What is the boiling point of a solution made by mixing 100.0 g ZnCl2 with 700.0 grams of water (Kb for water is 0.512 oC/m)?

Molar mass of ZnCl₂ = (65.38 + 35.45×2) g/mol = 136.28 g/mol
No. of moles of ZnCl₂ = (100.0 g) / (136.28 g/mol) = 0.7338 mol
Molality of ZnCl₂, m = (0.7338 mol) / (700.0/1000 kg) = 1.048 m

1 mole of ZnCl₂ contains 3 moles of ions (1 mole Zn²⁺ ions and 2 moles of Cl⁻ ions).
van’t Hoff factor, i = 3

Kf for water = 0.512°C/m

Elevation of boiling point, ΔTb = i Kb m = 3 × (0.512°C/m) × (1.048 m) = 1.61°C

Normal boiling point of water = 100°C
Boiling point of the solution = (100 + 1.61)°C = 101.61°C