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At 90 degree celsius the vapour density of N2O4 is 24.8(H=1).calculate the degree of dissociation at this temperature.


G. Pawan Verma 12th Aug, 2019
Answer (1)
NAVEEN MEENA Student Expert 12th Aug, 2019

Dear Pawan,

N2O4 ---------NO2.        

Initially N2O4 has conc.= 1, then at equilibrium N2O4 has conc.1-X.

And NO2 has conc. =2x

 initially we have taken N2O4 =1

Molar mass =92 .V.D at eq.=24.8 M.w.=60.

n initial/ n final =M final/M initial 

1/1-X+2X=60/92

1/1+X=60/92

1+X=92/60

X=92/60 -1 

X=32/60

X=0.53          

percentage dissociation=0.53 X 100=53

All the best 

Hope this helps 

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