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Surface Chemistry · TOPIC LESSON

Colloids and Emulsions

Connect particle size, interfacial structure and stability to emulsions, micelles, coagulation and surface-tension calculations.

JEE Main · JEE Advanced · NEET UG48 min study guide10 indexed sections

BY THE END, YOU CAN

  • Classify colloids using dispersed phase and continuous medium.
  • Explain light scattering, Brownian motion and selective membrane purification.
  • Distinguish electrostatic and steric stabilization and apply counterion reasoning cautiously.
  • Recognize oil-in-water and water-in-oil structures and micelle orientation.
  • Derive surface work, Laplace pressure and capillary rise with correct interface counts.
  • Use CMC, aggregation number and interfacial area in original quantitative problems.

BEFORE YOU START

  • Intermolecular forces, ionic charge and molecular amphiphilicity.
  • Area and volume of a sphere; force and energy units.
  • Molar concentration, particle counting and qualitative equilibrium.

Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.

CONNECT THE CONCEPT TO THE EXAM

Selected exam questions & worked solutions

These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.

No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.

01 / UNDERSTAND & APPLY

1. Classify the dispersed phase and the continuous medium separately

A colloidal dispersion contains structures large compared with ordinary small molecules but small enough that interfacial and thermal effects are important. A common introductory size range is roughly 1–1000 nm, but the boundaries are conventional and behaviour matters. A colloid can look uniform to the eye while remaining heterogeneous on a microscopic scale.

The dispersed phase consists of the particles, droplets or bubbles. The dispersion medium is the continuous surrounding material. Naming both prevents many classification errors: fog is liquid droplets in gas, smoke contains solid particles in gas, and an oil-in-water emulsion contains oil droplets in continuous water. A gas–gas mixture is molecularly mixed rather than a conventional colloidal dispersion.

A sol commonly means solid particles dispersed in a liquid. An emulsion is one liquid dispersed in another immiscible liquid. A foam has gas bubbles in a liquid or solid medium. An aerosol has liquid droplets or solid particles in a gas. A gel is a liquid-containing network with a solid-like mechanical response; treating every gel as isolated liquid spheres inside a rigid solid oversimplifies its structure.

Classification by phase differs from classification by how colloidal entities are formed. Multimolecular colloids contain aggregates of many small units, macromolecular colloids can consist of individual large polymer molecules, and associated colloids arise from reversible self-assembly such as micelles. The same word “particle” may therefore refer to a solid crystallite, a droplet or a molecular aggregate.

A molecular solution, colloid and coarse suspension form a continuum of sizes and behaviours rather than three materials separated by an exact visible boundary. Ordinary paper filtration often retains a coarse suspension but not a colloid; a suitable membrane can distinguish sizes more finely. Do not claim that every colloid is permanently stable or that every molecular solution scatters absolutely no light.

ORIGINAL WORKED EXAMPLE

Identify the continuous phase

A milky dispersion dilutes uniformly with water but not with oil under the stated test conditions. Oil-soluble dye colours its separated droplets. Identify the emulsion type.

  1. Uniform dilution is consistent with water being the continuous medium.
  2. The dye identifies the droplets as oil-rich.
  3. Together the observations support oil dispersed in water.

Answer Oil-in-water emulsion, under the specified tests.

02 / UNDERSTAND & APPLY

2. Light scattering and Brownian motion reveal different aspects of a dispersion

The Tyndall effect is visible scattering of light by dispersed structures with sufficient size and refractive-index contrast. A beam can become visible from the side because some light is redirected toward the observer. The observed intensity depends on particle size, concentration, wavelength and refractive-index contrast, so lack of a dramatic visible beam does not alone prove a sample is a molecular solution.

Brownian motion is irregular motion of suspended particles driven by fluctuating molecular impacts from the surrounding fluid. It is not self-propulsion and is not the same as the motion of every solvent molecule. Thermal motion can help small particles remain distributed, but it does not by itself prevent them from sticking together after collision.

