Calculating Kb and Kf from Latent Heat of Phase Change

By Aniket Bhardwaj · 11 November 2026 · Solutions

Colligative properties are a key scoring area in Class 12 board exams and entrance tests like JEE and NEET. Most students remember the standard equations for boiling point elevation, ΔT_b = K_b × m, and freezing point depression, ΔT_f = K_f × m. However, competitive exams often ask students to find the molal elevation constant (ebullioscopic constant, K_b) and molal depression constant (cryoscopic constant, K_f) directly from the latent heat of vaporisation or fusion of the solvent. Understanding the thermodynamic formulas connecting these constants prevents confusion in multi-step problems.

Thermodynamic Origin of Kb and Kf

The constants K_b and K_f depend exclusively on the properties of the pure solvent, not on the solute. Dissolving a non-volatile solute lowers the solvent chemical potential and vapour pressure. This shift alters the phase equilibrium temperatures where liquid, vapour, and solid phases coexist.

By combining Raoult's law with the Clausius-Clapeyron equation for dilute ideal solutions, physical chemistry yields direct equations linking K_b and K_f to the solvent boiling point, freezing point, gas constant, and enthalpy of phase change.

Formulas for Kb and Kf

Two equivalent forms are used depending on whether latent heat is given per mole or per gram of solvent.

Form 1: Using molar enthalpy (ΔH in J mol⁻¹ or cal mol⁻¹):

K_b = (M × R × T_b²) / (1000 × ΔH_vap)

K_f = (M × R × T_f²) / (1000 × ΔH_fus)

Form 2: Using specific latent heat (l in J g⁻¹ or cal g⁻¹):

Because l = ΔH / M, the molar mass M cancels:

K_b = (R × T_b²) / (1000 × l_v)

K_f = (R × T_f²) / (1000 × l_f)

Variables and standard units:

R: Universal gas constant = 8.314 J K⁻¹ mol⁻¹ (or 1.987 cal K⁻¹ mol⁻¹, often rounded to 2.0 cal K⁻¹ mol⁻¹ in older textbooks). Both unit systems are accepted.
M: Solvent molar mass in g mol⁻¹ (water = 18.015 g mol⁻¹, benzene = 78.11 g mol⁻¹).
T_b, T_f: Solvent boiling and freezing points in Kelvin. In standard calculations T(K) = T(°C) + 273.15 (or 273 in simplified questions). Both are used across boards.
ΔH_vap, ΔH_fus: Molar enthalpy of vaporisation and fusion in J mol⁻¹.
l_v, l_f: Latent heat of vaporisation and fusion per gram in J g⁻¹.
1000: Converts solvent grams to kilograms, giving K_b and K_f in K kg mol⁻¹.

Exam details

Source: practice problem written for this article (not a past-paper question)
Relevant exams: CBSE Class 12, State Boards, JEE Main, JEE Advanced, NEET, CUET
Topic: Colligative Properties and Thermodynamic Derivation of Molal Constants
Difficulty: Medium

Worked Numerical Examples

Question: Calculate the molal elevation constant (K_b) of pure water given that its normal boiling point is 100.0 °C and its specific latent heat of vaporisation is 2256 J g⁻¹. Take R = 8.314 J K⁻¹ mol⁻¹.

Solution:

Step 1: Convert boiling point to Kelvin: T_b = 100.0 + 273.15 = 373.15 K.

Square of boiling point: T_b² = (373.15)² = 139240.92 K².

Step 2: Note the given values: R = 8.314 J K⁻¹ mol⁻¹ and l_v = 2256 J g⁻¹.

Step 3: Apply the specific latent heat equation: K_b = (R × T_b²) / (1000 × l_v).

Numerator = 8.314 × 139240.92 = 1157649.01 J K mol⁻¹.

Denominator = 1000 × 2256 = 2256000.

Division: K_b = 1157649.01 / 2256000 = 0.5131 K kg mol⁻¹.

Step 4: Verify by a second route using molar enthalpy:

Molar mass of water M = 18.015 g mol⁻¹.

ΔH_vap = l_v × M = 2256 × 18.015 = 40641.84 J mol⁻¹.

K_b = (M × R × T_b²) / (1000 × ΔH_vap) = (18.015 × 1157649.01) / (1000 × 40641.84) = 20855046.9 / 40641840 = 0.5131 K kg mol⁻¹.

Both calculation methods give 0.513 K kg mol⁻¹.

Answer: 0.513 K kg mol⁻¹

Question: Calculate the molal depression constant (K_f) of water if the molar enthalpy of fusion of ice is 6008 J mol⁻¹ and the normal freezing point is 0.0 °C. Take R = 8.314 J K⁻¹ mol⁻¹ and solvent molar mass M = 18.015 g mol⁻¹.

Solution:

Step 1: Convert freezing point to Kelvin: T_f = 0.0 + 273.15 = 273.15 K.

Square of freezing point: T_f² = (273.15)² = 74610.92 K².

Step 2: Note the given values: R = 8.314 J K⁻¹ mol⁻¹, M = 18.015 g mol⁻¹, and ΔH_fus = 6008 J mol⁻¹.

Step 3: Apply the molar enthalpy equation: K_f = (M × R × T_f²) / (1000 × ΔH_fus).

Numerator = 18.015 × 8.314 × 74610.92 = 1117506.1 J g K mol⁻².

Denominator = 1000 × 6008 = 6008000.

Division: K_f = 1117506.1 / 6008000 = 1.8600 K kg mol⁻¹.

Step 4: Verify by a second route using specific latent heat:

Specific latent heat l_f = ΔH_fus / M = 6008 / 18.015 = 333.50 J g⁻¹.

K_f = (R × T_f²) / (1000 × l_f) = (8.314 × 74610.92) / (1000 × 333.50) = 620315.20 / 333500 = 1.8600 K kg mol⁻¹.

Both routes confirm 1.86 K kg mol⁻¹.

Answer: 1.86 K kg mol⁻¹

Common Mistakes to Avoid

Confusing latent heat per gram with enthalpy per mole: When latent heat is per gram (l), do not multiply by molar mass M. When enthalpy is per mole (ΔH), include M in the numerator.
Using Celsius instead of Kelvin: Temperatures must be converted to Kelvin before squaring. Using Celsius values leads to huge errors because temperature is squared.
Forgetting kilo prefix conversion: Enthalpy is often given in kJ mol⁻¹. Convert to J mol⁻¹ before using R = 8.314 J K⁻¹ mol⁻¹.
Using solute data: K_b and K_f depend only on the solvent. Never substitute solute properties into these equations.

Exam Relevance and Question Patterns

Examination	How the Topic Is Tested	Typical Question Structure
CBSE & State Boards	Direct formula substitution and derivations	Deriving the formula or calculating K_b from latent heat of vaporisation
JEE Main	Two-step numerical calculations	Calculating K_f from latent heat, then finding freezing point depression
JEE Advanced	Thermodynamic linking problems	Combining Clausius-Clapeyron relation with colligative property shifts
NEET	Direct numerical problem or formula choice	Selecting the correct formula for K_b or calculating K_f for water
CUET	Conceptual questions and unit checks	Identifying standard units (K kg mol⁻¹) and solvent dependence

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