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Question

138gm of N2O4 (g) is placed in 8.2L container at 300K. The equilibrium vapour density of mixture was found to be 30.67. Then: (R=0.082L atm mol1K1)

A
α= degree of dissociation ofN2O4=0.25
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B
Kp of N2O42NO2(g) will be 9atm.
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C
Total pressure at equilibrium =6.75atm
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D
The density of equilibrium mixture will be 16.83gm/litre.
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Solution

The correct options are
B Kp of N2O42NO2(g) will be 9atm.
C Total pressure at equilibrium =6.75atm
D The density of equilibrium mixture will be 16.83gm/litre.
138 gm of N2O4 (g) is placed in 8.2L container.
N2O42NO2.
t=0a
ta(1α)2aα
Vapour density 461+α=30.67
So 1+α=1.5,α=0.5=50%.
Total pressure=1.5×1.5×0.082×3008.2=6.75atm.
So Kp=4α21α2p=9atm.
And for density of mixture =1388.2gm/L=16.83gm/L

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