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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 mol−1 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=(D−d)(n−1)d .....(1) where, α is degree of dissociation & n is number of moles of product formed. The dissociation reaction is given by N2O4⇌2NO2 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=(46−30)(2−1)30 =1630=0.533=53.3 hence, the percentage dissociation of N2O4 is 53.3%.

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