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20%N(2)O(4) molecules are dissociated in...

`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.

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Total number of moles after dissociation =0.8 +0.4 =1.2 mol
Volume of the equilibrium mixture `=(nRT)/(P)`
P= 1 atm , n=1.2 mole R = 0.082 L atm `K_(-1) " mol"^(-1) , T =27^(@) C = 300K`
Substituting the values in the above relation:
`V=((1.2 "mol")xx (0.082 L " atm " K_(-1) " mol"^(-1)) (300K))/(1( " atm ")) =29 .52 L`
Mass of the equilibrium mixture `=0.8 xx 92 + 0.4 xx 46 =92 0g`
`:.` Density of the equilibrium mixture `=("Mass")/("Volume") =((92 g))/((29. 52 L)) =3.12 L^(-1)`
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