Properties of Solutions

Solution Properties

Dissolving a solute hinders solvent escape, so the boiling point rises and the freezing point falls—these colligative properties. Boiling-point elevation and freezing-point depression scale with molality, and osmotic pressure scales with molarity and absolute temperature. Electrolytes split into ions, so the van ’t Hoff factor i multiplies the effect by the number of particles. Here you change molality to watch boiling and freezing shifts and the osmotic membrane setup.

What Happens When You Add a Solute?

💡 Analogy: Adding Salt to Water

①Pure water freezes at 0°C, boils at 100°C

②With salt: doesn't freeze at 0°C, doesn't boil at 100°C

③Solute particles block solvent molecules from 'escaping' (evaporation)

④Boiling needs higher T, freezing needs lower T

⑤This is colligative property — proportional to particle count, not type

Boiling-Point Elevation & Freezing-Point Depression

Molality m1 m

Boiling-Point Elevation

ΔT_b = K_b × m

K_b: ebullioscopic constant (water: 0.52 °C/m)

Freezing-Point Depression

ΔT_f = K_f × m

K_f: cryoscopic constant (water: 1.86 °C/m)

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①Higher concentration: boiling↑, freezing↓

②Water K_f (1.86) ≫ K_b (0.52)

③Freezing-point depression ~3.6× larger effect

④Salt (CaCl₂) on icy roads: depresses freezing point

Osmotic Pressure

π = MRT

M: molarity, R: gas constant, T: absolute temperature (K)

💡 Principle of Osmosis

①Semipermeable membrane: solvent passes, solute does not

②Solvent flows from dilute → concentrated side (osmosis)

③Min pressure to prevent osmosis = osmotic pressure

④Plant roots absorb water; isotonic IV solutions

⑤π = MRT — same form as ideal gas PV = nRT!

Electrolytes and van't Hoff Factor

van't Hoff Factor

ΔT = i × K × m

i = (actual particle count) / (moles of solute)

💡 Why Electrolytes Are Stronger?

①NaCl → Na⁺ + Cl⁻ (2 ions) → i ≈ 2

②CaCl₂ → Ca²⁺ + 2Cl⁻ (3 ions) → i ≈ 3

③Glucose is non-electrolyte → i = 1

④Same m: electrolyte effect is i× larger

⑤That is why CaCl₂ is more effective than NaCl as a road de-icer!

Worked Examples

Example 1

A 2 m solution is made by dissolving a non-electrolyte in water. What is its freezing point? (water K_f = 1.86 °C/m)

Find the freezing-point depression ΔT_f = K_f × m.

ΔT_f = K_f × m = 1.86 × 2 = 3.72°C

It drops by ΔT_f from the 0°C freezing point of pure water.

freezing point = 0 − 3.72 = −3.72°C

▸ −3.72°C

Freezing-point depression is proportional to the number of solute particles (molality). For a non-electrolyte, i = 1.

Example 2

A 1 m NaCl solution in water. What is its boiling point? (K_b = 0.52, i = 2 for NaCl)

For an electrolyte, include the van’t Hoff factor: ΔT_b = i × K_b × m.

ΔT_b = 2 × 0.52 × 1 = 1.04°C

Add ΔT_b to the 100°C boiling point of pure water.

boiling point = 100 + 1.04 = 101.04°C

▸ 101.04°C

NaCl → Na⁺ + Cl⁻ gives 2 ions (i=2), so the effect is twice that of a non-electrolyte at the same m.

Summary

Colligative Properties Core

BP elevation, FP depression, osmotic pressure, vapor-pressure lowering

Four properties proportional to particle count, not solute type

CSAT-style

What is the osmotic pressure of a 0.1 M glucose solution at 27°C? (R = 0.0821)

①1.23 atm

②2.46 atm

③0.82 atm

④24.6 atm

⑤8.21 atm

▸ ② 2.46 atm

Use the osmotic pressure formula π = MRT. T is absolute temperature.

π = MRT, T = 27 + 273 = 300 K

Substitute M = 0.1, R = 0.0821, T = 300.

π = 0.1 × 0.0821 × 300 ≈ 2.46 atm

🎯 Exam Points

①Colligative = depends on particle count (not type)

②ΔTb = Kb·m, ΔTf = Kf·m — memorize K

③Electrolytes (NaCl → 2 particles) effect × i

⑤Osmotic π = MRT (similar to ideal gas)

⑤Molality m = mol solute / kg solvent — independent of T!

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Gas Laws

Reaction Enthalpy

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