At equilibrium N2O4=2No2 the observed molecular weight of N2O4is 80g/mol at 350k .The Percentage dissociation of No2 at 350k is?
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At equilibrium N2O4(g)⇌2NO2(g)the observed molecular weight of N2O4is 80g mol−1 at 350 K. The percentage dissociation of N2O4(g) at 350 K is:
A .
10%
B .
15%
C .
20%
D.
25%
ANSWER
N2O4⇌2NO2
t=0, 1 0
t=teq 1−α 2α
Normal molecular mass of N2O4=92g/mol
Observed molecular mass= 80g/mol
∴ Van’t Hoff factor i =8092=1.15
⇒i=11−α+2α=Calculated colligative propertyObserved colligative property
⇒1.15=11+α
⇒α=0.15
∴ Percentage dissociation of N
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