Phase Separations

Chapter Notes

All content from CE and SG texts aligned with CAPE Specific Objectives 9.1–9.7. Expand each topic to study.

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Introduction & Phase Separation Techniques

SO 9.5

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The word 'phase' is synonymous with 'state of matter'. The three fundamental states of matter are solid, liquid and gas. The concept of phase separation involves how different types of mixtures can be separated — most techniques involve a change of state: melting, subliming, evaporating or dissolving.

Mixtures are impure substances. Examples of separation in everyday life:

Home: water separated from pasta; tap water filtered to remove impurities.
Laboratory: chemists prepare and purify compounds.
Petroleum industry: components of crude oil are separated and selectively purified.

Separation Techniques Covered (CAPE)

Technique	Basis of Separation	Best Used For
Simple distillation	Widely differing bp (>25°C)	Two miscible liquids with large bp difference; liquid from involatile solid
Fractional distillation	Small differences in bp	Miscible liquids with similar bp; crude oil fractions
Vacuum distillation	bp decreases with reduced pressure	High bp or thermally decomposable liquids
Steam distillation	Immiscible mixture boils below individual bps	Temperature-sensitive organics immiscible with water
Solvent extraction	Different solubilities in immiscible solvents	Organic product dissolved in aqueous solution
Separating funnel	Density difference of immiscible liquids	Two immiscible liquids

✓ Key Principle

The separation technique chosen depends on the differences in the physical and chemical properties of the components of the mixtures.

🫧

Simple Distillation

SO 9.1 · 9.6 · 9.7

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When to Use Simple Distillation

Liquid solutions of substances with widely differing boiling points (more than 25°C difference) — both components are obtained.
Liquids from involatile solids or oils — the solvent is collected rather than the solute.

💡 Principle

Distillation is based on the differences in volatilities of the components of the mixture. The more volatile component (lower bp) evaporates preferentially.

How It Works

When the solution in the flask boils, pure vapour rises upwards and enters the inner tube of the Liebig condenser. The outer jacket contains cool running water that condenses the vapour. The condensed liquid — the distillate — is collected at the end of the condenser.

Apparatus Components

Round-bottomed flask — contains the mixture; heated by Bunsen burner over wire gauze on tripod
Boiling chips (anti-bumping granules) — prevents superheating
Still head — connects flask to condenser; holds thermometer
Thermometer — bulb positioned at the junction where vapour enters condenser side arm
Liebig condenser — inner tube (vapour path) surrounded by outer water jacket
Conical flask / receiver — collects the distillate

⚠️ Critical Detail — Condenser Water Flow

Water to cool the vapour enters the condenser from the end closer to the receiver and leaves from the end close to the distilling flask. This countercurrent arrangement ensures the vapour meets the coldest water first — maximising condensation efficiency. Reversing the flow would reduce efficiency.

Industrial Applications (SO 9.7)

Purification of sea water (salts remain, water distils)
Obtaining distilled spirits (simple, though fractional is preferred)
Separation of volatile solvents from dissolved solids

🏭

Fractional Distillation

SO 9.1 · 9.6 · 9.7

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Fractional distillation separates miscible liquids whose components have boiling points close together (usually <25°C apart) or are chemically similar, such as ethanol and water.

Key Feature: The Fractionating Column

A fractionating column is added between the flask and the condenser. It is a vertical tube packed with inert fragments (glass beads or glass rod) providing a large cool surface for repeated condensation and vaporisation. Once in equilibrium:

The column is hotter at the base than at the top — a temperature gradient exists.
Vapours travel upward and condense at various heights. The condensing liquid is richer in the more volatile component at higher positions.
Most condensed liquid runs back down. Imagine a series of 'condensing traps' — at each trap less volatile component accumulates.
If the column is long enough, only the most volatile component reaches the top and exits to the condenser.

💡 Why It Works (SO 9.1)

A series of successive liquid–vapour equilibria occur up the column. Each equilibrium stage enriches the vapour in the more volatile component. The fractionating column is equivalent to many simple distillations in sequence. The longer the column and the slower the heating, the better the separation.

Reading the Boiling Point / Composition Diagram

For an ideal binary mixture A (lower bp) + B (higher bp), the T-x,y diagram has two curves:

Bubble point curve (liquid line) — shows the boiling point of the liquid at each composition.
Dew point curve (vapour line) — shows the composition of the first vapour formed when a liquid of given composition is heated.
The vapour is always richer in the more volatile component (lower bp) than the liquid in equilibrium with it.

