Answer:
100.52°C is the boiling point of this solution
Explanation:
In 100 g of solution, we have 8 g of cesium chloride.
Mass of solvent in our solution is 92 g, of water.
Let's determine the molality → mol/kg
Mol of solute → 8 g . 1 mol/ 168.35 g = 0.0475 moles
Mass of solvent → from g to kg → 92 g . 1kg / 1000g = 0.092 kg
Molality → 0.0475 mol/0.092 kg = 0.516 m
According to the colligative property, we assume 100 % of ionization in the salt, boiling point for the solution will be:
T° boiling point for the solution = (0.512°C/ m . 0.516 m . 2) + 100°C
T° boiling point for the solution = 100.52°C
Van't Hoff factor = 2
CsCl → Cs⁺ + Cl⁻
