Grade 11-12 (age 16-18)

high school Properties of Solutions

Solution Properties

Dissolve salt or sugar in water and the solution boils at a higher temperature and freezes at a lower one than pure water. Dissolved solute particles get in the way of solvent molecules evaporating and crystallizing, and properties like these that depend only on the number of particles, not on what they are, are called colligative properties. Boiling-point elevation and freezing-point depression are proportional to molality, and the osmotic pressure across a semipermeable membrane is proportional to molarity and absolute temperature. Electrolytes split into ions, so their effect grows with the extra particles (the van 't Hoff factor i). Raising the molality in the diagrams makes both temperature changes grow together on the same scale and widens the difference in level across the membrane.

What Changes When a Solute Dissolves?

💡 Adding Salt to Water

①At 1 atm, pure water freezes at 0 °C and boils at 100 °C

②Salt water does not freeze at 0 °C and does not boil at 100 °C

③Dissolved solute particles take up part of the surface, making it harder for water molecules to evaporate, so a higher temperature is needed to boil

④Solute particles also get in the way of water molecules lining up into ice crystals, so a lower temperature is needed to freeze

⑤The effect depends on how many particles are dissolved, not on what they are: a colligative property

Boiling-Point Elevation and Freezing-Point Depression

Molality m1.0 mol/kg

Boiling-Point Elevation

ΔT_b = K_b × m

K_b: molal elevation constant (water: 0.52 °C·kg/mol)

Freezing-Point Depression

ΔT_f = K_f × m

K_f: molal depression constant (water: 1.86 °C·kg/mol)

🔍 What the Diagram Shows

①Raising the concentration raises the boiling point and lowers the freezing point, both in direct proportion to molality

②Water's K_f (1.86) is about 3.6 times its K_b (0.52), so on the same scale the freezing-point bar is 3.6 times longer

③Spreading calcium chloride on roads in winter uses freezing-point depression to melt snow and ice

④The osmotic-pressure panel takes 25 °C and treats molarity as equal to molality

Osmotic Pressure

π = MRT

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

💡 How Osmotic Pressure Works

①A semipermeable membrane lets water molecules through but blocks solute particles

②Water moves from the dilute side to the concentrated side, raising the level on the concentrated side: this is osmosis

③Stopping that flow takes extra pressure on the concentrated side, and the least pressure that does it is the osmotic pressure

④Plant roots taking up water and IV fluids being matched to the osmotic pressure of blood both rest on this principle

⑤π = MRT has the same form as the ideal gas equation PV = nRT

Electrolytes and the van 't Hoff Factor

van 't Hoff Factor

ΔT = i × K × m

i = actual number of particles in solution / moles of solute dissolved

💡 Why Electrolytes Have a Bigger Effect

①NaCl splits into two ions, Na⁺ and Cl⁻, so i ≈ 2

②CaCl₂ splits into one Ca²⁺ and two Cl⁻, so i ≈ 3

③Glucose (C₆H₁₂O₆) does not split into ions; it is a nonelectrolyte with i = 1

④So at the same molality, an electrolyte's colligative effect is i times larger

⑤That is also why CaCl₂ is widely used for de-icing: mole for mole it releases more particles than NaCl

Worked Examples

Example 1

A nonelectrolyte is dissolved in water to make a 2 mol/kg solution. What is its freezing point? (K_f of water = 1.86 °C·kg/mol)

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

ΔT_f = 1.86 × 2 = 3.72 °C

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

freezing point = 0 − 3.72 = −3.72 °C

▸ −3.72 °C

It is a nonelectrolyte, so i = 1. Set the concentration to 2.0 in the diagram and the freezing-point bar shows the same value.

Example 2

NaCl is dissolved in water to make a 1 mol/kg solution. What is its boiling point? (K_b = 0.52 °C·kg/mol, i = 2 for NaCl)

It is an electrolyte, so multiply by the van 't Hoff factor: ΔT_b = i × K_b × m.

ΔT_b = 2 × 0.52 × 1 = 1.04 °C

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

boiling point = 100 + 1.04 = 101.04 °C

▸ 101.04 °C

NaCl splits into two ions, Na⁺ and Cl⁻, doubling the number of particles, so the boiling point rises twice as much as for a nonelectrolyte at the same concentration.

Summary

Colligative Properties

boiling-point elevation, freezing-point depression, osmotic pressure, vapor-pressure lowering

Four properties proportional to the number of solute particles, not their kind

CSAT-style problem, adapted

What is the osmotic pressure of a 0.1 M glucose solution at 27 °C? (R = 0.0821 L·atm/(mol·K))

①1.23 atm

②2.46 atm

③0.82 atm

④24.6 atm

⑤8.21 atm

▸ ② 2.46 atm

Use π = MRT and convert the temperature to kelvins.

π = MRT, T = 27 + 273 = 300 K

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

π = 0.1 × 0.0821 × 300 ≈ 2.46 atm

🎯 Exam Points

①Colligative properties depend only on the number of solute particles, not their kind

②ΔT_b = K_b · m and ΔT_f = K_f · m, with K_b = 0.52 and K_f = 1.86 for water

③Electrolytes multiply the effect by i, according to how many ions they form

④Osmotic pressure π = MRT has the same form as the ideal gas equation

⑤Molality (mol of solute per kg of solvent) does not change with temperature

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Last updated 2026-10-17

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