Answer :
The freezing point of the solution is **-2.76°C**, and the boiling point of the solution is **101.42°C**.
The freezing point depression and boiling point elevation are colligative properties of a solution, which depend on the number of solute particles (ions or molecules) in the solution. To calculate these properties, we can use the formulas:
Freezing point depression: ΔTf = -Kf * m
Boiling point elevation: ΔTb = Kb * m
Where Kf is the freezing point depression constant, Kb is the boiling point elevation constant, and m is the molality of the solution (moles of solute per kilogram of solvent).
Given that the molecular weight of MgCl2 is 95.21 g/mol and it dissociates into three ions in solution (Mg²⁺ and two Cl⁻ ions), the number of moles of MgCl2 is:
Number of moles of MgCl2 = 112 g / 95.21 g/mol = 1.177 mol
The mass of the solvent (water) is 0.8 kg, so the molality (m) is:
Molality (m) = Number of moles of solute / Mass of solvent
Molality (m) = 1.177 mol / 0.8 kg = 1.471 mol/kg
Using the freezing point depression constant (Kf) for water (1.86°C·kg/mol), we can calculate the freezing point depression:
ΔTf = -Kf * m
ΔTf = -1.86°C·kg/mol * 1.471 mol/kg = -2.74°C
The freezing point of the solution is the freezing point of pure water (0°C) minus the freezing point depression:
Freezing point of solution = 0°C - 2.74°C = -2.76°C
Similarly, using the boiling point elevation constant (Kb) for water (0.512°C·kg/mol), we can calculate the boiling point elevation:
ΔTb = Kb * m
ΔTb = 0.512°C·kg/mol * 1.471 mol/kg = 0.75°C
The boiling point of the solution is the boiling point of pure water (100°C) plus the boiling point elevation:
Boiling point of solution = 100°C + 0.75°C = 100.75°C
Therefore, the freezing point of the solution is -2.76°C, and the boiling point of the solution is 100.75°C.
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