CHEMISTRY · LESSON 09 OF 12

Properties of solutions

A solution’s composition is described by concentration units (molarity, molality, mass percent, mole fraction). Dilution conserves solute. Dissolved particles lower vapour pressure and freezing point and raise boiling point and osmotic pressure.

What this lesson explains

Chemical engineers prepare, mix and separate solutions constantly: making reagents to a specified concentration, diluting stock solutions, predicting how much gas dissolves in a liquid under pressure, protecting engines with antifreeze, and understanding osmosis in membranes and water purification. Choosing the right concentration unit for the job avoids serious errors.

Before you begin

Moles and molar mass

n = m/M; M(NaCl) = 58.44 g/mol, M(glucose, C₆H₁₂O₆) = 180.16 g/mol.

Intermolecular attractions

Hydrogen bonding, dipole–dipole and dispersion forces decide which substances mix (see Intermolecular forces, phases and solubility).

The idea, made visible

A solution is a homogeneous mixture: a solute dissolved in a solvent. Whether something dissolves depends on the balance between the attractions broken (solute–solute, solvent–solvent) and those formed (solute–solvent): “like dissolves like” — polar and ionic substances dissolve in polar solvents such as water; non-polar substances in non-polar solvents.

A saturated solution holds the maximum amount of solute at that temperature and is in equilibrium with undissolved solid. Stirring speeds up dissolving but does not change how much can dissolve. Most solids become more soluble in hot water (not all); gases become less soluble as temperature rises, but more soluble at higher partial pressure (Henry’s law) — the reason carbonated drinks fizz when opened.

Diluting 0.20 L of 1.5 M solution to 0.60 L: the 0.30 mol of solute spreads through three times the volume, so the concentration drops to one third, 0.50 M.

Concentration can be expressed in several ways. Molarity (mol per litre of solution) is convenient for measuring volumes in the lab, but changes slightly with temperature because volume does. Molality (mol per kilogram of solvent) does not change with temperature and is used for freezing- and boiling-point calculations. Mass percent and mole fraction are also common.

When a solution is diluted, solvent is added but the moles of solute do not change: c₁V₁ = c₂V₂.

Colligative properties depend on the number of dissolved particles, not their identity. A non-volatile solute lowers the solvent’s vapour pressure, so the solution boils at a higher temperature and freezes at a lower one; it also creates osmotic pressure across a semipermeable membrane. Ionic solutes give more particles: NaCl gives about 2 per formula unit (the van ’t Hoff factor i ≈ 2), CaCl₂ about 3, glucose 1. Real values of i are slightly lower than ideal because ions interact.

Freezing-point depression of water for different solutes (m = 0.10 mol/kg, ideal i)
solute	particles per formula	i	ΔT_f = iK_fm (°C)	freezing point (°C)
glucose C₆H₁₂O₆	1 molecule	1	0.186	−0.186
NaCl	Na⁺ + Cl⁻	2	0.372	−0.372
CaCl₂	Ca²⁺ + 2Cl⁻	3	0.558	−0.558

Measured depressions for NaCl and CaCl₂ are slightly smaller than these ideal values, because oppositely charged ions associate in solution.

Key terms

Molarity c (M): Moles of solute per litre of solution: c = n/V.
Molality m: Moles of solute per kilogram of solvent: m = n_solute/m_solvent (mol/kg).
Mass percent: mass of solute / mass of solution × 100%.
Mole fraction χ: χ_A = n_A/n_total; the mole fractions of all components add to 1.
Saturated solution: A solution in equilibrium with undissolved solute; its concentration is the solubility at that temperature.
Colligative property: A property depending on the number of solute particles: vapour-pressure lowering, boiling-point elevation, freezing-point depression, osmotic pressure.
van ’t Hoff factor i: Number of particles produced per formula unit dissolved (ideal: glucose 1, NaCl 2, CaCl₂ 3).

The formulas and what they mean

Symbols, meanings and units
Symbol	Meaning	Unit
c or M	molarity	mol/L
m	molality (not mass!)	mol/kg solvent
χ	mole fraction	—
K_f, K_b	freezing and boiling constants (water: 1.86 and 0.512)	°C·kg/mol
Π	osmotic pressure	atm (or Pa)
k_H	Henry’s law constant	mol/(L·atm)
R	gas constant, 0.08206	L·atm/(mol·K)

Molarity and dilution

c = n / V; c₁V₁ = c₂V₂

Conditions and limits: V is the volume of solution. Dilution: only solvent added, no reaction.

Molality

m = n_solute / mass of solvent (kg)

Conditions and limits: Divide by the solvent’s mass, not the solution’s.

Henry’s law

c_gas = k_HP_gas

Conditions and limits: Dilute gas solubility at constant temperature; P is the partial pressure of that gas. Some books define the constant the other way round (P = k c); check units.

Raoult’s law

P_solvent = χ_solventP°_solvent

Conditions and limits: Ideal solution; non-volatile solute.

Boiling and freezing points

ΔT_b = iK_bm; ΔT_f = iK_fm

Conditions and limits: Dilute solutions; m is molality; ΔT is the size of the change (boiling point rises, freezing point falls).

Osmotic pressure

Π = iMRT

Conditions and limits: Dilute solutions; M molarity, T in kelvin, R = 0.08206 L·atm/(mol·K).

Why it works: Why dilution obeys c₁V₁ = c₂V₂

Take V₁ litres of a solution of molarity c₁ and add water to a total volume V₂.

