The weight of 350 mL of a diatomic gas at 0°C and 2 atm pressure is 1g. The weight of one atom is (N is the Avogadro's number):

Explanation

To find the weight of one atom of the diatomic gas, you can use the ideal gas law: PV = nRT Where: P = Pressure (2 atm) V = Volume (350 mL = 0.35 L) n = Number of moles R = Ideal gas constant T = Temperature (0°C = 273.15 K) First, you need to calculate the number of moles of the gas: PV = nRT (2 atm) * (0.35 L) = n * (0.0821 L·atm/(mol·K)) * (273.15 K) 0.7 = n * 22.413 n ≈ 0.0313 moles Now, you can calculate the molar mass of the diatomic gas by using the weight and the number of moles: Molar mass = Weight / Moles Molar mass = 1 g / 0.0313 moles ≈ 31.95 g/mol Since the gas is diatomic, its molar mass is approximately 2 times the molar mass of a single atom: The molar mass of one atom = Molar mass of the gas / 2 Molar mass of one atom = 31.95 g/mol / 2 ≈ 15.98 g/mol Now, we need to convert this molar mass into atomic mass units (amu). To do this, you divide by Avogadro's number (N, which is approximately 6.022 x 10^23): Molar mass of one atom (in amu) ≈ (15.98 g/mol) / N So, the correct answer is approximately: Molar mass of one atom ≈ (15.98 g/mol) / N