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Question

The vapour density of a mixture containing NO2 and N2O4 is 27.6. Mole fraction of NO2 in the mixture is:


A
0.2
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B
0.4
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C
0.6
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D
0.8
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Solution

The correct option is D 0.8

Let the moles of NO2 be x.
So, moles of N2O4 will be 1x.
We know that, Molar mass = 2×vapour density
So, Molar mass = 2×27.6 = 55.2 g
Molar mass of NO2 = 46 g and N2O4 = 92 g
x×46+(1x)×921 = 55.2
= 46x+9292x=55.2
On solving, x=36.846=0.8
mole fraction of NO2 will also be 0.8 and mole fraction of N2O4 will be 0.2.


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