At `627^(@)C` and `1` atm `SO_(3)` is partially dissociated into `SO_(2)` and `O_(2)` by the reaction
`SO_(3)(g)hArrSO_(2)(g)+1//2O_(2)(g)`
The density of the equilibrium mixture is `0.925 g L^(-1)`. What is the degree of dissociation?

Explanation

Observed molar mas can be aclculated from the given density as follows :
`PV = nRT = w/M RT`
or `M _(obs) = (wRT)/(VP) = (d RT)/P= (0*925 gL^(-1) xx 0*0821"
L atm "K^(-1) mol^(-1) xx (627 + 273) K )/(1 " atm")`
` 68*35 " g mol"^(-1)`
Theoretical molar mass of `SO_(3) (g) hArr ( M_("theoretical"))=32 + 48 = 80" g mol"^(-1)`
If `alpha` is the degree of dissociation,
`{:(,SO_(3)(g),hArr,SO_(2)(g),+,1/2O_(2)(g)),(" Initial moles",1,,0,,0),("At equilibrium ",1-alpha,,alpha,,1/2alpha):}"Total"=1+alpha/2`
Theoretical `V.D. (D) alpha 1/V(V=" Molar volume ")`
` "Volume of " (1+alpha/2) " moles i.e.", " volume after dissociation "=(1+alpha/2) V`
`:. "Observed V.D. (d) " alpha1/((1+alpha/2)V)`
`:. D/d = 1 + alpha/2 or alpha = 2 ((D-d)/d)`
or `alpha = 2 ((M_("theoretical" )-M_("observed"))/(M_("observed")))= 2((80-68*35)/(68*35))= 0*3409`