Understanding the Concepts
* Freezing Point Depression: The freezing point of a solution is lower than the freezing point of the pure solvent. This depression is proportional to the molality of the solute.
* Boiling Point Elevation: The boiling point of a solution is higher than the boiling point of the pure solvent. This elevation is also proportional to the molality of the solute.
Calculations
1. Determine the van't Hoff factor (i):
* CaCl₂ dissociates into three ions in solution (one Ca²⁺ and two Cl⁻):
* CaCl₂ → Ca²⁺ + 2Cl⁻
* Therefore, the van't Hoff factor (i) for CaCl₂ is 3.
2. Calculate the molality (m):
* We need to convert molarity to molality. Assuming the density of the solution is approximately the same as water (1 g/mL), we can approximate the molality as:
* molality (m) ≈ molarity (M)
* m ≈ 0.1 mol/kg
3. Calculate the freezing point depression (ΔT_f):
* ΔT_f = i * K_f * m
* K_f (freezing point depression constant for water) = 1.86 °C/m
* ΔT_f = 3 * 1.86 °C/m * 0.1 mol/kg = 0.558 °C
4. Calculate the freezing point of the solution:
* Freezing point of pure water = 0 °C
* Freezing point of solution = 0 °C - 0.558 °C = -0.558 °C
5. Calculate the boiling point elevation (ΔT_b):
* ΔT_b = i * K_b * m
* K_b (boiling point elevation constant for water) = 0.512 °C/m
* ΔT_b = 3 * 0.512 °C/m * 0.1 mol/kg = 0.154 °C
6. Calculate the boiling point of the solution:
* Boiling point of pure water = 100 °C
* Boiling point of solution = 100 °C + 0.154 °C = 100.154 °C
Important Note:
* These calculations are based on the assumption that the solution is ideal. In reality, the van't Hoff factor may be slightly less than 3 due to ion pairing at higher concentrations.
* For more precise calculations, consider using the actual density of the solution to convert molarity to molality.
Therefore, the approximate freezing point of a 0.1 molar CaCl₂ solution in water is -0.558 °C, and the approximate boiling point is 100.154 °C.
