Question

An aqueous solution containing 12.48 g of barium chloride (BaCl₂) is 1000g of water, boils at 100.0832°C. Calculate the degree of dissociation of barium chloride. 
(K_b for water = 0.52 K kg mol^-1, at. wt. Ba = 137, Cl = 35.5)

Answer 1

K_b = 0.52 K kg mol^-1
W_B = 100 g = 0.01 kg
ΔT_b = 100.0832 – 100 = 0.0832°C = 0.0832 K
Molality of BaCl2

`=(1.248xx1000)/(208.34xx100)`

`=5.99xx10^-2`

BaCI₂ ⇌ Ba² + 2CI^-

 

Molality after dissociation = 5.99 × 10^-2(1 + 2α)
ΔT_b = K_b × m
0.0832 = 0.52 × 5.99 × 10^-2 (1 + 2α)
0.0832 = 0.0312 + 0.0624α
α = 0.833 = 83.3%