Question

an aqueous solution containing 1.248 g of BaCl2(molar mass =208.34). in 100 g of water boils at 100.0832°c. calculate degree of dissociation of BaCl2

Answer 1

m(molality) = (wt / Mw) x (1000 / weight of solvent in gm)
m                = (1.248 / 208.34) x (1000 / 100)
m                = 0.0599 ≈ 0.06 m

∆Tb = T1 - T = 100.0832 - 100 = 0.0832°C or 0.0832 K(because here we find difference of temperature )
we know that Kb of water = 0.512 K / m 

∆Tb = i x Kb x m

i = ∆Tb / ( Kb x m) = 0.0832 / ( 0.512 x 0.06)
i = 2.7

α = (i-1) / (n-1)
for BaCl2 , n = 3

α = (2.7 - 1) / (3-1)
α = 1.7/2 = 0.85

and in % α = 85 %

i hope it will help you

regards