Question

An aqueous solution containing 1.248 g of barium chloride (molar mass = 208.34 g mol⁻¹) in 100 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)

Answer 1

Given: Mass of BaCl₂ = 1.248 g

Molar mass of BaCl₂ = 208.34 g/mol

Mass of water = 100 g = 0.100 kg

Boiling point of solution = 100.0832°C

Boiling point of pure water = 100.0000°C

Elevation in boiling point (ΔT_b) = 0.0832°C

K_b​ for water = 0.52 K kg mol⁻¹

Moles of BaCl₂ = `1.248/208.34`

= 0.00599 mol

Molality (m) = `0.00599/0.100`

= 0.0599 mol/kg

ΔT_b = iK_bm

⇒ `i = (Delta T_b)/(K_b * m)`

`i = 0.0832/(0.52 xx 0.0599)`

= `0.0832/0.03115`

= 2.67

For BaCl₂

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

Total particles after dissociation = 3

Van’t Hoff factor for partial dissociation (i) = 1 + α(n − 1)

= 1 + α(3 − 1)

= 1 + 2α

2.67 = 1 + 2α

⇒ `alpha = 1.67/2`

= 0.835

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