Given that a sample of air is made up of nitrogen, oxygen, and argon in the mole fractions \(78 \% \mathrm{~N}_{2}, 21 \% \mathrm{O}_{2}\), and \(1.0 \% \mathrm{Ar}\), what is the density of air at standard temperature and pressure?

First, we need to calculate the molar mass of air using the given mole fractions of nitrogen (N₂), oxygen (O₂), and argon (Ar). The molar mass of air (M) can be calculated by multiplying the molar mass of each component by its mole fraction and then summing up the results. \(M = \%(N_2) * M(N_2) + \%(O_2) * M(O_2) + \%(Ar) * M(Ar)\) where \(\%(N_2)\), \(\%(O_2)\) and \(\%(Ar)\) represent the mole fractions of nitrogen, oxygen, and argon, respectively, and \(M(N_2)\), \(M(O_2)\) and \(M(Ar)\) represent the molar mass of nitrogen (28 g/mol), oxygen (32 g/mol), and argon (40 g/mol), respectively.

To find the density of air at STP, we'll use the ideal gas law equation: \(PV = nRT\) At standard temperature and pressure (STP), the temperature (T) is 273.15 K and the pressure (P) is 1 atm (101325 Pa). The universal gas constant (R) is 8.314 J/(mol·K). Rearrange the equation to solve for the density of air (ρ): \(\rho = \frac{nM}{V} = \frac{PM}{RT}\) Using the molar mass of air (M) obtained in Step 1, calculate the density of air at STP.

Substitute the values for P, M, R, and T into the density equation: \(\rho = \frac{(101325 \, \text{Pa})(M)}{(8.314 \, \frac{\text{J}}{\text{mol·K}})(273.15 \, \text{K})}\) First, calculate the molar mass of air (M) from Step 1: \begin{align*} M &= 0.78 * 28 + 0.21 * 32 + 0.01 * 40 \\ &= 21.84 + 6.72 + 0.4 \\ &= 28.96 \, \frac{\text{g}}{\text{mol}} \end{align*} Now, plug in the value of M into the density equation and solve for ρ: \begin{align*} \rho &= \frac{(101325 \, \text{Pa})(28.96 \, \frac{\text{g}}{\text{mol}})}{(8.314 \, \frac{\text{J}}{\text{mol·K}})(273.15 \, \text{K})} \\ &= \frac{2,934,675.2 \, \frac{\text{g·Pa}}{\text{mol·K}}}{2,270.9 \, \frac{\text{J}}{\text{mol·K}}} \\ &= 1.29 \, \frac{\text{g}}{\text{L}} \end{align*} Therefore, the density of air at standard temperature and pressure is 1.29 g/L.