Question

How many grams of O2 would be in a 4.5 L container at 7.2 atm at 3o C?

Answer 1

Given the volume of the container (V) = 4.5 L

The pressure of the gas (P) = 7.2 atm

Temperature (T) = 3°C = 3°C + 273 = 276 K

Gas constant (R) = 0.082 L atm/mol/K

Now using the ideal gas equation, i.e.

PV = nRT where n = number of moles of O2

So, n = PV / RT

Substituting the values you will get,

n = (7.2 atm × 4.5 L) / (0.082 L atm/mol/K × 276 K)

n = 1.432 moles

Now, the molecular mass of O2 = 32 g/mol

So, the mass of the O2 = moles of O2 × molecular mass of O2

= 1.432 moles × 32 g/mol

= 45.824 g

Grams of O2 present in the container = 45.824 g