An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution? Kb = 0.51°C/m for water. Enter your answer with 2 decimal places and no units.

The boiling point of solution is 100.53

Explanation:

We are given:

8.00 wt % of CsCl

This means that 8.00 grams of CsCl is present in 100 grams of solution

Mass of solvent = (100 - 8) g = 92 grams

The equation used to calculate elevation in boiling point follows:

To calculate the elevation in boiling point, we use the equation:

Or,

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

= molal boiling point elevation constant = 0.51°C/m

= Given mass of solute (CsCl) = 8.00 g

= Molar mass of solute (CsCl) = 168.4  g/mol

= Mass of solvent (water) = 92 g

Putting values in above equation, we get:

Hence, the boiling point of solution is 100.53