Final answer:
The volume of 50.0 g of nitrogen is approximately 40.05 L at STP, and when the volume is halved, the number of moles present, which is 1.785 mol, remains the same.
Step-by-step explanation:
The question asks for the volume of 50.0 g of nitrogen gas, N₂, and the amount of moles present when the volume is halved. To answer this, we need to use the molar mass of nitrogen and the ideal gas law. The molar mass of molecular nitrogen, N₂, is 28.01 g/mol.
First, we calculate the number of moles of nitrogen in 50.0 g by using the molar mass:
Number of moles (n) = Mass (g) / Molar mass (g/mol)
n = 50.0 g / 28.01 g/mol = 1.785 mol
Now, we need to find the volume at standard conditions. Using the ideal gas law PV = nRT, and assuming standard temperature and pressure (STP) where P is 1 atm and T is 273 K, and R is 0.0821 L.atm.mol⁻¹.K⁻¹, we can calculate the volume:
V = nRT / P
V = (1.785 mol)(0.0821 L.atm.mol⁻¹.K⁻¹)(273 K) / 1 atm ≈ 40.05 L
When the volume is halved, we have 20.025 L. However, the number of moles present does not change by altering the volume, as long as temperature and pressure remain constant. Therefore, even after the volume is halved, there are still 1.785 moles of nitrogen present.