Question

In a fixed-volume container at 25°C, NO2 gas partially combines to form N2O4 gas. The mass percentage of N2O4 in the resulting mixture is 76.3%, and the total pressure is 0.500 atm. What is the partial pressure of N2O4?

Answer 1

Answer – Given, mass percent of N₂O₄ = 76.3 % , total pressure = 0.500 atm

We assume solution mass =100 g

So, mass of N₂O₄ = 76.3 g , mass of NO₂ = 100-76.3 = 23.7 g

Moles of N₂O₄ = 76.3 g / 92.01 g.mol^-1 = 0.829 moles

Moles of NO₂ = 23.7 g / 46.005 g.mol^-1 =0.515 moles

So, mole fraction of N₂O₄ = 0.829 / 0.829+0.515

                                          = 0.617

We know Raoult’s law

Partial pressure N₂O₄ = mole fraction * total pressure

                                   = 0.617 * 0.500 atm

                                   = 0.308 atm