An ideal gas has a density of \(0.0899 \mathrm{~g} / \mathrm{L}\) at \(20.00^{\circ} \mathrm{C}\) and \(101.325 \mathrm{kPa}\). Identify the gas.

Given density: \(0.0899 \mathrm{~g} / \mathrm{L}\) Temperature: \(20.00^{\circ} \mathrm{C}\) Pressure: \(101.325 \mathrm{kPa}\) First, let's convert the given values to the units compatible with the ideal gas law: 1. Convert density to \(\mathrm{kg/m^3}\): \(0.0899 \mathrm{~g} / \mathrm{L} \times 1000 \mathrm{~mg} / \mathrm{g} \times 1 \mathrm{~L} / 1000 \mathrm{~mL} = 89.9 \mathrm{~kg/m^3}\) 2. Convert temperature to kelvins (K): \(T_{K} = 20.00 + 273.15 = 293.15 \mathrm{K}\) 3. Convert pressure to pascals (Pa): \(101.325 \mathrm{kPa} \times 1000 \mathrm{~Pa} / \mathrm{kPa} = 101325 \mathrm{~Pa}\) Now we have the following values: Density: \(89.9 \mathrm{~kg/m^3}\) Temperature: \(293.15 \mathrm{K}\) Pressure: \(101325 \mathrm{~Pa}\)

The ideal gas law is written as \(PV = nRT\). We can rewrite the equation in terms of density and molar mass: \(n = m/M\), where m is the mass of the gas and M is the molar mass. \(m/V = \rho\), where \(\rho\) is the density of the gas. Therefore, \(\rho = nM/V\), and we can substitute this into the ideal gas law: \(P = \rho RT/M\)

Now, we will solve for the molar mass (M) using the given values for pressure, density, and temperature, as well as the ideal gas constant (R): \(M = \frac{\rho RT}{P} = \frac{89.9 \mathrm{~kg/m^3} \times 8.314 \mathrm{J} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \times 293.15 \mathrm{K}}{101325 \mathrm{~Pa}}\) \(M \approx 28.96 \mathrm{~g/mol}\)

Finally, we will identify the gas based on its molar mass. A gas with a molar mass of approximately \(28.96 \mathrm{~g/mol}\) is nitrogen (N₂), which has a molar mass of \(28.02 \mathrm{~g/mol}\). Our result is very close to the actual value, so it is reasonable to identify the gas as nitrogen (N₂).