Ch. 2 | Solutions0/13

Depression of Freezing Point

Why solutions freeze lower, cryoscopic constants, and anti-freeze chemistry

Quantitative ReasoningLevel 3 · Analysis

Road crews have a choice: spread NaCl (M = 58.5 g/mol, i = 2) or CaCl₂ (M = 111 g/mol, i = 3) on icy roads. Per 100 g of salt spread, which provides greater freezing point depression — and by how much? (Kf water = 1.86 K·kg/mol)

✦Real Life Hook

Antifreeze in a car radiator is typically 50% ethylene glycol (HOCHX2CHX2OH\ce{HOCH2CH2OH}, M=62M = 62) in water. At 50% by volume, this lowers the freezing point to about −37°C-37°\text{C} and raises the boiling point to about 108°C108°\text{C}. But why don't we use pure ethylene glycol? Because the freezing point of pure glycol is −13°C-13°\text{C} — worse than the mixture! Adding water to glycol initially lowers the freezing point (colligative effect). The mixture beats either pure component. Nature's own antifreeze strategy: Arctic fish produce glycoproteins that interact with ice crystals and prevent their growth — a different mechanism but the same goal.

The Equation

ΔTf=Tf∘−Tf=Kf⋅m\Delta T_f = T_f^\circ - T_f = K_f \cdot m

where:

  • ΔTf\Delta T_f = depression of freezing point (K or °C), always positive
  • Tf∘T_f^\circ = freezing point of pure solvent; TfT_f = freezing point of solution (<Tf∘< T_f^\circ)
  • KfK_f = cryoscopic constant (molal freezing point depression constant) — depends only on the solvent
  • mm = molality of solution

Expanded form:
ΔTf=Kf×w2×1000M2×w1\Delta T_f = K_f \times \frac{w_2 \times 1000}{M_2 \times w_1}

Solving for molar mass:
M2=Kf×w2×1000ΔTf×w1M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1}

Cryoscopic constants (Kf) for common solvents

SolventFreezing Point (°C)Kf (K·kg/mol)
Water0.01.86
Benzene5.55.12
Acetic acid16.63.90
Camphor179.037.7
Cyclohexane6.620.2
📖NCERT 2.10NCERT Intext

Problem

45 g of ethylene glycol (CX2HX6OX2\ce{C2H6O2}, M=62M = 62 g/mol) is mixed with 600 g of water. Calculate the freezing point of the solution. (KfK_f for water = 1.86 K·kg/mol)

📖NCERT 2.11NCERT Intext

Problem

1.00 g of an unknown non-electrolyte dissolved in 50.0 g of benzene lowered the freezing point of benzene from 5.48°C to 5.25°C. Find the molar mass of the unknown. (KfK_f for benzene = 5.12 K·kg/mol)

✦JEE / NEET Exam InsightJEE / NEET
◆Kf vs Kb: Kf>KbK_f > K_b for the same solvent. This makes freezing point depression a more sensitive method for molar mass determination — especially for camphor (Kf=37.7K_f = 37.7, Rast's method for large molecules).
◆Common trap: ΔTf=Tf∘−Tf\Delta T_f = T_f^\circ - T_f. The freezing point falls, so ΔTf\Delta T_f is positive even though TfT_f decreases. Students often get the sign wrong.
◆Electrolytes: ΔTf=i⋅Kf⋅m\Delta T_f = i \cdot K_f \cdot m. For NaCl: i≈2i \approx 2. For AlCl₃: i≈4i \approx 4 (Al³⁺ + 3Cl⁻). For weak electrolytes: ii between 1 and maximum based on degree of dissociation.
Quick Check

Q1.The freezing point of a solution is always _____ than the freezing point of the pure solvent, and the value of ΔTf is always _____: