Question

vapour density of Na2N2O4 mixture= 34.5 find the mole % of Na2

Answer 1

The equilibrium is as follows  2Na2N2O4⇔  (Na2N2O4)2

molar mass of monomer is 138g/mol
Vapour density of the mixture is 34.5
Molar mass of mixture =  2 × 34.5 = 69g/mol

Degree of association, α = molar mass of mixturemolar mass of monomer= 69138=0.5       (1)

Assume that the total moles of the reactant and product in the reaction are c
Moles of reactants and products at various times during the reactions are: 
Initial:    moles of monomer at the start of reaction = c  and  dimer = 0
At equilibrium:  
moles of monomer dissociated = cα 
Moles of monomer (Na2N2O4) is  c-cα  and of dimer is cα                   


Mole percent of Na2N2O4=moles of Na2N2O4  × 100total moles in the experiment=(c−cα)×100c=c(1−α) ×100c= (1−α) ×100  
Susbtituting the value of α  from (1),  mole percent of Na2N2O4 = 100 (1-0.5) = 50%

Answer 2

Hey mate......

here's ur answer.......

The answer is 50% .....

Hope it helps ❤️❤️