Question

 Suppose you have a container filled with air at 212 ^oF. The volume of the container 1.00 L, the pressure of air is 1.00 atm. The molecular composition of air is 79% N₂ and 21% O₂ for simplification. Calculate the mass of air and moles of O₂ in the container.

Answer 1

n = PV/RT

  = 1×1/(0.0821×373.15)    Where, P=1 atm, T = 212 ^oF =100 ^oC = 373.15K, V=1 L

  = 0.0326 mol

X_O2 = moles O₂ = 0.0326×0.21 = 0.006846 (Answer)

X_N2 = moles of N₂ = 0.0326×0.79 = 0.025754

M_air = X_O2×32 + X_N2×28     

      = (0.006846×32 + 0.025754×28)g

     = 0.939 g    (Answer)

The mass of air is 0.939 g

0.006846 moles of O₂ in the container.