(28.96)}{(0.0821)(273.15)}\) \(\rho \approx \frac{28.96}{22.42}\) \(\rho \approx 1.29~g/L\) The density of air at standard temperature and pressure is approximately 1.29 g/L.