Final answer:
The density of NH3 gas at 435 K and 1.00 atm can be determined using the ideal gas law and the molar mass of NH3, yielding a density of 0.477 g/L, which corresponds to option a.
Step-by-step explanation:
To determine the density of NH3 gas at 435 K and 1.00 atm, we can use the ideal gas law combined with the molar mass of NH3. The molar mass of ammonia (NH3) is 17.03 g/mol, and the ideal gas law is PV = nRT, where P is pressure, V is volume, n is amount of substance in moles, R is the ideal gas constant (0.0821 L.atm.mol⁻¹¹K⁻¹), and T is temperature in Kelvin.
First, we rearrange the ideal gas law to solve for the volume (V) that 1 mole of NH3 would occupy under the given conditions: V = nRT/P. For 1 mole (n = 1) at 435 K and 1.00 atm, the volume is calculated as follows: V = (1 mol) x (0.0821 L.atm.mol⁻¹¹K⁻¹) x (435 K) / (1.00 atm) = 35.71 L.
Next, we calculate the density by dividing the molar mass by the volume: Density = molar mass / V = (17.03 g/mol) / (35.71 L) = 0.477 g/L. Therefore, the correct answer is option a. 0.477 g/L.