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Question

The vapour density of N2O4 at a certain temperature is 30. What is the percentage dissociation of N2O4 at this temperature?

A
53.3%
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B
76.6%
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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 A 53.3%
We know that, Molecular weight = 2× Vapour density
Molecular weight of N2O4 is 92 g mol1
if D is the initial Vapour density of N2O4 then, D=molecular weight of N2O42=922=46
if d is the Vapour density of equilibrium mixture, then d = 30

we have the relation, α=(Dd)(n1)d .....(1)
where, α is degree of dissociation & n is number of moles of product formed.

The dissociation reaction is given by N2O42NO2
mean 2 mole of NO2 will form from 1 mole of N2O4 i.e. n=2
putting the values in equation (1) we get,
α=(4630)(21)30
=1630=0.533=53.3
hence, the percentage dissociation of N2O4 is 53.3%.

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