The composition of the equilibrium mixture (`Cl_(2)` `2Cl`) , which is attained at `1200^(@)C`, is determined by measuring the rate of effusion through a pin hole. It is observed that a `1.80 mm Hg` pressure, the mixture effuses `1.16 times` as fact as krypton effuses under the same conditions. Calculate the fraction of chlorine molecules dissociated into atoms (atomic weight of `Kr` is `84`).

Answer

`0.14`

Explanation

If `‘alpha ’` is the degree of dissociation, then at equilibrium
`Cl_2 " " 2Cl`
Moles `1- alpha " " 2 alpha` Total `= 1 + alpha`
From diffusion information `(r_("(mix)"))/(r_("(kr)")) = 1.16 = sqrt((84)/(M ("mix")))`
`rArr M_(("mix")) = 62.4 rArr M_(("mix")) = (71)/(1 + alpha) = 62.4 rArr alpha = 0.14`