Chapter 13 & 14 Study Guide: Solids, Solutions, and Colligative Properties

Chapter 13: Crystalline Solids and Unit Cell Calculations

Unit Cell Calculations

Unit cells are the smallest repeating units in a crystalline solid. Calculations involving unit cells are essential for understanding the structure and properties of solids.

• Mass of One Particle: The mass of a single particle (atom, ion, or molecule) in a unit cell is calculated by dividing the molar mass by Avogadro’s number. $\text{Mass of one particle} = \frac{\text{Molar mass}}{N_A}$ where $N_A$ is Avogadro’s number ($6.022 \times 10^{23}$ mol-1).

• Mass of a Unit Cell: Multiply the mass of one particle by the number of particles in the unit cell. Body-centered cubic (BCC): 2 particles per unit cell Face-centered cubic (FCC): 4 particles per unit cell $\text{Mass of unit cell} = \text{Mass of one particle} \times \text{Number of particles}$

• Length of Unit Cell: The length ($l$) of a unit cell can be calculated using geometric relationships based on the type of unit cell (BCC or FCC). Body-centered cubic: $l = \frac{4r}{\sqrt{3}}$ Face-centered cubic: $l = 2\sqrt{2}r$ where $r$ is the radius of the atom.

• Density Calculation: Density ($D$) is calculated as mass divided by volume. $D = \frac{m}{v}$ where $m$ is mass and $v$ is volume of the unit cell.

Classification of Crystalline Solids

Crystalline solids are classified based on the nature of their constituent particles and the forces holding them together.

• Molecular Solids: Composed of molecules held together by intermolecular forces (e.g., H2O, CO2). Properties: Soft, low melting points, poor conductors.

• Ionic Solids: Composed of ions held together by electrostatic forces (e.g., NaCl, MgO). Properties: Hard, high melting points, conduct electricity when molten or dissolved.

• Atomic Solids: Composed of atoms, which may be metallic (e.g., Mg), covalent (e.g., diamond), or network solids. Properties: Variable hardness and melting points, metallic solids conduct electricity.

Table: Classification of Crystalline Solids

Type	Constituent Particles	Examples	Properties
Molecular	Molecules	H2O, CO2	Soft, low melting point, non-conductive
Ionic	Ions	NaCl, MgO	Hard, high melting point, conductive in solution
Atomic	Atoms	Mg (metallic), Diamond (covalent)	Variable properties

Addition Polymerization

Addition polymerization is a process where monomers join together to form a polymer without the loss of any small molecules.

• Monomer: A small molecule that can join with others to form a polymer (e.g., ethylene).

• Polymer: A large molecule formed from repeating units of monomers (e.g., polyethylene).

• Example: Ethylene ($\text{CH}_2=\text{CH}_2$) polymerizes to form polyethylene ($[-\text{CH}_2-\text{CH}_2-]_n$).

Chapter 14: Solutions and Colligative Properties

Solution Calculations

Understanding solutions involves calculations related to concentration, solubility, and colligative properties.

• Heats of Solution, Hydration, and Lattice Energy: The energetics of dissolving a solute involve these three quantities. $\Delta H_{\text{solution}} = \Delta H_{\text{lattice}} + \Delta H_{\text{hydration}}$

• Solubility and Partial Pressure: The solubility of gases in liquids is proportional to their partial pressure (Henry’s Law). $C = k_P \cdot P$ where $C$ is concentration, $k_P$ is Henry’s constant, $P$ is partial pressure.

• Number of Moles: $\text{Moles} = \frac{\text{Mass}}{\text{Molar mass}}$

• Moles of Solute in Solution: Calculated using molarity and volume. $\text{Moles} = M \times V$ where $M$ is molarity, $V$ is volume (L).

• Molarity (M): $M = \frac{\text{moles of solute}}{\text{liters of solution}}$

• Molality (m): $m = \frac{\text{moles of solute}}{\text{kg of solvent}}$

• Mole Fraction ($x$): $x = \frac{\text{moles of component}}{\text{total moles}}$

• Percent by Mass: $\%\text{ by mass} = \frac{\text{mass of solute}}{\text{total mass}} \times 100$

Colligative Properties

Colligative properties depend on the number of solute particles, not their identity.

• Vapor Pressure Lowering: Addition of solute lowers the vapor pressure of a solvent. $P = x_{\text{solvent}} \cdot P^0_{\text{solvent}}$ where $x_{\text{solvent}}$ is mole fraction, $P^0_{\text{solvent}}$ is pure solvent vapor pressure.

• Freezing Point Depression: Solute lowers the freezing point. $\Delta T_f = K_f \cdot m$ where $K_f$ is freezing point depression constant, $m$ is molality.

• Boiling Point Elevation: Solute raises the boiling point. $\Delta T_b = K_b \cdot m$ where $K_b$ is boiling point elevation constant.

• Osmotic Pressure: Pressure required to prevent osmosis. $\Pi = MRT$ where $\Pi$ is osmotic pressure, $M$ is molarity, $R$ is gas constant, $T$ is temperature (K).

Energetics of Solution Formation

The formation of a solution involves breaking and forming intermolecular forces, which affects the enthalpy change.

• Endothermic or Exothermic: If the energy required to break bonds is greater than the energy released, the process is endothermic.

• Intermolecular Forces: The solubility of a solute depends on the similarity of intermolecular forces ("like dissolves like").

Factors Affecting Solubility

Solubility is influenced by temperature, pressure, and the nature of solute and solvent.

• Temperature: Solubility of solids generally increases with temperature; solubility of gases decreases.

• Pressure: Only affects gases; higher pressure increases solubility (Henry’s Law).

• Intermolecular Forces: Polar solutes dissolve in polar solvents; nonpolar in nonpolar.

Determining Solubility

To determine if a solute will dissolve in a solvent, compare their intermolecular forces.

• Polar vs. Nonpolar: Polar solutes dissolve in polar solvents due to dipole-dipole interactions.

• Hydrogen Bonding: Solutes capable of hydrogen bonding are more soluble in water.

Summary Table: Colligative Properties

Property	Equation	Effect of Solute
Vapor Pressure	$P = x_{\text{solvent}} \cdot P^0_{\text{solvent}}$	Lowers
Freezing Point	$\Delta T_f = K_f \cdot m$	Lowers
Boiling Point	$\Delta T_b = K_b \cdot m$	Raises
Osmotic Pressure	$\Pi = MRT$	Increases

Key Concepts and Examples

• Example (Unit Cell): Calculate the density of NaCl using FCC unit cell. Steps: Find mass of unit cell, calculate volume from cell length, apply $D = m/v$.

• Example (Colligative Property): Calculate freezing point depression for a solution with known molality. Steps: Use $\Delta T_f = K_f \cdot m$.

Additional info: Academic context and formulas have been expanded for completeness and clarity.