Question

Nitrogen dioxide dimerizes to give dinitrogen tetroxide: 2NO2(g)→ N2O4(g). At 298 K, 12.55 g of an NO2/N2O4 mixture exerts a pressure of 0.963 atm in a volume of 6.92 L. What are the mole fractions of the two gases in the mixture?

Answer 1

To answer this item, we assume that the gases are ideal for us to be able to use the ideal gas law.
                                       PV = nRT
where P is pressure, V is volume, n is the number of moles, R is the universal gas constant and T is the temperature (in K)

Substituting the known values,
                             (0.963 atm)(6.92 L) = n(0.0821 L.atm/mol.K)(298 K)
The value of n from the equation is n = 0.27237 moles

We let x and y be the number of moles of NO2 and N2O4, respectively. Given the calculated total moles above and the total mass,
                               x + y = 0.27237
                             30x + 60y = 12.55
The values of x and y are:
                             x = 0.126 moles
                             y = 0.146 moles

The mole fractions of each gases are therefore:
                    mole fraction of NO2 = 0.126/(0.126 + 0.146) = 0.46
                    mole fraction of N2O4 = 0.146/(0.126 + 0.146) = 0.54