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Vapour density of moxture of NO2 and N2O4 is 34.5, then % abundance of NO2 in mixture is ______?
a. 50%
b. 25%
c. 40%
d. 60%
Let total moles of mixture is 100.
let there be x moles of NO2 molecular mass is 46 g
thus there are (100-x) moles of N2O4 molecular mass is 92g .
total mass of mixture = x * 42 + (100-x) 92
molecular mass of mixture = 2 * V.D = 2* 34.5 = 69
(x*42 + (100-x)92) / 100 = 69
solving... x=46
percentage abundance = ( (46*42) / (46*42 + 64*92) ) *100 = 24.7 (approx 25)
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