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 oC/m.)

Boiling Temperature for solution is 100.2°C

Explanation:

To solve this we apply the boiling point elevation formula;

T° boiling of solution - T° boiling of pure solvent = Kb . m . i

i = Van't Hoff factor (number of ions dissolved in solution)

LaCl₃ → La³⁺  +  3Cl⁻          in this case, i = 4 (stoichiometry)

Let's determine the m (moles of solute in 1kg of solvent)

We convert the mass of solvent from g to kg

10 g. 1kg/1000g = 0.01 kg

Moles of solute → 0.2453 g. 1mol/245.3g = 0.001 moles

m → 0.001 mol / 0.01kg = 0.1 m

We replace data given:

Boiling T° for solution - 100°C = 0.512°C/m . 0.1 m . 4

Boiling T° for solution = 0.512°C/m . 0.1 m . 4  + 100°C → 100.2°C

b is the correct

1.43 x 10^-11 j the guy commented the awnser just making sure u see it .