The composition ofthe equilibrium mixture (Cl_(2)hArr2Cl), which is attained at 1200^(@)C is determined by measuring the rate of effusion through a pin-hole. It is observed that at 1.80 mm Hg pressure, the mixture effuses 1.16 times as fast as Krypton effuses under the same conditions. Calculate the fraction of chlorine molecules dissociated into atoms (Atomic weight of Kr=84)

The composition ofthe equilibrium mixture (Cl_(2)hArr2Cl), which is at

1.	The composition ofthe equilibrium mixture (Cl_(2)hArr2Cl), which is attained at 1200^(@)C is determined by measuring the rate of effusion through a pin-hole. It is observed that at 1.80 mm Hg pressure, the mixture effuses 1.16 times as fast as Krypton effuses under the same conditions. Calculate the fraction of chlorine molecules dissociated into atoms (Atomic weight of Kr=84)
Answer» Solution :`(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` WHICHGIVES x=0.137. Hence, fraction dissociated=0.137