The density of a noble gas is 5.35 g/L at 3.00 atm and 0°C. Identify the gas.?

Answer 1

PV = nRT
n = PV/RT
P = 3 atm, V = 1L, T = 273K, R = 0.082 L-atm/mol-K
n = (3)(1)/(273)(0.082) = 0.134 moles

5.35g/0.134mol = 39.9g/mol

The gas is argon

Answer 2

V = one million L P = 3.00 atm T = 0 + 273 ok = 273 ok n = (PV)/(RT) = (3.00 x one million)/(0.08206 x 273) = 0.134 Molar mass = 2.seventy one/0.134 g/mol = 20.2 g/mol This suits the molar mass of neon. Ans: Neon ______________.