How many oxygen molecules are in 3.36 L of oxygen gas at STP?
with work

Answer is: in 3.36 L of oxygen gas there is 0,15 mol (6·10²²) molecules of oxygen. 
V(O₂) = 3,36 L.
Vm(O₂) = 22,4 L/mol.
n(O₂) = V(O₂) ÷ Vm(O₂).
n(O₂) = 3,36 L ÷ 22,4 L/mol.
n(O₂) = 0,15 mol.
N(O₂) = n(O₂) · Na.
N(O₂) = 0,15 mol · 6·10²³ 1/mol.
N(O₂)  = 9 · 10 ²².
n - amount of substance.
Na - Avogadro number.