We’re being asked to **determine the boiling point of a solution**. The solution is prepared by dissolving **0.2453 g of LaCl3 in 10 g of water**.

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:

$\overline{){\mathbf{\u2206}}{{\mathbf{T}}}_{{\mathbf{b}}}{\mathbf{=}}{{\mathbf{T}}}_{\mathbf{b}\mathbf{,}\mathbf{}\mathbf{solution}}{\mathbf{-}}{{\mathbf{T}}}_{\mathbf{b}\mathbf{,}\mathbf{}\mathbf{pure}\mathbf{}\mathbf{solvent}}}$

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

$\overline{){\mathbf{\u2206}}{{\mathbf{T}}}_{{\mathbf{b}}}{\mathbf{=}}{{\mathbf{iK}}}_{{\mathbf{b}}}{\mathbf{m}}}$

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)

Recall that the ** Molality of a solution** is given by:

$\overline{){\mathbf{molality}}{\mathbf{\left(}}{\mathbf{m}}{\mathbf{\right)}}{\mathbf{=}}\frac{\mathbf{moles}\mathbf{}\mathbf{solute}}{\mathbf{kg}\mathbf{}\mathbf{solvent}}}$

Lanthanum(III) chloride dissolves in water according to: LaCl_{3}(s) → La^{3+}(aq) + 3 Cl^{-}(aq). What is the boiling point of the solution when 0.2453 g of LaCl_{3 }(molar mass 245.3 g/mol) is dissolved in 10.0 g of H_{2}O? (K_{b} of water is 0.512 °C/m.)

a. 100.051°C

b. 100.154°C

c. 100.102°C

d. 100.205°C

e. 100.512°C

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Boiling Point Elevation concept. If you need more Boiling Point Elevation practice, you can also practice Boiling Point Elevation practice problems.