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)

Approved