The vapour density of $N_2{O}_4$ at a certain temperature is 30.67. The percentage of dissociation of $N_2{O}_4$ at given temperature is ___?

$$\begin{gathered} N_2{O}_4\rightleftharpoons 2N{O}_2 \\ Initially 1mole 0 \\ -x +2x \\ \text{ at equilibrium } 1-x 2x \end{gathered}$$

According to law of conservation of mass, mass of equilibrium mixture is 92 grams.

Mass of 1mole of equilibrium mixture $=\frac{92}{1+x}$

$\frac{92}{1+x}=2\times 30.67$

x=0.5

$\text{ So, }{N}_2{O}_4\text{ has been }50\%\text{ dissociated at given temperature }$

or 

you can use 

where D = theoritical vapour density (GMW/2), d= experimentally determined vapour density, n= no of product ions or molecules formed from one molecule of reactant