Question

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.

Answer 1

Answer: The boiling point of solution is 100.53

Step-by-step 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:

`\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}`

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

`\Delta T_b=iK_bm`

Or,

`\text{Boiling point of solution}-\text{Boiling point of pure solution}=i* K_b* \frac{m_(solute)* 1000}{M_(solute)* W_(solvent)\text{ (in grams)}}`

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

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

`m_(solute)` = Given mass of solute (CsCl) = 8.00 g

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

`W_(solvent)` = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

`\text{Boiling point of solution}-100=2* 0.51^oC/m* (8.00* 1000)/(168.4g/mol* 92)\\\\\text{Boiling point of solution}=100.53^oC`

Hence, the boiling point of solution is 100.53