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.