For dilute spherical particles in a simple continuum liquid, the Stokes–Einstein model gives D=k_BT/(6πηr), where D is translational diffusion coefficient, η viscosity and r hydrodynamic radius. At the same temperature and viscosity, smaller particles diffuse faster. A hydrated or polymer-coated particle’s hydrodynamic radius can exceed the size of its solid core.

For low-Reynolds-number settling of a dilute sphere, Stokes’ model gives speed v=2Δρgr²/(9η), with Δρ the density difference. Its r² dependence explains why aggregation can greatly accelerate settling. These formulas require their stated idealizations; concentrated, nonspherical or strongly interacting dispersions need more careful treatment.

A colloid may be kinetically long-lived even if aggregation would lower interfacial free energy. Slow settling and a barrier to particle attachment are distinct contributions to that persistence. Centrifugation increases the effective body force and can separate particles that settle very slowly under ordinary gravity.

ORIGINAL WORKED EXAMPLE

Separate diffusion from settling

Two dilute spherical particles have equal material density but radii r and 3r in the same liquid at the same temperature. Compare diffusion coefficients and Stokes settling speeds.

  1. Stokes–Einstein gives D proportional to 1/r, so D_large/D_small=1/3.
  2. Stokes settling gives v proportional to r² at fixed density contrast and viscosity.
  3. Thus v_large/v_small=9.

Answer The larger particle diffuses one third as fast and settles nine times as fast within the models.

03 / UNDERSTAND & APPLY

3. Preparation changes size; purification removes selected small species

Dispersion methods break larger material into colloidal-sized structures, while condensation methods assemble small molecular or ionic units into larger particles. Mechanical dispersion and controlled particle formation are broad concepts, not guarantees that every chemical system will yield a stable sol. A stabilizing surface environment is often needed to keep newly formed particles apart.

Peptization is conversion of a freshly formed aggregate or precipitate into a colloidal dispersion by a suitable adsorbing species or change in medium. Surface charging can help separate particles. It is not identical to dissolving the material into individual ions, even though both processes can make a coarse precipitate disappear from view.

Dialysis uses a membrane permeable to small ions or molecules but retaining the relevant colloidal species. Diffusion down a concentration gradient removes diffusible impurities from the retained compartment. The membrane cutoff and the sizes of all species matter; not every ion-selective or semipermeable membrane performs the same separation.

Electrodialysis uses an electric field with suitable membranes to enhance ion transport. Ultrafiltration uses a pressure-driven membrane to retain larger entities while allowing smaller components and solvent through. These are separations based on transport and selectivity, not chemical transformations of every impurity.

Removing too much electrolyte can sometimes change a sol’s stability if particular adsorbed ions help maintain its surface charge. Conversely, excess electrolyte can destabilize a charge-stabilized sol by screening repulsion. Purification therefore requires the actual colloid chemistry to be considered rather than an assumption that “less salt is always better”.

04 / UNDERSTAND & APPLY

4. Surface charge is surrounded by compensating ions

Colloidal particles can acquire charge through surface-group ionization, selective adsorption of ions or other interfacial processes. A negatively charged particle is accompanied by an excess of positive counterions in its surrounding liquid, preserving overall electroneutrality. The charge is not an isolated uncompensated creation of electricity.

A simplified electrical double layer contains charge near the particle surface and a more diffuse compensating ionic atmosphere. Two particles with like-sign charge can repel as these environments overlap. Electrolyte can shorten the range of this repulsion, while specifically adsorbed ions can also change the surface charge itself.

The zeta potential describes the potential at a hydrodynamic slipping plane relative to the bulk liquid, rather than necessarily the exact potential at the solid surface. Its sign and magnitude can help characterize electrokinetic behaviour, but it is not a universal standalone stability test for every sterically stabilized or complex dispersion.

