At 627^(@) and one atmosphere pressure, SO_(3) is partially dissociated into SO_(2) and O_(2) as SO_(3) (g) hArr SO_(2) (g) + 1/2 O_(2) (g) The density of the equilibrium mixture is found to be 0*925 g L^(-1) . Calculate the degree of dissociation of SO_(3)under the given conditions.

At 627^(@) and one atmosphere pressure, SO_(3) is partially dissociate

1.	At 627^(@) and one atmosphere pressure, SO_(3) is partially dissociated into SO_(2) and O_(2) as SO_(3) (g) hArr SO_(2) (g) + 1/2 O_(2) (g) The density of the equilibrium mixture is found to be 0*925 g L^(-1) . Calculate the degree of dissociation of SO_(3)under the given conditions.
Answer» Solution :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= (0925 gL^(-1) xx 00821" L atm "K^(-1) mol^(-1) xx (627 + 273) K )/(1 " atm")` ` 6835 " gmol"^(-1)` Theoretical molar mass of `SO_(3) (g) HARR ( M_("theoretical"))=32 + 48 = 80" gmol"^(-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((80-6835)/(6835))= 03409`