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`).

Explanation

`(r_("mix"))/(r_(Kr))=sqrt((M_(Kr))/(M_("mix"))),` i.e., `1.16=sqrt((84)/(M_("mix")))" or " M_("mix")=62.43`
If x moles of `Cl_(2)` dissociate at equilibrium, then
`{:(,Cl_(2),hArr,2Cl),("At.eqm".,1-x,,2x):}`
Average , molecular mass of the mixture`=((1-x)xx71+2x xx35.5)/((1-x)+2x)=(71)/(1+x)`
`:.(71)/(1+x)=62.43` which gives x=0.137. Hence, fraction dissociated=0.137