Solution:

Given: Ice floats on liquid water Find: Explain why and why this is unusual

Step 1: Recall density principles

General rule: Objects float if less dense than the liquid

For ice to float on water: $\rho_{ice} < \rho_{water}$

This means: Ice has larger volume than same mass of water

Step 2: Typical behavior of substances

Most substances:

• Solid is MORE dense than liquid
• Particles pack closer in solid (ordered structure)
• Solid sinks in its own liquid
• Example: Solid paraffin sinks in liquid paraffin

Step 3: Explain water's unusual behavior

In liquid water:

• H₂O molecules in constant motion
• H-bonding exists but constantly breaking/reforming
• Average of ~3.4 H-bonds per molecule
• Molecules relatively close but not in fixed positions
• Density at 4°C: 1.000 g/cm³ (maximum density)

In ice (solid water):

• H₂O molecules in fixed crystalline structure
• Each molecule forms 4 H-bonds (maximum)
• Tetrahedral arrangement around each molecule
• Creates hexagonal crystal structure with lots of empty space
• Open, cage-like structure
• Density at 0°C: 0.917 g/cm³

Step 4: Why the open structure?

Hydrogen bonding geometry:

• Each O has 2 H atoms (H-bond donors)
• Each O has 2 lone pairs (H-bond acceptors)
• Total: Can form 4 H-bonds per molecule

In ice:

• All 4 H-bonds form simultaneously
• Tetrahedral geometry (109.5° angles)
• This specific geometry requires more space
• Molecules held at fixed distances
• Creates channels and cavities in structure

In liquid water:

• H-bonds constantly breaking/reforming
• Not all 4 H-bonds present at once
• Molecules can pack more closely
• Less open space

Step 5: Quantitative comparison

Ice: 0.917 g/cm³ Water: 1.000 g/cm³

Ratio: Ice is about 8-9% less dense than water

Volume increase: When water freezes, it expands by ~9%

• This is why pipes burst in winter
• Why ice cubes expand in freezer

Step 6: Why is this unusual?

For most substances:

• Freezing brings particles closer together
• Solid is 10-20% MORE dense than liquid
• Increased order = tighter packing

For water:

• Freezing creates more H-bonds
• More H-bonds = more rigid, open structure
• Increased order = LESS dense

Only a few substances behave this way:

• Water (most important)
• Silicon
• Bismuth
• Antimony

Answer:

Ice floats because it is less dense than liquid water (0.917 g/cm³ vs. 1.000 g/cm³).

Explanation:

In ice, water molecules form a hexagonal crystalline structure where each H₂O molecule forms 4 hydrogen bonds in a tetrahedral arrangement. This creates an open, cage-like structure with large empty spaces between molecules.

In liquid water, H-bonds are constantly breaking and reforming, so molecules are not locked into this rigid structure and can pack more closely together, resulting in higher density.

Why unusual:

For most substances, the solid phase is MORE dense than the liquid because particles pack closer in the ordered solid structure. Water is unusual because the specific geometry required by hydrogen bonding (tetrahedral, 4 H-bonds per molecule) forces molecules APART, creating an open structure that is less dense than the liquid.

Importance:

1. Aquatic life: Ice forms on surface of lakes/oceans, insulating water below and allowing life to survive winter
2. Climate: Ice caps reflect sunlight (albedo effect)
3. Weathering: Water expanding when freezing can crack rocks
4. Plumbing: Pipes can burst when water freezes and expands

Without this property, bodies of water would freeze from bottom up, potentially killing aquatic ecosystems.

Solution:

Given:

• Hexane (C₆H₁₄): MW = 86 g/mol, BP = 69°C, low viscosity, low surface tension
• Water (H₂O): MW = 18 g/mol, BP = 100°C, higher viscosity, higher surface tension

Find: Explain property differences using IMF concepts

Step 1: Identify intermolecular forces in each liquid

Hexane (C₆H₁₄):

Structure: CH₃-CH₂-CH₂-CH₂-CH₂-CH₃ (saturated hydrocarbon)

Lewis structure analysis:

• Only C-C and C-H bonds
• C-H bonds slightly polar (ΔEN = 0.4)
• Molecular geometry: Nonpolar overall (symmetrical)
• No N-H, O-H, or F-H bonds

IMFs present:

• London Dispersion Forces (LDF) ONLY
• 42 total electrons (more polarizable than water)
• Linear structure → good surface contact

Water (H₂O):

Lewis structure analysis:

• O-H bonds (highly polar, ΔEN = 1.4)
• Bent geometry (2 lone pairs on O)
• Molecular geometry: Polar

IMFs present:

• Hydrogen bonding (strongest)
• Dipole-dipole forces
• London Dispersion Forces (weakest, only 10 electrons)

Step 2: Explain boiling point difference

Expected based on molar mass:

• Hexane: 86 g/mol → more electrons → stronger LDF
• Water: 18 g/mol → fewer electrons → weaker LDF
• Prediction: Hexane should have higher BP

Actual result: Water has higher BP (100°C vs. 69°C)

Explanation:

Type of IMF matters more than size!

