Solutions · Lesson 8 of 11
Depression of Freezing Point
“Explain solvent freezing equilibrium and use freezing-point depression to determine concentrations and molar masses.”
• Explain why dissolved solute lowers the solvent’s freezing temperature. • Derive the dilute freezing relation and the cryoscopic constant. • Calculate a final freezing temperature and distinguish it from its depression. • Combine freezing measurements with molar-mass and simultaneous-equation calculations.
Freezing is an equilibrium between solid and liquid solvent
At the freezing point of a pure solvent, solid and liquid can coexist with no net transfer from one phase to the other. Their molecular escaping tendencies are equal. Adding a solute that stays mainly in the liquid lowers the liquid solvent’s escaping tendency, while the pure solid solvent is largely unaffected. Equality is restored only at a lower temperature. Thus the liquid solution can remain unfrozen below the freezing temperature of the pure solvent.
A vapour-pressure diagram offers another view: the vapour-pressure curve over the solution lies below that over pure liquid solvent. It meets the pure solid-solvent curve at a lower temperature. This reasoning assumes that the solid that first separates is essentially pure solvent and that the solute is excluded from that solid. It is not a complete description of every multicomponent solid phase or of a solution as progressively more ice forms.
The dilute law and its physical assumptions
The relationship is proportional to molality for small solute concentrations. Double the number of independent solute particles in the same solvent mass and the predicted depression doubles. Changing solvent changes K_f: identical molalities can produce very different temperature shifts in water and benzene. The chemical identity of a non-reacting solute does not enter this first-order particle-count relation.
Deriving the solvent constant
At equilibrium with pure solid solvent, the change in the pure solvent’s fusion tendency must balance the mixing effect in the liquid. A small-temperature-shift equilibrium expression gives −ln x₁≈[ΔfusH/(R(T_f°)²)]ΔT_f. Here ΔfusH is the molar enthalpy of melting, and the fusion enthalpy is treated as nearly constant over the small interval. This form also follows by comparing the solid and liquid vapour-pressure slopes near their intersection.
Equate the expressions for −ln x₁ and rearrange to isolate ΔT_f. The coefficient of molality is K_f. The square of the pure freezing temperature must use kelvin, because it comes from a thermodynamic temperature relation. This derivation explains both the solvent dependence and the restriction to small depressions.
| Solvent | Boiling point / K | K_b / K kg mol⁻¹ | Freezing point / K | K_f / K kg mol⁻¹ |
|---|---|---|---|---|
| Water | 373.15 | 0.52 | 273.15 | 1.86 |
| Ethanol | 351.5 | 1.20 | 155.7 | 1.99 |
| Cyclohexane | 353.74 | 2.79 | 279.55 | 20.00 |
| Benzene | 353.3 | 2.53 | 278.6 | 5.12 |
| Chloroform | 334.4 | 3.63 | 209.6 | 4.79 |
| Carbon tetrachloride | 350.0 | 5.03 | 250.5 | 31.8 |
| Carbon disulfide | 319.4 | 2.34 | 164.2 | 3.83 |
| Diethyl ether | 307.8 | 2.02 | 156.9 | 1.79 |
| Acetic acid | 391.1 | 2.93 | 290.0 | 3.90 |
These are approximate reference values near ordinary pressure; use the numerical constant specified in a particular problem consistently. Some experimental examples use a different rounded benzene K_f, such as 4.9 or 5.1, rather than the table’s 5.12. Such differences must not be mixed within one calculation. Pure water’s usual freezing reference is 273.15 K; a table rounded to 273 K should not force premature rounding of a calculated temperature.
From concentration to freezing temperature
Problem
Mix 45 g ethylene glycol, molar mass 62 g mol⁻¹, with 600 g water. Use K_f=1.86 K kg mol⁻¹. Estimate the depression and final freezing point.
- 1.Glycol amount=45/62=0.72580645 mol. Water mass=0.600 kg.
- 2.Molality=(45/62)/0.600=1.20967742 mol kg⁻¹. Retain intermediate digits.
- 3.ΔT_f=1.86×1.20967742=2.250 K. T_f=273.15−2.250=270.900 K, or −2.250 °C.
