The vapour density of N2O4 at a certain temp is 30. What is the percentage dissociation of N2O4 At this temp?
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Answers (3)
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
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
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18th Sep, 2017
Solving related questions ask on Byjus, Meritnation etc. It's not a educational site, it's carrer counselling site
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18th Sep, 2017
Only Career related questions should be asked..,
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