Solutions, Colloids, Suspensions, and Colligative Properties: Molality, Freezing Point Lowering, and Boiling Point Elevation

Solutions, Colloids, and Suspensions

Types of Mixtures

Mixtures in chemistry can be classified based on the size and behavior of their particles. The three main types are solutions, colloids, and suspensions.

• Solution: A homogeneous mixture where solute particles are small (atoms, ions, or small molecules) and uniformly dispersed in the solvent. Solutions appear transparent and cannot be separated by filters or semipermeable membranes.

• Colloid: A homogeneous mixture with larger particles (such as proteins or groups of molecules/ions) than those in solutions. Colloidal particles pass through filters but not semipermeable membranes and do not settle out.

• Suspension: A heterogeneous, nonuniform mixture with very large particles that can often be seen with the naked eye. Suspended particles settle out rapidly and can be separated by filters and semipermeable membranes.

Properties of Solutions

• Solute particles are small and uniformly dispersed.

• Solutions are transparent and homogeneous.

• Particles pass through filters and semipermeable membranes.

• Semipermeable membrane: Allows solvent and very small solute particles to pass, but blocks large molecules.

Properties of Colloids

• Colloidal particles are much larger than solute particles in solutions.

• Colloids are homogeneous and do not settle out.

• Particles pass through filters but not semipermeable membranes.

Table: Examples of Colloids

Colloid	Substance Dispersed	Dispersing Medium
Fog, clouds, hair sprays	Liquid	Gas
Dust, smoke	Solid	Gas
Shaving cream, whipped cream, soapsuds	Gas	Liquid
Styrofoam, marshmallows	Gas	Solid
Mayonnaise, homogenized milk	Liquid	Liquid
Cheese, butter	Liquid	Solid
Blood plasma, paints (latex), gelatin	Solid	Liquid

Properties of Suspensions

• Suspensions are heterogeneous and nonuniform.

• Particles are large and visible; they settle out after mixing.

• Particles are trapped by filters and semipermeable membranes.

• Examples: Muddy water, some medications (Kaopectate, calamine lotion, antacid mixtures, liquid penicillin).

• Water-treatment plants use suspensions to purify water by forming and filtering out floc (large suspended particles).

Table: Comparison of Solutions, Colloids, and Suspensions

Type of Mixture	Type of Particle	Settling	Separation
Solution	Small particles (atoms, ions, small molecules)	Particles do not settle	Cannot be separated by filters or semipermeable membranes
Colloid	Larger molecules or groups of molecules/ions	Particles do not settle	Can be separated by semipermeable membranes but not by filters
Suspension	Very large particles (may be visible)	Particles settle rapidly	Can be separated by filters

Colligative Properties: Freezing Point Lowering and Boiling Point Elevation

Introduction to Colligative Properties

Colligative properties are physical properties of solutions that depend only on the concentration of solute particles, not their identity. Adding a solute to a solvent (such as water) changes its vapor pressure, lowers its freezing point, and raises its boiling point.

• Freezing point lowering: The freezing point of a solution is lower than that of the pure solvent.

• Boiling point elevation: The boiling point of a solution is higher than that of the pure solvent.

Example: Ethylene Glycol (Antifreeze)

• Ethylene glycol (HO—CH2—CH2—OH) is an organic compound with two —OH groups, making it highly soluble in water.

• Used in car radiators to lower freezing point and raise boiling point, preventing freezing in cold weather and boiling in hot weather.

• A 50–50% by mass mixture of ethylene glycol and water freezes at about –30 °F and boils at 225 °F.