In simple electrophoresis, charged particles move relative to the liquid in an applied electric field. A negative sol moves toward the positive electrode, the anode, in the usual externally powered arrangement. Electro-osmosis instead refers to liquid motion relative to a stationary charged surface. Confusing which component moves leads to a wrong identification.

The sign of a sol can depend on preparation conditions. For a standard silver iodide illustration, adsorption of excess iodide is associated with a negative particle surface, whereas excess silver ions can produce a positive surface. The compound formula AgI alone therefore does not uniquely specify the sol charge.

ORIGINAL WORKED EXAMPLE

Choose the counterion, not the salt name

A negative sol is compared with NaCl, CaCl₂ and AlCl₃ in a simplified counterion-valence coagulation question. Which ions are the relevant counterions, and what qualitative order is expected under otherwise comparable conditions?

  1. The sol is negative, so the relevant counterions are the cations Na⁺, Ca²⁺ and Al³⁺.
  2. The conventional Hardy–Schulze trend gives greater coagulating effectiveness for larger counterion charge magnitude.
  3. Thus the expected simple-model order is Al³⁺>Ca²⁺>Na⁺. Specific chemistry or pH changes can modify real behaviour.

Answer Relevant counterions: Na⁺, Ca²⁺, Al³⁺; expected order Al³⁺>Ca²⁺>Na⁺.

05 / UNDERSTAND & APPLY

5. Stability can be electrostatic, steric or both

Coagulation or aggregation occurs when particles attach to form larger structures. Flocculation often denotes formation of loose clusters, while coalescence is fusion of liquid droplets into larger droplets. Terminology varies somewhat by field, so identify the structural event rather than using every word as an exact synonym.

For many charge-stabilized lyophobic sols, increasing electrolyte concentration screens repulsion and allows attractive interactions to dominate. The critical coagulation concentration is the concentration associated with a defined rapid-aggregation criterion under a specified protocol. A smaller value indicates stronger coagulating action in that comparison.

The Hardy–Schulze rule is a qualitative trend that higher-charge counterions are often much more effective at coagulating oppositely charged lyophobic particles. It does not say that the co-ion controls the trend. Nor is a universal exact inverse-sixth-power concentration law guaranteed across all surfaces, ionic species and pH values.

Lyophilic means favourable interaction with the dispersion medium; hydrophilic is the water-specific case. Solvation and adsorbed or attached polymer layers can help stabilize particles. Steric stabilization occurs when approaching particles would compress or overlap solvated chains in an unfavourable way. Such a dispersion need not rely exclusively on a large surface charge.

A protective colloid can stabilize an otherwise sensitive sol by forming an appropriate interfacial layer. The traditional gold number compares the minimum protective-colloid mass required to prevent a specified gold-sol colour change in a standardized electrolyte test; smaller values indicate greater protective action only within the same test protocol. It is not an intrinsic number independent of the assay.

Polymer addition is not invariably protective. Too little coverage can promote bridging between particles, and changes in solvent quality can collapse a protective layer. Protein dispersions can aggregate when pH reduces net charge or heating changes structure. A stability explanation should name the particular barrier being strengthened or removed.

06 / UNDERSTAND & APPLY

6. Distinguish droplets, their interface and the surrounding phase

In an oil-in-water emulsion, oil is dispersed and water is continuous. In a water-in-oil emulsion, those roles reverse. Milk is commonly treated as an oil-in-water example, while butter has a water-in-fat continuous structure. Real foods can contain proteins, crystals and other dispersed components as well, so these labels describe their principal emulsion topology.

An emulsifier accumulates at the oil–water interface and can lower interfacial tension while providing electrostatic or steric barriers to droplet coalescence. A familiar amphiphile has a water-compatible head group and an oil-compatible hydrocarbon region. In an oil droplet surrounded by water, heads generally face the water and tails point into the oil-rich region.

