Answer:
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
Boiling Point of the solution is 100.2°C.
The extent of boiling-point elevation is directly proportional to the molal concentration (amount of substance per mass) of the solution. It is given by:
T° boiling of solution - T° boiling of pure solvent = Kb . m . i
i = Van-t Hoff factor (number of ions dissolved in solution)
According to reaction:
LaCl₃ → La³⁺ + 3Cl⁻ in this case, i = 4 (stoichiometry)
Calculation for number of moles:
Moles of solute → 0.2453 g. 1mol/245.3g = 0.001 moles
m → 0.001 mol / 0.01kg = 0.1 m
On substituting the values:
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
Thus, the boiling point of solution is 100.2°C.
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