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Question

The vapour density of undecomposed N2O4 is 46. When heated, vapour density decreases to 24.5 due to its dissociation to NO2. The % dissociation of N2O4 is:

A
40
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B
57
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C
67
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D
87
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Solution

The correct option is D 87
The vapour density of undissociated N2O4 and dissociated N2O4 are 46 and 24.5 respectively.

Molecular weight is twice vapour density.

For undissociated N2O4, molecular weight =2×46=92

For dissociated N2O4, molecular weight =2×24.5=49

N2O4(g)100100x2NO2(g)02x

Assume initially 100 moles of N2O4 are present and x is the percent dissociation.

x moles of N2O4 dissociate to form 2x moles of
NO2.
Total number of moles after dissociation =(100x)+2x=100+x
The molecular weights of pure N2O4 and pure NO2 are 92 g/mol and 46 g/mol respectively. The molecular weight of dissociated N2O4 is 49 g/mol.
The average molecular weight after dissociation

92(100x)+46(2x)100+x=49

92(100x)+46(2x)=49100+x

920092x+92x=4900+49x

4300=49x

x=87

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