Question

The vapour density of partially dissociated iodine vapour at a temperature was found to be 80. What is its degree of dissociation?

Answer 1

Vapour density of any substance is defined as fraction of molecules dissoaciating at a given time.
Given:
Molecular weight of Idoine=254
Let "α " be the degree of dissociation under equilibrium.
equation:
I₂₍g₎      ⇄        2I₂₍g₎
1-α                  2α
VApour density of undissociated vapour
d₀ =254/2 =127
Number of moles at equilibrium=1-α +2α⇒1+α
Vapour density of undissociated I₂
______________________________        = d/d₀  =1/1+α
vapour density of equilibrium molecule

degree of dissociation =α=d₀-d/d  =127=80/80 =0.587