What is difference between density of the equilibrium and vapour density of the mixture
Let's start from $m = 0.925 g˙\ce{SO3}$ that will be heated at $900 K$. This is the amount of $\ce{SO3}$ that will occupy a volume of $1.00 L$ after heating it at $900 K$ at $1 atm$. and after partial decomposition. In moles it is also  $n_o$ moles $\ce{SO3}$,  $$\ce{ $n_o = \frac{m}{80 g/mol} = \frac{0.925 g}{80 g/mol} = 0.01156 mol$}$$ At $900 K$,  a fraction $\alpha$ of these moles are decomposed into $\alpha$ moles $\ce{SO2}$ plus $\alpha/2 $ moles $\ce{O2}$. Then the total number of moles at equilibrium will be : $$\ce{$n$ = $n_o$˙($1$ - $\alpha$ + $\alpha$ +    \alpha/2}) = n_o(1 + \alpha/2) = 0.01156 mol (1 + \alpha /2)$$ The total volume is : $$\ce{$V = nRT/p= 0.01156 mol$(1 + \alpha/2)$RT/p = 0.001 m^3$}$$  Simple rearrangement of this formula gives : $$\ce{1 + \alpha /2  = \frac{0.001 $m^3·p$}{0.01156 $mol·RT$} = \frac{0.001 $m^3$·101325 $Pa$}{0.01156 $mol$·8.314 $J mol^{-1}K^{-1}·900 K$} = 1.172 }$$
The degree of dissociation $\alpha$ is equal to $\alpha = 2 ·0.172 = 34.4$ %