For the reaction 2NH₃(g) $\to$N₂(g) + 3H₂(g) if the observed molar mass of the mixture is 12.75.The degree of dissociation is 

$\mathrm{Degree} \mathrm{of} \mathrm{dissociation} = \frac{\mathrm{Theoretical} \mathrm{molar} \mathrm{mass} - \mathrm{observed} \mathrm{molar} \mathrm{mass}}{(\mathrm{n}-1)\mathrm{Observed} \mathrm{molar} \mathrm{mass}}$

n = number of product particles formed per molecule of reactant

= $\frac{17-12.75}{(2-1)\times 12.75}=\frac{1}{3}$