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
