Question

An aqueous solution containing 12.48 g of barium chloride (BaCl₂) in 1000 g 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

Given: Mass of BaCl₂ = 12.48 g

Mass of water = 1000 g = 1 kg

Boiling point of solution = 100.0832°C

Boiling point of pure water = 100.0°C

ΔT_b = 100.0832 − 100 = 0.0832°C

K_b = 0.52 K kg mol⁻¹

Molar mass of BaCl₂ = 137 + 2 × 35.5 = 208 g/mol

Moles of BaCl₂ = `12.48/208` 

= 0.06 mol

Molality (m) = `(0.06  mol)/(1  kg)`

= 0.06 mol/kg

We know that

ΔT_b = i K_b m

`i = (Delta T_b)/(K_b * m)`

= `0.0832/(0.52 xx 0.06)`

= `0.0832/0.0312`

= 2.67

BaCl₂ dissociates as

\[\ce{BaCl2 -> Ba^2+ + 2Cl-}\]

Total particles = 3

i = 1 + α(n − 1)

= 1 + α(3 − 1)

i = 1 + 2α

2.67 = 1 + 2α

`alpha = (2.67 - 1)/2`

= `1.67/2`

= 0.835

∴ Degree of dissociation (α) is 0.835 or 83.5%.