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Question

# The vapour density of N2O4 at a certain temperature is 30. What is the % dissociation of N2O4 at this temperature:

A
53.3%
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B
106.6%
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C
26.7%
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D
None of the above
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Solution

## The correct option is B 53.3%Vapour Density = MolecularMass2,If D is the initial vapour density of N2O4,D = 922=46,If d is the vapour density at equilibrium,d=30,Also, if α is the degree of dissociation,then we have the relation, α=D−d(n−1)d = 46−3030 = 1630 = 0.53n=2 for N2O4 as 2 moles of NO2 are formed for its 1 mole.Substituting the values, we get α=0.533,Hence the % dissociation is 53.33.

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