Colligative Properties and Determination of Molar Mass

Colligative properties depend only on the NUMBER of solute particles, not on their nature. The four colligative properties are relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure. All are used to determine molar masses.

Elevation of boiling point

Adding a non-volatile solute raises the boiling point: ΔTb = Kb·m, where Kb is the molal elevation (ebullioscopic) constant and m is molality. Molar mass M₂ = (Kb·w₂·1000)/(ΔTb·w₁).

Depression of freezing point

The freezing point falls: ΔTf = Kf·m, where Kf is the molal depression (cryoscopic) constant. Molar mass M₂ = (Kf·w₂·1000)/(ΔTf·w₁). This is why salt is spread on icy roads and antifreeze is added to radiators.

Osmotic pressure

Osmosis is the flow of solvent through a semipermeable membrane from lower to higher solute concentration. The osmotic pressure Π = C R T = (n₂/V)RT, so M₂ = w₂RT/(ΠV). Osmotic pressure is the PREFERRED method for macromolecules (proteins, polymers) because it is measurable at room temperature and gives large, easily measured values.

Isotonic, hypertonic and hypotonic

Solutions with the same osmotic pressure are isotonic (e.g. 0.9% NaCl with blood). Reverse osmosis (applying pressure greater than Π) is used for desalination of sea water.

Exam Tricks & Tips

• 🎯 Colligative properties depend on the number of particles, not their identity — the core definition.
• 🎯 Osmotic pressure is best for polymers/proteins because it is large and measured at room temperature (no heating that could denature them).
• 🎯 Kb and Kf are properties of the SOLVENT only, independent of the solute.
• 🎯 Blood is isotonic with 0.9% (m/V) NaCl — remember this classic value.
• 🎯 Reverse osmosis needs pressure GREATER than the osmotic pressure and is used to desalinate sea water.
• ❌ Common mistake: forgetting the factor 1000 (unit conversion g→kg) in the molar-mass formulas, or mixing up w₁ (solvent) and w₂ (solute). w₂ is always the solute mass.

Board exam focus

At least one numerical to find molar mass from ΔTb, ΔTf or Π is almost guaranteed, plus reasoning on why osmotic pressure suits macromolecules.

Cover the answer, recall, then check. 12 cards on colligative properties.

Q1. What are colligative properties?
A1. Properties that depend only on the number of solute particles, not their chemical nature.

Q2. Name the four colligative properties.
A2. Relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure.

Q3. Write the elevation of boiling point relation.
A3. ΔTb = Kb·m, where Kb is the molal elevation constant and m is molality.

Q4. Write the depression of freezing point relation.
A4. ΔTf = Kf·m, where Kf is the molal depression constant and m is molality.

Q5. What do Kb and Kf depend on?
A5. Only the solvent — they are independent of the solute.

Q6. Give the osmotic pressure equation.
A6. Π = CRT = (n₂/V)RT, where C is molar concentration of solute.

Q7. Why is osmotic pressure best for measuring molar masses of macromolecules?
A7. It gives large, easily measured values at room temperature, avoiding heating that could denature proteins.

Q8. What is osmosis?
A8. Net flow of solvent through a semipermeable membrane from a dilute (low solute) to a concentrated (high solute) solution.

Q9. Define isotonic solutions.
A9. Solutions having the same osmotic pressure; e.g. 0.9% NaCl is isotonic with blood.

Q10. What is reverse osmosis and one use?
A10. Applying pressure greater than the osmotic pressure to force solvent back through the membrane; used for desalination of sea water.

Q11. Why is salt spread on icy roads?
A11. It depresses the freezing point of water, melting the ice.

Q12. Give the molar-mass formula from freezing-point depression.
A12. M₂ = (Kf·w₂·1000)/(ΔTf·w₁), where w₂ is solute mass and w₁ is solvent mass in grams.

Why we salt icy roads and freeze ice cream with rock salt

Spreading salt on winter roads stops ice forming, and rock salt around an ice-cream churn drops the temperature below 0 °C. Both tricks use colligative properties — properties that depend only on the number of solute particles, not their nature.

Core idea: A non-volatile solute changes four measurable properties of the solvent — vapour pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure — and each change is proportional to the amount of solute, letting us "count" particles and thus find molar masses.

