KBr is 80% dissociated in aqueous solution of 0.5 m concentration. (Given, Kf for water = 1.86 K kg mol-1). The solution freezes at

∆Tf=i×Kf×molality Here, degree of dissociation = 80%                KBr⇌K++Br-∵   i=1-α+nα1;i=1-0.8+2×0.81=1.8 ∴ ∆Tf=1.8×1.86×0.5⇒∆Tf=1.674 K       ∆Tf=Tof-Tf             ⇒1.674=273-Tf           Tf=273-1.674     ⇒Tf=271.326 K