An aqueous CsCl solution is 8.00 percent by mass CsCl. What is the boiling point of this solution? Kb = 0.51°C/m for water.

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⁻

at the bottom

explanation: