What this chapter is about
This chapter introduces the quantitative study of solutions — homogeneous mixtures where one or more substances (solutes) dissolve uniformly in another substance (solvent). You learn how to express the amount of solute present using different concentration units, and why the choice of unit matters in different contexts such as laboratory work, industrial processes and medical applications.
A major focus is on the colligative properties of dilute solutions: lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure. These properties depend only on the number of solute particles, not on what those particles are. Understanding them lets you calculate molar masses experimentally, explain why salt is spread on icy roads, and see how plants draw water from soil.
By the end of the chapter you should be able to convert between concentration units, apply Raoult's law to ideal and non-ideal solutions, predict deviations, and solve problems on colligative properties for both non-electrolytes and electrolytes using the van 't Hoff factor.
Key ideas
- A solution is homogeneous at the molecular level; the solute cannot be separated by filtration or settling.
- Concentration can be expressed as molarity (mol/L), molality (mol/kg solvent), mole fraction, mass percentage, or parts per million (ppm).
- Raoult's law states that the partial vapour pressure of each volatile component equals its mole fraction in solution multiplied by its pure-component vapour pressure: p₁ = x₁ × p₁°.
- An ideal solution obeys Raoult's law at all compositions; mixing produces no enthalpy change and no volume change. Real solutions show positive or negative deviations.
- Colligative properties — relative lowering of vapour pressure, boiling-point elevation, freezing-point depression, osmotic pressure — depend on the number of solute particles, not their identity.
- For electrolytes, the van 't Hoff factor i accounts for dissociation or association: i = (observed colligative property) / (calculated for non-electrolyte).
- Osmotic pressure π = iCRT, where C is molarity in mol/L, R = 0.0821 L atm K⁻¹ mol⁻¹, and T is temperature in kelvin.
Formulas and facts to remember
- Quantity: Molarity (M) · Formula: M = n(solute) / V(solution in L) · Meaning: Moles of solute per litre of solution
- Quantity: Molality (m) · Formula: m = n(solute) / mass of solvent in kg · Meaning: Moles of solute per kilogram of solvent; independent of temperature
- Quantity: Mole fraction · Formula: x₁ = n₁ / (n₁ + n₂) · Meaning: Fraction of total moles that is component 1
- Quantity: Raoult's law · Formula: p₁ = x₁ × p₁° · Meaning: Partial pressure equals mole fraction times pure vapour pressure
- Quantity: Relative lowering of vapour pressure · Formula: (p° − p) / p° = x₂ · Meaning: Equals mole fraction of non-volatile solute
- Quantity: Boiling-point elevation · Formula: ΔT_b = i × K_b × m · Meaning: K_b is molal elevation constant (K kg mol⁻¹)
- Quantity: Freezing-point depression · Formula: ΔT_f = i × K_f × m · Meaning: K_f is molal depression constant
- Quantity: Osmotic pressure · Formula: π = iCRT · Meaning: C in mol/L, R = 0.0821 L atm K⁻¹ mol⁻¹, T in K
- Quantity: van 't Hoff factor · Formula: i = (observed property) / (expected for non-electrolyte) · Meaning: i > 1 for dissociation, i < 1 for association
Worked examples
Example 1 — Converting molality to mole fraction
A solution contains 5.85 g of sodium chloride (NaCl, molar mass 58.5 g/mol) dissolved in 500 g of water. Find the mole fraction of NaCl assuming complete dissociation is not considered at this stage.
Step 1: Moles of NaCl = 5.85 / 58.5 = 0.100 mol.
Step 2: Moles of water = 500 / 18 = 27.78 mol.
Step 3: Mole fraction of NaCl = 0.100 / (0.100 + 27.78) = 0.100 / 27.88 ≈ 0.00359.
Example 2 — Boiling-point elevation
6.0 g of urea (molar mass 60 g/mol, non-electrolyte) is dissolved in 250 g of water. K_b for water is 0.52 K kg mol⁻¹. Find the boiling point of the solution.
Step 1: Moles of urea = 6.0 / 60 = 0.10 mol.
Step 2: Molality = 0.10 / 0.250 = 0.40 mol/kg.
Step 3: ΔT_b = K_b × m = 0.52 × 0.40 = 0.208 K ≈ 0.21 K.
Step 4: Boiling point = 100 + 0.21 = 100.21 °C.
Example 3 — Osmotic pressure of an electrolyte
Calculate the osmotic pressure at 300 K of a 0.05 M aqueous solution of MgCl₂, assuming complete dissociation. R = 0.0821 L atm K⁻¹ mol⁻¹.
Step 1: MgCl₂ → Mg²⁺ + 2 Cl⁻ gives 3 ions per formula unit, so i = 3.
Step 2: π = iCRT = 3 × 0.05 × 0.0821 × 300 = 3.69 atm ≈ 3.7 atm.
Common mistakes
- Confusing molarity and molality → Molarity uses solution volume; molality uses solvent mass — they differ especially for concentrated solutions.
- Forgetting to use the van 't Hoff factor for salts → Strong electrolytes produce more particles than formula units; always multiply by i.
- Using ΔT in Celsius where Kelvin is needed → For the elevation and depression formulae the change is the same numerically, but osmotic-pressure calculations need absolute temperature in kelvin.
- Applying Raoult's law to highly non-ideal mixtures without noting deviations → Mixtures of ethanol–water or acetone–chloroform show significant deviations; ideal-solution equations give only approximate answers.
- Taking mole fraction of solute as moles solute divided by moles solvent alone → Mole fraction uses total moles (solute + solvent) in the denominator.
Quick revision
- Molality is temperature-independent; molarity changes with temperature because volume changes.
- Raoult's law: partial pressure = mole fraction × pure vapour pressure.
- Colligative properties depend on particle count, not particle identity.
- Boiling point rises and freezing point falls when a non-volatile solute is added.
- π = iCRT — osmotic pressure is the most sensitive colligative property for molar-mass determination.
- van 't Hoff factor i corrects for dissociation (i > 1) or association (i < 1).