Vapour density of  $N_2{O}_4$ at  ${60}^{0}C$  is found to be 30.6.  The degree of dissociation of  $N_2{O}_4$  is approximately 

Degree of Dissociation of N₂O₄ at 60°C

The vapour density of N₂O₄ at 60 °C is found to be 30.6.
Calculate the degree of dissociation.

For A_n ⇄ nB:

α = (D - d) / [d (n-1)]
D = Initial vapour densityd = Equilibrium vapour density

M_N2O4 = 92
D = M_N2O4 / 2 = 92 / 2 = 46
d = 30.6
n = 2

α = (46 - 30.6) / [30.6 (2 - 1)]
α = 15.4 / 30.6 = 0.5032

Degree of dissociation = 0.5032 × 100% = 50.32%

Explanation: The calculation uses the change in vapour densities and the relationship with number of molecules at equilibrium.
Approximately half of the N₂O₄ is dissociated at 60 °C.