Forming many small droplets creates more interfacial area than maintaining two large separated liquid layers. For a positive interfacial tension, creating this area has a free-energy cost. Ordinary emulsions are typically kinetically stabilized rather than automatically thermodynamically stable. Equilibrium microemulsions are a distinct class requiring suitable compositions; the prefix “micro” alone does not establish their nature.

Creaming is migration of droplets upward when they are less dense than the continuous phase; sedimentation is downward migration for denser droplets. Neither necessarily means the droplets have fused. Coalescence destroys the separating films and produces larger droplets. A creamed emulsion may redisperse, while a fully broken one may not recover by the same mild action.

Dilution and conductivity tests can help identify a continuous phase when their conditions are suitable: an oil-in-water emulsion often dilutes with water and conducts through its aqueous phase if ions are present. Conductivity alone is not infallible, because composition and additives influence it. Use several observations rather than one memorized label.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Oil-in-water and water-in-oil emulsion interfacesTop panel shows an oil droplet in continuous water with emulsifier polar heads outside and tails into oil. Bottom panel shows a water droplet in continuous oil with polar heads facing the internal water and tails into the external oil. Background labels identify the continuous phase.The continuous phase changes the topologyOil-in-water emulsionOilWaterWaterOilHeads face external water; tails enter oil.Water-in-oil emulsionHeads face the water droplet; tails face external oil.An emulsifier coats an interface between two liquids.This is a droplet cross-section, not a molecular micelle.Circles = polar heads; red lines = hydrocarbon regions.
Idealized orientation of an amphiphilic emulsifier at a liquid-liquid interface. The drawings represent droplets much larger than individual molecules; relative sizes are deliberately schematic. Emulsions and micelles are distinct structures.

07 / UNDERSTAND & APPLY

7. Surface tension is force per length and reversible work per area

For a fluid interface at fixed temperature and appropriate composition, surface tension γ is the reversible free-energy cost per unit additional area. It also appears as a tangential force per unit contact-line length. Its SI units N m⁻¹ equal J m⁻². A liquid interface is not a literal solid skin, even though an elastic-sheet analogy can help visualize area reduction.

If γ is effectively constant, the reversible work to create additional total interfacial area is W=γΔA. A soap film has two liquid–air interfaces, so pulling a straight movable wire of length L requires force 2γL in the simple geometry. A single liquid interface contributes γL for the corresponding contact line. Counting surfaces is essential.

At an air–water interface, many amphiphiles arrange water-compatible heads in water while hydrophobic portions avoid water. Their adsorption can lower the interfacial free-energy cost, reducing surface tension. An ionic head can be anionic or cationic; nonionic and zwitterionic surfactants also exist. “Surfactant” is a functional class, not a synonym for one specific salt.

Sodium dodecyl sulfate contains a sulfate ester head, CH₃(CH₂)₁₁–O–SO₃⁻ with Na⁺. A sulfonate instead has a direct carbon–sulfur connection such as R–SO₃⁻. Those names and connectivities must not be interchanged. This distinction matters in interpreting structural formulas, even when both substances can act as surfactants.

Surface tension depends on temperature and composition. For many pure liquids it decreases as temperature rises toward the critical point. Surfactant adsorption can also take time, so a newly formed interface may have a different dynamic surface tension from its long-time equilibrium value. A single quoted γ should be used only for the state or approximation specified.

ORIGINAL WORKED EXAMPLE

Pull a two-sided film

A thin soap film is held by a movable wire of length 4.00 cm. With γ=0.0300 N m⁻¹ and two equivalent interfaces, find the force and reversible work for moving the wire 2.00 cm.

  1. L=0.0400 m; the force is 2γL=2×0.0300×0.0400=0.00240 N.
  2. The total new area is 2LΔx=2×0.0400×0.0200=0.00160 m².
  3. Work is γΔA=0.0300×0.00160=4.80×10⁻⁵ J.

