At `627^(@)C` and 1 atm pressure, `SO_(3)` undergoes partial dissociation into `SO_(2)` and `O_(2)`
`SO_(3) hArr SO_(2) + (1)/(2)O_(2)`
if the observed density of the equilibrium mixture is 0.925 g/L, calculate degree of dissociation of `SO_(3)`.

Explanation

Let the initial no. of moles of `SO_(3)` be 1 and its degree of dissociation x.
`{:(1,,0,,0,"Initial no. of moles"),(SO_(3),+,SO_(2),+,(1)/(2)O_(2),),(1-x,,x,,x//2,"Moles at equilibrium"):}`
`therefore` total no. of moles at eqb. `= 1 - x + x + (x)/(2) = 1 + (x)/(2)`
Thus applying pV = nRT
`1 xx V = (1+(x)/(2)) xx 0.0821 xx (627 + 273)`
`V = (1+(x)/(2)) xx 73.89` litres.
Now w.t of 1 mole of `SO_(3) = 80g` and therefore,
from the law of conservation of mass, we have,
wt. of gases at eqb. = 80 g.
`therefore "density" = ("wt. in g")/("vol. in litre") = (80)/((1+(x)/(2)) xx 73.89) = 0.925` (given)
or x = 0.34.