Moles of solute before: n = c₁V₁.
Definition of molarity.
Adding solvent adds no solute: n is unchanged.
Conservation of the solute.
After: n = c₂V₂. So c₁V₁ = c₂V₂.
Equate the two expressions for n.

V₁ and V₂ may be in any volume unit, as long as it is the same on both sides.

A first worked example

Solution calculations

Identify which concentration unit the problem needs (molarity for volumes and reactions; molality for ΔT_b, ΔT_f; mole fraction for Raoult).
Convert masses to moles and volumes to litres (or solvent masses to kg).
For dilution use c₁V₁ = c₂V₂; for colligative properties decide i from the formula of the solute.
Substitute, keeping units, and check whether the answer is a change (ΔT) or a final value (T).

Molarity

Problem. Find the molarity of 0.30 mol of solute in 0.50 L of solution, and of 5.85 g NaCl in 250 mL of solution.

c = 0.30 / 0.50 = 0.60 M.
c = n/V.
n(NaCl) = 5.85 / 58.44 = 0.100 mol; c = 0.100 / 0.250 = 0.400 M.
Convert grams to moles and mL to L.

Result: 0.60 M and 0.400 M.

What it means: To prepare the NaCl solution, dissolve 5.85 g and make up to 250 mL in a volumetric flask (not add 250 mL of water).

A different case

Dilution

Problem. A 0.20 L sample of 1.5 M solution is diluted to 0.60 L. Find the final concentration.

c₂ = c₁V₁ / V₂ = 1.5 × 0.20 / 0.60 = 0.50 M.
Moles of solute are conserved.

Result: 0.50 M.

What it means: Tripling the volume divides the concentration by three.

More worked cases

Each case below uses a different skill. Every step and result is shown.

Preparing a dilute solution from a stock

Problem. What volume of 12.0 M HCl is needed to make 500 mL of 0.600 M HCl?

V₁ = c₂V₂ / c₁ = 0.600 × 500 mL / 12.0 = 25.0 mL.
Same unit (mL) on both sides.
Add the 25.0 mL of acid to water, then make up to 500 mL.
Safety: always add acid to water.

Result: 25.0 mL of the stock solution.

What it means: Diluting concentrated acid releases heat; adding it slowly to water avoids splattering.

Antifreeze: freezing-point depression

Problem. Dissolve 0.10 mol glucose in 0.50 kg of water (K_f = 1.86 °C·kg/mol). Find the freezing point.

m = 0.10 mol / 0.50 kg = 0.20 mol/kg.
Molality: divide by the solvent mass.
ΔT_f = iK_fm = 1 × 1.86 × 0.20 = 0.372 °C.
Glucose does not ionise: i = 1.
T_f = 0 − 0.372 = −0.372 °C.
The freezing point is lowered from 0 °C.

Result: Freezing point ≈ −0.37 °C.

What it means: Using 0.10 mol NaCl instead would give about −0.74 °C (i ≈ 2).

Henry’s law

Problem. At 25 °C, k_H for CO₂ in water is 3.4 × 10⁻² mol/(L·atm). Find the CO₂ solubility at a CO₂ partial pressure of 2.5 atm (a sealed soft drink), and at 4.0 × 10⁻⁴ atm (open air).

Sealed: c = 3.4 × 10⁻² × 2.5 = 0.085 M.
c = k_H P.
Open air: c = 3.4 × 10⁻² × 4.0 × 10⁻⁴ = 1.4 × 10⁻⁵ M.
Much lower partial pressure.

Result: 0.085 M sealed; 1.4 × 10⁻⁵ M in open air.

What it means: Opening the bottle lowers the CO₂ pressure, so gas comes out of solution as bubbles.

Osmotic pressure

Problem. Find the osmotic pressure of 0.10 M glucose at 25 °C.

Π = iMRT = 1 × 0.10 × 0.08206 × 298 = 2.45 atm.
T in kelvin: 25 + 273 = 298 K.

Result: Π ≈ 2.4 atm.

What it means: Even dilute solutions produce large osmotic pressures; reverse osmosis must exceed them to push water out.

Common misunderstandings

Misunderstanding: Confusing molality with molarity.
Correct idea: Molality divides by kg of solvent; molarity by litres of solution.
Misunderstanding: Dividing by the volume of water added rather than the final solution volume.
Correct idea: Molarity uses the total solution volume.
Misunderstanding: Forgetting i for ionic solutes.
Correct idea: NaCl gives about twice, CaCl₂ about three times, the effect of a non-electrolyte.
Misunderstanding: Reporting ΔT_f as the freezing point.
Correct idea: The freezing point is 0 °C − ΔT_f for water.
Misunderstanding: “Stirring increases solubility.”
Correct idea: It speeds dissolving but does not change the equilibrium amount.

Keep in mind

c = n / V; c₁V₁ = c₂V₂Molarity and dilution
m = n_solute / mass of solvent (kg)Molality
c_gas = k_HP_gasHenry’s law
P_solvent = χ_solventP°_solventRaoult’s law
ΔT_b = iK_bm; ΔT_f = iK_fmBoiling and freezing points
Π = iMRTOsmotic pressure

Scope of this lesson

Colligative formulas are for dilute, near-ideal solutions; concentrated and non-ideal solutions need activity coefficients.
Solubility equilibria (K_sp) are not covered in these lessons; check your outline.

Next: Properties of gases. An ideal gas obeys PV = nRT with T in kelvin. The simple gas laws, gas densities, mixtures (Dalton) and gas stoichiometry all follow, and kinetic-molecular theory explains why.

Original study text. Sources and credits.

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