# Problem: Lanthanum(III) chloride dissolves in water according to: LaCl3(s) → La3+(aq) + 3 Cl-(aq). What is the boiling point of the solution when 0.2453 g of LaCl3 (molar mass 245.3 g/mol) is dissolved in 10.0 g of H2O? (Kb of water is 0.512 °C/m.)a. 100.051°Cb. 100.154°Cc. 100.102°Cd. 100.205°Ce. 100.512°C

###### FREE Expert Solution

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 (ΔT­b) is given by:

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

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

where:

i = van’t Hoff factor

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

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

Recall that the Molality of a solution is given by: ###### Problem Details

Lanthanum(III) chloride dissolves in water according to: LaCl3(s) → La3+(aq) + 3 Cl-(aq). What is the boiling point of the solution when 0.2453 g of LaCl(molar mass 245.3 g/mol) is dissolved in 10.0 g of H2O? (Kb 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