Answer F=2.40 mN; W=4.80×10⁻⁵ J.

08 / UNDERSTAND & APPLY

8. Derive curvature pressure and capillary rise with the correct geometry

For a spherical liquid droplet of radius r with one interface, a small expansion gives dV=4πr²dr and dA=8πrdr. Balancing reversible pressure work ΔP dV with γdA gives ΔP=2γ/r. Here ΔP is the pressure inside minus outside for the convex droplet. A spherical gas bubble within a liquid also has one interface and the corresponding excess interior pressure is 2γ/r.

A thin soap bubble in air has inner and outer liquid–air interfaces of nearly equal radius. Their area contributions give approximately ΔP=4γ/r between the inner and outer gases, assuming the same surface tension on both faces and negligible film thickness. A rubber balloon has an elastic membrane with different mechanics, so one cannot automatically apply a constant-γ soap-bubble formula to it.

For a wetting liquid rising in a narrow cylindrical tube, the vertical surface-tension force is 2πrγ cosθ, with contact angle θ measured through the liquid. Balancing it against the weight πr²hρg gives h=2γ cosθ/(ρgr), neglecting gas density and the small meniscus-volume correction. If θ exceeds 90°, cosθ is negative and the formula describes depression rather than rise.

Breaking one spherical droplet of radius R into N equal spherical droplets conserves liquid volume: NR_small³=R³. The total final area is N4πR_small²=4πR²N^(1/3). Thus making more droplets raises area and requires reversible interfacial work if γ remains positive and constant. Surfactants can reduce that work but do not make the geometric increase disappear.

Use radius rather than diameter in these formulas, convert millimetres to metres, and distinguish total area from the area of one droplet. Check whether one or two interfaces are present before multiplying by a factor of two. These are model calculations for well-defined shapes, not a complete prediction of every emulsion’s flow or breakup.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

One-interface droplet and two-interface soap bubbleA liquid droplet in gas is drawn as one circular interface with excess interior pressure2gamma/r. A soap bubble has gas inside and outside a thin liquid film, drawn with two concentric interfaces, giving approximately4gamma/r between the gases when the film is thin and the two surface tensions equal.Count the interfaces before using ΔPLiquid droplet in gas: one interfaceliquidinsideΔP = P_inside − P_outside = 2γ/rrgasThin soap bubble: two interfacesgasinsidegasThin liquid film between the two circlesΔP between gases ≈ 4γ/rInner and outer radii treated as nearly equal.A gas bubble in bulk liquid has only ONE interface.Film thickness is exaggerated to show both surfaces.
Cross-sections of spherical objects. The soap film is exaggerated in thickness for clarity; the4γ/r formula uses nearly equal inner and outer radii. A gas bubble in bulk liquid has one interface and uses2γ/r, not4γ/r.

ORIGINAL WORKED EXAMPLE

A droplet and a soap bubble

A spherical liquid droplet and a thin soap bubble both have radius 1.00 mm and γ=0.0400 N m⁻¹. Find each excess interior pressure.

  1. The droplet has one interface: ΔP=2γ/r=0.0800/0.00100=80.0 Pa.
  2. The soap bubble has two interfaces: ΔP=4γ/r=160 Pa.
  3. The difference is interface count, not a different choice of radius.

Answer Droplet: 80.0 Pa; soap bubble:160 Pa.

ORIGINAL WORKED EXAMPLE

Fragment one droplet into eight

One spherical droplet of radius 2.00 mm divides into eight equal droplets at constant γ=0.0500 N m⁻¹. Find the new radius and reversible area-creation work.

  1. Volume conservation gives r=R/8^(1/3)=1.00 mm.
  2. Final total area is twice the initial area because 8^(1/3)=2.
  3. The area increase equals the initial area 4πR²=4π(0.00200)²=5.0265×10⁻⁵ m².
  4. W=γΔA=0.0500×5.0265×10⁻⁵.

