# Problem: 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.a. 101.72°Cb. 100.30°Cc. 100.53°Cd. 103.91°Ce. 107.45°C

###### FREE Expert Solution

We’re being asked to calculate for the boiling point of an aqueous solution containing 8% (wt/wt) CsCl

Recall that the boiling point of a solution is higher than that of the pure solvent and the change in boiling point (ΔT­b) is given by:

The change in boiling point is also related to the molality of the solution:

$\overline{){{\mathbf{\Delta T}}}_{{\mathbf{b}}}{\mathbf{=}}{{\mathbf{iK}}}_{{\mathbf{b}}}{\mathbf{m}}}$

where:

i = van’t Hoff factor

m = molality of the solution (in m or mol/kg)

Kb = boiling point elevation constant (in ˚C/m)

We need to convert the 8% wt of CsCl to molality. Recall that the molality of a solution is given by:

We will calculate the boiling point of the solution using the following steps:

Step 1. Determine the composition of the solution.
Step 2. Calculate the moles of the solute.
Step 3. Calculate the mass of the solvent (in kg).
Step 4. Calculate the molality of the solution.
Step 5. Calculate the boiling point of the solution

Step 1. Determine the composition of the solution.

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###### Problem Details

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.

a. 101.72°C

b. 100.30°C

c. 100.53°C

d. 103.91°C

e. 107.45°C

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