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 (ΔT_{b})** is given by:

$\overline{){{\mathbf{\Delta T}}}_{{\mathbf{b}}}{\mathbf{=}}{{\mathbf{\Delta T}}}_{\mathbf{b}\mathbf{,}\mathbf{}\mathbf{solution}}{\mathbf{-}}{{\mathbf{\Delta T}}}_{\mathbf{b}\mathbf{,}\mathbf{}\mathbf{pure}\mathbf{}\mathbf{solvent}}}$

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)

**K _{b}** = 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:

$\overline{){\mathbf{Molality}}{\mathbf{}}{\mathbf{\left(}}{\mathbf{m}}{\mathbf{\right)}}{\mathbf{=}}\frac{\mathbf{moles}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solute}}{\mathbf{Kilograms}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solvent}}}$

**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.

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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