Question

Calculate the van’t Hoff factor of CdSO₄ (molecular mass = 208.4) if the dissociation of 5.21 g of CdSO₄ in half litre water gives a depression in freezing point of 0.168°C. (K_f of water is 1.86 K kg mol⁻¹)

Answer 1

Given: Mass of solute (CdSO₄) = 5.21 g

Molar mass of CdSO₄ = 208.4 g/mol

Volume of water = 0.5 L → Mass of water ≈ 500 g = 0.500 kg

ΔT_f = 0.168°C

K_f = 1.86 K kg mol⁻¹

Moles of CdSO₄ = `5.21/208.4` mol

Molality (m) = `0.025/0.500` = 0.05 mol/kg

ΔT_f ​= i ⋅ K_f ​⋅ m

⇒ `i = (Delta T_f)/(K_f * m)`

= `0.168/(1.86 xx 0.05)`

= `0.168/0.093`

= 1.806

∴ The van’t Hoff factor iii for CdSO₄ is approximately 1.806.