At $627°\mathrm{C}$  and 1 atm ${\mathrm{SO}}_3$  is partially dissociated into ${\mathrm{SO}}_2$ and  ${\mathrm{O}}_2$ by the reaction  ${\mathrm{SO}}_3\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{SO}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)$
The density of the equilibrium mixture is 0.925 g ${\mathrm{L}}^{-1}$ . What is the percentage degree of dissociation?(Round off to the nearest integer)
(Use $\text{R}=0.0821{\text{\hspace{0.17em}\hspace{0.17em}L\hspace{0.17em}\hspace{0.17em}atm\hspace{0.17em}\hspace{0.17em}K}}^{-1}{\text{\hspace{0.17em}mol}}^{-1}$ , Atomic Mass : S = 32, O = 16)

Let the molecular mass of the mixture at equilibrium be  ${\mathrm{M}}_{\mathrm{mix}}$.
Applying the relation:
${\mathrm{M}}_{\mathrm{mix}}=\frac{\mathrm{dRT}}{\mathrm{P}}=\frac{0.925\times 0.0821\times 900}{1}=68.348$
Molecular mass of  ${\mathrm{SO}}_3=80$
Vapour density of  ${\mathrm{SO}}_3$,  $\mathrm{D}=\frac{80}{2}=40$
Vapour density of mixture,  $\mathrm{d}=\frac{68.348}{2}=34.174$
Let the degree of dissociation be  $\mathrm{x}$.
$\mathrm{x}=\frac{\mathrm{D}-\mathrm{d}}{(\mathrm{n}-1)\mathrm{d}}=\frac{40-34.174}{(\frac{3}{4}-1)\times 34.174}=\frac{5.826\times 2}{34.174}=0.34$