`20%N_(2)O_(4)` molecules are dissociated in a sample of gas at `27^(@)C` and 760 torr. Calculate the density of the equilibrium mixture.

Explanation

` {:(,N_(2)O_(4)(g),hArr,2NO_(2)(g),),("Intial",1 " mole",,, ),("At.eqm.",1-0*2 = 0*8" mole",,0*4 "mole,",Total = 1*2 " moles" ):}`
If V is the volume of the vapour per mole, volume of vapour before dissociation = V
` " Hence density " (D) propto1/V `
But density after dissociation
`D= (" Mol.wt.of " N_(2)O_(4))/2 = 92/2=46" " ("Theoretical density")`
Volume after dissociation = `1*2` V
` :. "Density (d)" propto 1/(1*2 V) `
` :. D/d = 1/V xx 1*2V =1*2 or d=D/(1*2)= 46/(1*2)= 38*3`
Alternatively, use the formula directly,
` alpha = (D-d)/d`