Deep explanation

Beginner: the four colligative properties

1. Relative lowering of vapour pressure.
2. Elevation of boiling point (ΔT_b): solute raises boiling point.
3. Depression of freezing point (ΔT_f): solute lowers freezing point.
4. Osmotic pressure (π): pressure needed to stop solvent flowing across a semipermeable membrane.

Intermediate: the working equations

• ΔT_b = K_b · m, where K_b is the molal elevation (ebullioscopic) constant.
• ΔT_f = K_f · m, where K_f is the molal depression (cryoscopic) constant.
• π = C R T, where C is molar concentration (mol/L), R the gas constant, T the temperature in kelvin.

Each K depends only on the solvent (K_b for water = 0.52 K·kg/mol; K_f for water = 1.86 K·kg/mol).

Advanced: why osmotic pressure is the method of choice for macromolecules

Osmotic pressure is measured at room temperature (no boiling/freezing that could damage proteins), and even a tiny molar concentration gives a measurable π because R·T is large. So for polymers and proteins with huge molar masses, osmotic pressure is the preferred colligative method. Solutions with equal osmotic pressure are isotonic (e.g. 0.9% NaCl with blood cells). A hypertonic solution shrinks cells (plasmolysis); a hypotonic one swells and bursts them. Reverse osmosis — applying pressure greater than π — pushes solvent backwards through the membrane and is the basis of desalination.

Worked example

1.00 g of a non-volatile, non-electrolyte solute dissolved in 50 g of benzene lowers the freezing point by 0.40 K. K_f for benzene = 5.12 K·kg/mol. Find the molar mass of the solute.

Molality m = ΔT_f / K_f = 0.40 / 5.12 = 0.0781 mol/kg.
Moles of solute in 50 g (0.05 kg) benzene = 0.0781 × 0.05 = 3.906 × 10⁻³ mol.
Molar mass = mass / moles = 1.00 / 3.906 × 10⁻³ = ≈ 256 g/mol.

Real-world / exam application

Antifreeze (ethylene glycol) in radiators lowers the coolant's freezing point and raises its boiling point; salt melts road ice; intravenous fluids are made isotonic with blood; desalination plants use reverse osmosis; food is preserved in brine/sugar by osmosis pulling water out of microbes. Numericals asking for K_b, K_f, ΔT, or molar mass are guaranteed exam material.

Exam tricks & shortcuts

• Molar mass M₂ = (K · w₂ × 1000)/(ΔT × w₁) — memorise this combined form (w₂ = solute mass g, w₁ = solvent mass g).
• Osmotic pressure is best for high molar masses and heat-sensitive solutes.
• Mnemonic — "BE FRee OSmosis": Boiling Elevation, FReezing depression, OSmotic pressure — the countable four.

Key formulas

• Relative lowering of VP: (p° − p)/p° = x_solute.
• Elevation of boiling point: ΔT_b = K_b·m.
• Depression of freezing point: ΔT_f = K_f·m.
• Osmotic pressure: π = CRT = (n/V)RT.
• Molar mass: M = (w_B·K_b)/(ΔT_b·w_A) [and analogous forms from each property]; from osmosis M = w_B RT/(πV).

Colligative properties let us determine molar mass; osmotic pressure is best for macromolecules (large, measurable π).

Worked Example

Problem: When 1.8 g of a non-volatile, non-electrolyte solute is dissolved in 100 g of water, the freezing point of the water is lowered by 0.186°C. Find the molar mass of the solute. (K_f for water = 1.86 K·kg/mol.)

Solution:
Step 1 — Relate the freezing-point depression to molality:
ΔT_f = K_f × m ⟹ m = ΔT_f / K_f.

Step 2 — Substitute ΔT_f = 0.186 and K_f = 1.86:
m = 0.186 / 1.86 = 0.1 mol/kg.

Step 3 — Find the moles of solute. Molality = moles of solute per kg of solvent, and the solvent is 100 g = 0.1 kg:
moles = m × (kg of solvent) = 0.1 × 0.1 = 0.01 mol.

Step 4 — Compute the molar mass:
M = mass / moles = 1.8 / 0.01 = 180 g/mol.

Answer: The molar mass of the solute is 180 g/mol.