Question

the vapour density of a mixture containing No2 and N2o4 is 27.6. the mole fraction of N2O4 in the mixture is

Answer 1

Molar mass of the mixture = 27.6 x 2 = 55.2 g
Let x moles of N2o4 be present in the 100 mole mixture

So,(100-x) x 46 + 92x = 100 x 55.2
=> 100 - x + 2x = 5520/46 = 120
=> 100 + x = 120
=> x = 20

Hence,mole fraction of N2O4 in the mixture = 20/100 = 0.2

Answer 2

Answer:0.2

Explanation:Density =mass avg. /2

27.6*2=55.2

M avg=total mass /total mole

Let total mole be 100

And mole of N2O4 be x

Mavg=(100-x)46+92x/100

55.2*100=(100-x)46+92x

5520=4600-46x+92x

5520-4600=46x

920=46x

920/46=x

20=x(mole of N2O4)

Mole fraction =20/100

0.2