How many liters of space are in 4.00x10^22 of O2 occupy at STP

V O2 = 1.623 L

Explanation:

1 mol ≡ 6.022 E23 molecules

∴ molecules O2 = 4.00 E22 molecules

⇒ moles O2 = (4.00 E22 molecules O2) * (mol O2/6.022 E23 molecules)

⇒ moles O2 = 0.0664 moles

at STP:

∴ T = 25°C ≅ 298 K

∴ P = 1 atm

assuming ideal gas:

∴ V = RTn/P

⇒ V O2 = ((0.082 atm. L/K. mol) (298 K) (0.0664 mol)) / (1 atm)

⇒ V O2 = 1.623 L