Water:

• Hydrogen bonding strength: 10-40 kJ/mol per H-bond
• Each H₂O can form up to 4 H-bonds (2 donors, 2 acceptors)
• Extensive H-bonding network
• Total IMF energy per molecule: ~40 kJ/mol
• Requires significant energy to break H-bonds and boil

Hexane:

• LDF strength: ~30 kJ/mol total (for all LDF in molecule)
• Even with 42 electrons (strong for LDF)
• LDF still weaker than H-bonding
• Requires less energy to overcome IMFs and boil

$\text{H-bonding (H₂O)} > \text{LDF (hexane)}$

Result: Water boils at higher temperature

Step 3: Explain viscosity difference

Viscosity = resistance to flow

Factors:

1. Strength of IMFs
2. Molecular size/shape
3. Temperature

Water:

• Higher viscosity (relatively)
• Strong H-bonding between molecules
• Molecules must "unstick" from neighbors to flow
• H-bonds constantly breaking/reforming as molecules move

Hexane:

• Lower viscosity
• Only weak LDF
• Molecules slide past each other easily
• Less resistance to flow
• Despite being larger molecule, weaker IMFs dominate

However, note: Water actually has relatively LOW viscosity for a H-bonding liquid because it's so small. Glycerol (3 OH groups) has much higher viscosity.

Step 4: Explain surface tension difference

Surface tension = energy to increase surface area

Cause: Molecules at surface experience net inward pull

Water:

• High surface tension (~73 mN/m at 25°C)
• Strong H-bonding pulls surface molecules inward
• Surface molecules "miss" the H-bonds above them
• Strong cohesive forces
• Creates strong "skin" effect
• Allows insects to walk on water, water drops to bead up

Hexane:

• Low surface tension (~18 mN/m at 25°C)
• Only weak LDF
• Less "pulling" of surface molecules inward
• Weaker cohesive forces
• Spreads out easily (wets surfaces better)

Ratio: Water has ~4× higher surface tension than hexane

Step 5: Summary of IMF effects

Property relationships:

All these properties depend on IMF strength:

Solution:

Given:

• T₁ = 25°C = 298 K
• P₁ = 23.8 mmHg
• T₂ = 50°C = 323 K
• P₂ = 92.5 mmHg
• R = 8.314 J/(mol·K)

Find: ΔHvap for water

Step 1: Write Clausius-Clapeyron equation

$\ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

Step 2: Identify known and unknown values

Known:

• P₁ = 23.8 mmHg
• P₂ = 92.5 mmHg
• T₁ = 298 K
• T₂ = 323 K
• R = 8.314 J/(mol·K)

Unknown:

• ΔHvap = ?

Step 3: Calculate pressure ratio

$\frac{P_2}{P_1} = \frac{92.5}{23.8} = 3.887$

$\ln\left(\frac{P_2}{P_1}\right) = \ln(3.887) = 1.358$

Step 4: Calculate temperature term

$\frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{323} - \frac{1}{298}$

$= 0.003096 - 0.003356$

$= -0.000260 \text{ K}^{-1}$

Or more precisely: $= -2.60 \times 10^{-4} \text{ K}^{-1}$

Step 5: Substitute into Clausius-Clapeyron equation

$1.358 = -\frac{\Delta H_{vap}}{8.314} \times (-2.60 \times 10^{-4})$

$1.358 = \frac{\Delta H_{vap}}{8.314} \times 2.60 \times 10^{-4}$

Step 6: Solve for ΔHvap

$\Delta H_{vap} = \frac{1.358 \times 8.314}{2.60 \times 10^{-4}}$

$\Delta H_{vap} = \frac{11.29}{2.60 \times 10^{-4}}$

$\Delta H_{vap} = 43,400 \text{ J/mol}$

$\Delta H_{vap} = 43.4 \text{ kJ/mol}$

Step 7: Check reasonableness

Literature value: ΔHvap for water ≈ 40.7 kJ/mol at 100°C

Our calculated value: 43.4 kJ/mol

Why slightly higher?

• ΔHvap decreases slightly with temperature
• At 25-50°C (lower than boiling point), ΔHvap is slightly higher
• Our value is reasonable!

Step 8: Verify with unit analysis

$\Delta H_{vap} = \frac{\ln(P_2/P_1) \times R}{(1/T_2 - 1/T_1)}$

Units: $= \frac{\text{(dimensionless)} \times \text{J/(mol·K)}}{\text{K}^{-1}}$

$= \frac{\text{J/(mol·K)}}{\text{K}^{-1}} = \text{J/mol}$ ✓

Answer:

$\boxed{\Delta H_{vap} = 43.4 \text{ kJ/mol}}$

Step-by-step summary:

1. Converted temperatures to Kelvin
2. Calculated ln(P₂/P₁) = 1.358
3. Calculated (1/T₂ - 1/T₁) = -2.60 × 10⁻⁴ K⁻¹
4. Substituted into Clausius-Clapeyron equation
5. Solved for ΔHvap = 43.4 kJ/mol

Physical interpretation:

43.4 kJ/mol is the energy required to convert 1 mole of liquid water to water vapor at constant temperature (around 25-50°C).

This energy is needed to:

• Break hydrogen bonds between water molecules
• Separate molecules from liquid to gas phase
• Overcome intermolecular attractions

Comparison:

• ΔHvap of water: 43.4 kJ/mol (calculated)
• ΔHfus of water: 6.01 kJ/mol (literature)

Ratio: ΔHvap/ΔHfus ≈ 7.2

Much more energy required for vaporization than melting because:

• Melting: Only partially breaks IMFs (molecules still close)
• Vaporization: Completely breaks all IMFs (molecules far apart)

Alternative approach - Linear form:

Could also use: ln P = -ΔHvap/R × (1/T) + C

Plot ln P vs. 1/T → slope = -ΔHvap/R

This problem uses the two-point form, which is more common for single calculations.