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Question

The vapour density of a mixture of NO2 and N2O4 at 27C is 38.3. How many moles of NO2 will be present in 500 g of the mixture?

A
1.825 mol
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B
3.485 mol
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C
8.125 mol
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D
2.185 mol
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Solution

The correct option is D 2.185 mol
We know that:
Vapour density of the mixture=average molar mass2 (i)
Let the mass of NO2=x
Moles of NO2+moles of N2O4=total moles of the mixture
Average molar mass=Total massTotal moles
Substituting in equation (i)
2×38.3=500x46+500x92
x46+500x92=5002×38.3 ; 2x+500x92=50076.6
(x+500)=500×9276.6=600.52
x=600.52500=100.52
Moles of NO2=100.5246=2.1852 mol

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