Colligative Properties and Freezing Point Depression Experiment Notes

Experiment Overview

• Objective: To use the concept of freezing point depression to determine the molar mass of a compound by measuring the freezing point of a pure solvent and a solution using a colligative property.

Apparatus and Chemicals

• Apparatus:

  • 25 mm x 200 mm test tube
  • Thermometer
  • 600 mL beaker

• Chemicals:

  • Acetic Acid
  • Ice
  • Unknown solid assigned by instructor
  • 0.0001 g analytical balance
  • Spatula

Key Concepts

• Solutions:

  • Homogeneous mixtures of two or more substances.
  • Major component -> Solvent
  • Minor component -> Solute

• Colligative Properties:

  • Depend on the number of solute particles, not their identity.
  • Include:
  • Boiling point elevation
  • Freezing point lowering
  • Vapor pressure lowering
  • Osmotic pressure

Freezing Point Depression

• When a solute is added to a solvent:
  • Freezing point is lowered compared to the pure solvent.
  • The magnitude depends on the number of solute particles, represented by the equation:
    $\Delta T<em>f = T</em>f(solvent) - T<em>f(solution) = -iK</em>f m$
  • Where:
  • $\Delta T_f$ = change in freezing point
  • $T_f$ = freezing point
  • $i$ = van't Hoff factor (1 for nonelectrolytes, varies for electrolytes)
  • $K_f$ = molal freezing point depression constant (°C/m)
  • $m$ = molality of the solution (moles of solute per kg of solvent)

Example Calculations

1. Calculating Freezing Point Depression:

   • For a solution of ethylene glycol and water (volume ratio 25:75):
   • Density calculations to find moles:
     • $\text{Mass ethylene glycol} = 1.50 \text{ L} \times 1.11 \text{ g/mL} = 1665 \text{ g}$
   • Moles of ethylene glycol:
     $\text{Moles} = \frac{1665 \text{ g}}{62.07 \text{ g/mol}} = 26.9 \text{ mol}$
   • Mass of solvent (water): 4.50 kg
   • Molality:
     $m = \frac{26.9 \text{ mol}}{4.50 \text{ kg}} = 5.96 m$
   • Freezing point depression calculation:
     $\Delta T_f = -1(1.86 \text{ °C/m})(5.96 m) = -11.1 °C$
   • Final freezing point of solution: $T_f(solution) = 0.0 °C + (-11.1 °C) = -11.1 °C$

2. Using NaCl to Understand Depression:

   • Mass of NaCl = 360.0 g / 1.00 kg of water.
   • Calculate molality:
     • Moles NaCl: $\text{Moles} = \frac{360.0 \text{ g}}{58.44 \text{ g/mol}} = 6.16 \text{ mol}$
   • Molality:
     $\text{Molality} = \frac{6.16 \text{ mol}}{1.00 \text{ kg H2O}} = 6.16 m$
   • Freezing point depression:
     $\Delta T_f = -2(1.86 \text{ °C/m})(6.16 m) = -22.9 °C$

3. Determining Molar Mass of Unknown:

   • Example: Dissolving 0.100 g of an unknown in 20 g of cyclohexane, resulting in a freezing point depression of 1.06 °C.
   • Calculate molality:
     • From freezing point:
       $m = \frac{1.06 °C}{1.86 °C/m} = 0.0530 m$
   • Calculate moles of solute: 0.0530 mol / (20.0 g/1000 g) = 0.0530 m[0.0530 mol/kg]
   • Determine molar mass:
     $\text{Molar mass} = \frac{0.100 g}{0.0530 mol} = 94.3 g/mol$

Experiment Procedure

Freezing Point Depression

1. Prepare glacial acetic acid and perform initial volume measurements.
2. Create an ice-water bath; immerse test tube and monitor temperature until freezing occurs.
3. Add unknown mass and record freezing point depression.
4. Calculate molar mass from the recorded data based on freezing point changes.

Crystallization and Temperature Observations

• Observing crystallization from solutions and identifying shapes.
• Performing solubility tests at varying temperatures to observe compound behaviors.

Final Thoughts

• Differentiating Molarity and Molality: Molarity is dependent on volume, while molality is dependent on mass.
• Crystal Shapes: Crystallization can indicate compound identity, reference materials can offer additional clarity.