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Grade: 12th pass

                        

The vapour density of N2O4 at certain temperature is 30. Calculate the percentage dissociation of N2O4 at this temperature. N2O4(g) gives 2NO2(g)

2 years ago

Answers : (2)

Arun
24742 Points
							
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.=30 .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
2 years ago
Anurag kumar
13 Points
							
Consider the total mole =100 initially Let ued be x Then, N2O4-----2NO2 Left(100-x) 2x average mass =2×v.d=2×35=70 So, total mass/total moles=70 =100×92/100-x+2x=70 31.4%
one year ago
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