Answer r=1.00 mm; W=2.51×10⁻⁶ J.

09 / UNDERSTAND & APPLY

9. Micelles are dynamic molecular aggregates, not tiny soap bubbles

Above a suitable critical micelle concentration, CMC, many surfactants in water assemble into aggregates with hydrocarbon-rich interiors and water-compatible head groups exposed to the aqueous surroundings. A spherical micelle has no gas core and is not a thin liquid film enclosing air. Its molecules exchange with the surrounding monomer population.

The CMC is an operational concentration range associated with changes in properties such as surface tension or conductivity. It depends on surfactant structure, temperature, added electrolyte and solvent composition. For many ionic surfactants, a Krafft temperature marks a solubility-related threshold below which ordinary micelle formation is restricted; it is distinct from the concentration threshold CMC.

Above the CMC, a useful elementary approximation treats free monomer concentration as nearly fixed near the CMC and places most added surfactant into micelles. This is not exact at arbitrary concentration. Aggregation number is the number of surfactant molecules per micelle under the stated conditions, and it can change with environment and aggregate shape.

In a nonpolar solvent, reverse micelles can have polar head groups inward and hydrocarbon portions outward. Bilayers use a different arrangement: two molecular leaflets place hydrophobic tails against each other and expose polar heads on both outer faces. A vesicle encloses an aqueous compartment with such a bilayer; it is not structurally identical to a normal micelle.

Micellar solubilization can incorporate hydrophobic material into aggregate interiors, while emulsification stabilizes larger droplets with an interfacial layer. Both involve amphiphiles but are not the same size or structural process. Cleaning also involves wetting, detachment, dispersion and rinsing, so it should not be reduced to one static sphere drawing.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Amphiphile orientation in micelles, reverse micelles and bilayersNormal aqueous micelle has polar circular heads on the outer circumference and hydrocarbon lines directed inward. Reverse micelle has polar heads at the inner circumference and tails facing the external nonpolar solvent. Bilayer consists of two opposing leaflets with polar heads facing water on both sides and tails in the middle.Normal micelle in waterPolar heads outside; hydrocarbon tails insideHydrophobicinteriorWater surrounds the head groups.Reverse micelle in nonpolar solventPolarinteriorHydrocarbon portions face the nonpolar medium.Bilayer: two leaflets, tails face tailswaterwaterCircles = hydrophilic heads; red lines = hydrophobic tails.Schematic cross-sections; not to molecular scale.
Topology schematics, not atom-resolved structures or simulated motion. Tail count and shape are simplified; many bilayer-forming amphiphiles have two hydrocarbon chains. Normal micelles do not enclose a gas bubble.

ORIGINAL WORKED EXAMPLE

Count micelles above the CMC

A 1.00 L solution contains total surfactant concentration 0.0100 mol L⁻¹. Assume free monomer remains at a CMC of 0.00200 mol L⁻¹ and each micelle contains 80 surfactant molecules. Find micelle amount and number using N_A=6.022×10²³.

  1. Surfactant in aggregates is (0.0100−0.00200)×1.00=0.00800 mol of molecules.
  2. Divide by 80 to find 0.000100 mol of micelles.
  3. The number is 1.00×10⁻⁴×6.022×10²³=6.022×10¹⁹.

Answer 1.00×10⁻⁴ mol of micelles;6.02×10¹⁹ micelles under the assumptions.

10 / UNDERSTAND & APPLY

10. Solved practice: classify, count and check the interface

The most useful final check is to name the continuous phase, the stabilizing mechanism and the object being counted. A counterion is selected by particle charge; a micelle count requires aggregation number; surface work requires total new area. The original exercises below keep these decisions explicit.

ORIGINAL PRACTICE · 1

Classify fog and smoke by dispersed phase and medium.

Reveal answer & reasoning

Fog: liquid in gas. Smoke: solid particles in gas.