Particles in Solution: Electrolytes and Nonelectrolytes

• Nonelectrolytes dissolve as molecules (e.g., glucose, C6H12O6): 1 mole of C6H12O6(l) = 1 mole of C6H12O6(aq)

• Strong electrolytes dissolve as ions (e.g., NaCl, CaCl2): 1 mole of NaCl(s) = 2 moles of particles (Na+(aq) and Cl–(aq)) 1 mole of CaCl2(s) = 3 moles of particles (Ca2+(aq) and 2 Cl–(aq))

Molality and Calculations

Definition of Molality (m)

Molality is a unit of concentration used in colligative property calculations. It is defined as:

• Formula: $\text{Molality}~(m) = \frac{\text{moles of solute particles}}{\text{kilograms of solvent}}$

• Molality is preferred over molarity for colligative properties because it does not change with temperature.

Sample Problem 1: Calculating Molality

• Given: 35.5 g of glucose (C6H12O6), 0.400 kg of water

• Step 1: Calculate moles of glucose: $\text{Moles of glucose} = \frac{35.5~\text{g}}{180.16~\text{g/mol}} = 0.197~\text{mol}$

• Step 2: Calculate molality: $m = \frac{0.197~\text{mol}}{0.400~\text{kg}} = 0.493~m$

Colligative Property Calculations

Freezing Point Lowering

• The decrease in freezing point ($\Delta T_f$) is calculated using:

$\Delta T_f = m \times K_f$

• $K_f$ (freezing point constant) for water = 1.86 °C/m

• New freezing point: $T_{\text{solution}} = T_{\text{water}} - \Delta T_f$

Boiling Point Elevation

• The increase in boiling point ($\Delta T_b$) is calculated using:

$\Delta T_b = m \times K_b$

• $K_b$ (boiling point constant) for water = 0.51 °C/m

• New boiling point: $T_{\text{solution}} = T_{\text{water}} + \Delta T_b$

Table: Effect of Solute Concentration on Freezing and Boiling Points

Substance/kg water	Type of Solute	Molality of Particles	Freezing Point	Boiling Point
Pure water	None	0	0.00 °C	100.00 °C
1 mol of C6H12O6	Nonelectrolyte	1 m	–1.86 °C	100.51 °C
1 mol of NaCl	Strong electrolyte	2 m	–3.72 °C	101.02 °C
1 mol of CaCl2	Strong electrolyte	3 m	–5.58 °C	101.53 °C

Sample Problem 2: Freezing Point of a Solution

• Given: 225 g CaCl2 in 500 g water

• Step 1: Calculate moles of CaCl2: $\text{Moles of CaCl}_2 = \frac{225~\text{g}}{110.98~\text{g/mol}} = 2.03~\text{mol}$

• Each mole of CaCl2 produces 3 moles of particles: $2.03~\text{mol} \times 3 = 6.08~\text{mol particles}$

• Step 2: Calculate molality: $m = \frac{6.08~\text{mol}}{0.500~\text{kg}} = 12.2~m$

• Step 3: Calculate freezing point lowering: $\Delta T_f = 12.2~m \times 1.86~\frac{^\circ\text{C}}{m} = 22.7~^\circ\text{C}$

• New freezing point: $T_{\text{solution}} = 0.00~^\circ\text{C} - 22.7~^\circ\text{C} = -22.7~^\circ\text{C}$

Sample Problem 3: Boiling Point of a Solution

• Given: 4.6 mol propylene glycol in 1.55 kg water

• Step 1: Calculate molality: $m = \frac{4.6~\text{mol}}{1.55~\text{kg}} = 3.0~m$

• Step 2: Calculate boiling point elevation: $\Delta T_b = 3.0~m \times 0.51~\frac{^\circ\text{C}}{m} = 1.5~^\circ\text{C}$

• New boiling point: $T_{\text{solution}} = 100.0~^\circ\text{C} + 1.5~^\circ\text{C} = 101.5~^\circ\text{C}$

Summary Review

• Colloids contain particles that pass through most filters but do not settle out or pass through semipermeable membranes.

• Suspensions have very large particles that settle out.

• Particles in a solution lower vapor pressure, freezing point, and raise boiling point.

• Molality is the moles of solute per kilogram of solvent, usually water.