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20 % of N_(2)O_(4) " molecules are disociated in a sample of a gas at " 27^(@)C and 760 " torr . Calculate the density of the equilibrium mixture ". |
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Answer» Solution :` {:(,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*2OR d=D/(1*2)= 46/(1*2)= 38*3` Alternatively, USE theformula directly, ` alpha = (D-d)/d` |
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