Rate of diffucion of ozonized oxygen is `0.4sqrt5` times that of pure oxygen what is the per cent degreeof association of oxygen assuming pure `O_(2)` in the sample initially ?

Options

A. 20
B. 40
C. 60
D. 80

Answer

C

Explanation

`(r_("mix"))/(r_(o_(2)))=sqrt((32)/(M_("mix")))=0.4 sqrt5`
`rArr M_("mix")=40g.mol`
`O_(2) (g) Leftrightarrow 2/3 O_(3) (g)`
`{:(,"Initial moles",1,0),(,"at eqm",1-x,2/3x):}`
`(32)/(40)=(1-x+2/3x)/(1)`
x=0.6 i.e 60%