Question

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

Answer 1

Answer:

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