If 64 grams of o2 gas occupy a volume of 56 liters at a temperature or 25°c, what is the pressure of the gas, in atmospheres (atm)?

Using Ideal Gas Laws :
PV = nRT
Where,
P is Pressure = Not given
V is the Volume = 56L
n is number of moles of gas = Not given
R is the proportionality constant = 0.08206 L.atm/K.mol
T is the temperature = 25 + 273.15 = 298 K

Step 1:
Calculate n,
n =  2 x 15.999  = 31.998 g/mole
n (oxygen) = 64/31.998
                  = 2 moles of oxygen
Step 2:
Calculate Pressure,
P = nRT/V
Substitute the values,
P = 2 x 0.08206 x 298/56
P = 0.88 atm

What do you mean by that? like the contents? the thc level?