Both are aerosols; gas is the continuous medium, not the dispersed phase in these examples.

ORIGINAL PRACTICE · 2

A negatively charged sol migrates in a simple externally powered electrophoresis setup. Toward which electrode does it move?

Reveal answer & reasoning

The positive electrode, the anode.

Negative particles are attracted toward the positive electrode. This is particle motion relative to liquid.

ORIGINAL PRACTICE · 3

For a positive lyophobic sol, compare Cl⁻, SO₄²⁻ and PO₄³⁻ under the conventional counterion-valence approximation.

Reveal answer & reasoning

PO₄³⁻>SO₄²⁻>Cl⁻ in coagulating effectiveness.

The anions oppose the positive sol charge. The comparison is conditional; real phosphate speciation and pH can matter.

ORIGINAL PRACTICE · 4

Two protective colloids have gold numbers 0.10 and 1.0 in the same standardized assay. Which is more protective?

Reveal answer & reasoning

The one with gold number 0.10.

A smaller mass is required to prevent the defined coagulation-related colour change in that same assay.

ORIGINAL PRACTICE · 5

What does dialysis remove from a retained colloid when its membrane passes small ions but not the colloidal particles?

Reveal answer & reasoning

Diffusible small-ion impurities.

The concentration gradient drives ion diffusion through the membrane; the larger retained particles are not converted chemically.

ORIGINAL PRACTICE · 6

A spherical droplet’s radius halves while γ stays constant. How does its Laplace excess pressure change?

Reveal answer & reasoning

It doubles.

For one interface, ΔP=2γ/r, so pressure is inversely proportional to radius.

ORIGINAL PRACTICE · 7

A wetting liquid has γ=0.0500 N m⁻¹, θ=0°, ρ=1000 kg m⁻³ and g=10.0 m s⁻² in a tube of radius 0.500 mm. Find ideal capillary rise.

Reveal answer & reasoning

0.0200 m, or 2.00 cm.

h=2γ/(ρgr)=0.100/(1000×10.0×0.000500)=0.0200 m.

ORIGINAL PRACTICE · 8

For a capillary with contact angle 120°, is the ideal height positive or negative relative to the reservoir surface?

Reveal answer & reasoning

Negative: a capillary depression.

cos 120° is negative, so the vertical interfacial force gives a depression in the simple formula.

ORIGINAL PRACTICE · 9

One droplet becomes 27 equal droplets. Find the final total area divided by initial area.

Reveal answer & reasoning

3.

The ratio is N^(1/3)=27^(1/3)=3 by volume conservation.

ORIGINAL PRACTICE · 10

At fixed viscosity and particle radius, the temperature rises from 300 to 330 K. What happens to D in the Stokes–Einstein model?

Reveal answer & reasoning

It increases by 10%.

D is proportional to absolute temperature when η and r are explicitly held fixed; the ratio is 330/300=1.10.

ORIGINAL PRACTICE · 11

In an ordinary aqueous micelle, where are hydrophilic heads and hydrocarbon tails?

Reveal answer & reasoning

Heads face water; tails form the interior.

This reduces unfavourable hydrocarbon–water contact while preserving interactions of the polar or ionic head groups with water.

ORIGINAL PRACTICE · 12

Does creaming prove that emulsion droplets have fused?

Reveal answer & reasoning

No.

Creaming is gravitational separation of droplets into a concentrated region. Coalescence is droplet fusion; the two processes are distinct.

KEEP THE BIG IDEAS

Quick revision

  • Classify both dispersed phase and continuous medium.
  • Thermal diffusion, slow settling and resistance to aggregation are different effects.
  • Use counterion charge for the conventional coagulation trend, with real-chemistry qualifications.
  • Surfactant orientation depends on the surrounding phases.
  • Count one interface for a droplet and two for a thin soap bubble.
  • CMC, aggregation number and monomer balance govern simple micelle-count calculations.
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The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.

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