- 4.Given the input precision and dilute-model limitation, report about a 2.3 K depression and freezing near 270.9 K. Premature rounding of molality to 1.2 changes the later displayed result unnecessarily.
Problem
A 1.00 g solute in 50.0 g benzene lowers freezing temperature by 0.40 K. Use K_f=5.12 K kg mol⁻¹. Find molar mass.
- 1.Molality=0.40/5.12=0.078125 mol kg⁻¹.
- 2.Solute amount=0.078125×0.0500=0.00390625 mol.
- 3.M₂=1.00/0.00390625=256 g mol⁻¹. This is about 2.6×10² g mol⁻¹ at the precision of 0.40 K.
- 4.The interpretation assumes independent solute molecules. Association would make the inferred molar mass greater than the monomer value.
Problem
Compounds AB₂ and AB₄ each have 1.00 g dissolved separately in 20 g benzene. Depressions are 2.3 K and 1.3 K; K_f=5.1 K kg mol⁻¹. Estimate atomic molar masses A and B, assuming i=1.
- 1.Calculate formula molar masses separately: M(AB₂)=1000×5.1×1/(2.3×20)=110.8696 g mol⁻¹; M(AB₄)=196.1538 g mol⁻¹.
- 2.Let atomic molar masses be a and b. Then a+2b=110.8696 and a+4b=196.1538.
- 3.Subtract equations to remove a: 2b=85.2843, so b=42.6421 g mol⁻¹. Substitute back: a=25.5853 g mol⁻¹.
- 4.The data have limited precision, so report approximate values, around 26 and 43 g mol⁻¹. No periodic-table identity needs to be invented from these hypothetical data.
ΔT_f is a positive difference; the solution freezing point is lower. Also, use the initial solvent mass when applying the simple law to the initial composition. As solvent freezes out, the remaining liquid becomes more concentrated and the freezing condition changes.
Quick Quiz
Quiz
Why does a non-volatile solute lower freezing temperature in the stated model?
Explanation: Mixing lowers the liquid solvent tendency; a lower temperature is needed to restore equality with pure solid solvent.
If T_f°=273.15 K and ΔT_f=1.86 K, what is T_f?
Explanation: Freezing depression is subtracted: 273.15−1.86=271.29 K.
What does K_f depend on in the dilute non-electrolyte law?
Explanation: The solvent’s pure phase properties determine K_f.
Which solid phase is assumed in the simple freezing derivation?
Explanation: The liquid is a solution while the first solid is treated as pure solvent, excluding solute.
At equal solvent masses and equal moles of two independent non-electrolyte solutes, their predicted freezing depressions are:
Explanation: The same number of independent particles per solvent mass gives equal molality and therefore equal K_fm.
Practice Problems
- What mass of ascorbic acid, 176 g mol⁻¹, in 75 g acetic acid produces a 1.5 K depression if K_f=3.9?
- A 5% by mass sucrose solution freezes at 271 K; pure water freezes at 273.15 K. Estimate the freezing point of a 5% by mass glucose solution using molar masses 342 and 180 g mol⁻¹ and the same dilute proportional model.
- Calculate K_f for a solvent with T_f°=300 K, M₁=80 g mol⁻¹ and ΔfusH=12000 J mol⁻¹. Use R=8.314.
- Why might an antifreeze freezing calculation be an estimate rather than an exact temperature?
- A 0.60 g independent solute in 100 g water depresses freezing by 0.186 K. Find molar mass with K_f=1.86. Then explain how halving solvent mass would affect the predicted depression.
Required molality=1.5/3.9=0.384615 mol kg⁻¹. Solute amount=0.384615×0.075=0.0288462 mol. Mass=0.0288462×176=5.07692 g≈5.1 g. The 75 g refers to solvent, and the melting-point depression equals the freezing-point depression at equilibrium.
Key Takeaways
• A solute retained in the liquid lowers the solvent’s freezing temperature. • The simple model assumes the solid that separates is essentially pure solvent. • Depression is defined as T_f°−T_f and is a positive quantity. • For dilute ideal non-electrolytes, ΔT_f=K_fm. • K_f is a solvent property with unit K kg mol⁻¹. • Retain intermediate numerical precision and round the final answer appropriately. • Freezing measurements can determine molar mass and, with multiple compounds, atomic-mass relationships.