Industrial Applications (SO 9.7)

Petroleum refinery — crude oil separated into fractions: LPG, petrol, kerosene, diesel, fuel oil, lubricants, bitumen
Rum and whisky production — ethanol concentrated from fermented wash by fractional distillation
Liquid air separation — oxygen (bp −183°C) and nitrogen (bp −196°C) obtained

⚠️ Ethanol–Water Azeotrope Limitation

Fractional distillation of ethanol–water can produce a maximum of 95.6% ethanol by mass (78.2°C). This azeotrope composition cannot be exceeded by distillation. Anhydrous ethanol requires a drying agent (e.g. silica gel, CaO) to remove the remaining water.

Comparison: Simple vs Fractional Distillation (SO 9.6)

Feature	Simple	Fractional
Boiling point difference	>25°C	<25°C
Fractionating column	No	Yes (glass beads)
Purity of distillate	Lower	Higher
Efficiency	Lower	Higher (many equilibria)
Example	Water from salt	Ethanol from wine/beer

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Raoult's Law & Vapour Pressure

SO 9.1

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Vapour Pressure of a Pure Liquid

In a closed container, molecules at the liquid surface with enough kinetic energy escape into the vapour phase, while vapour molecules re-enter the liquid. Equilibrium is reached when the rate of evaporation = rate of condensation:

liquid ⇌ vapour

The equilibrium vapour is saturated and the pressure is the saturated vapour pressure (SVP) at that temperature. SVP increases with temperature. In a mixture, each component exerts its own partial vapour pressure.

Raoult's Law — Statement (SO 9.1)

📜 Raoult's Law

"The partial vapour pressure of a component in a volatile mixture is equal to the vapour pressure of the pure component at that temperature multiplied by its mole fraction in the mixture."
— François-Marie Raoult (1830–1901)

P_A = P°_A × x_A P_B = P°_B × x_B P_total = P_A + P_B = (P°_A × x_A) + (P°_B × x_B)

Mole Fraction

x_A = n_A / (n_A + n_B) and x_A + x_B = 1

Ideal Solutions

Raoult's law applies only to ideal solutions. Conditions for ideality:

Intermolecular forces A–A = B–B = A–B
ΔH_mix = 0 (no heat change on mixing)
ΔV_mix = 0 (no volume change on mixing)

Near-ideal examples (hydrogen bonds):

Water + methanol
Propan-1-ol + propan-2-ol
Butan-1-ol + butan-2-ol

Near-ideal examples (van der Waals):

Hexane + heptane
Benzene + methylbenzene
Benzene + toluene

Mole Fraction in Vapour Phase

For gases, the mole fraction in the vapour is the ratio of partial pressure to total pressure (Dalton's Law):

y_A = P_A / P_total = (P°_A × x_A) / P_total

Since P°_A > P°_B (A is more volatile), y_A > x_A — the vapour is always enriched in the more volatile component.

Reference Data — Vapour Pressures at 25°C

Substance	bp (°C, 1 atm)	VP at 25°C (mmHg)
Water (H₂O)	100.0	23.8
Ethanol (C₂H₅OH)	78.5	59.3
Methanol (CH₃OH)	64.7	127.2
Benzene (C₆H₆)	80.1	95.1
Toluene (C₇H₈)	110.6	28.4
Hexane (C₆H₁₄)	68.7	150.6
Heptane (C₇H₁₆)	98.4	45.8
Acetone (CH₃COCH₃)	56.1	231.4
Chloroform (CHCl₃)	61.2	199.0

📈

Deviations from Raoult's Law & Azeotropes

SO 9.1

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When two dissimilar liquids mix, the A–A, B–B and A–B intermolecular forces differ. This causes deviation from Raoult's law — such solutions are called non-ideal solutions.

Positive Deviation

The actual vapour pressure is higher than predicted by Raoult's law.

Cause

A–B forces are weaker than A–A or B–B. Endothermic mixing. Molecules escape more easily. ΔH_mix > 0, volume expands.

Examples

Ethanol + water / hexane / CHCl₃
Methanol + benzene
Acetone + CS₂
Ethanol + cyclohexane

Large positive deviations → positive azeotrope (minimum boiling point):

📌 Ethanol–Water Azeotrope

bp = 78.2°C (below pure ethanol 78.5°C and water 100°C)
Composition: 95.6% by mass ethanol (mole fraction ≈ 0.894)
Cannot obtain pure ethanol by distillation alone.

Negative Deviation

The actual vapour pressure is lower than predicted by Raoult's law.

Cause

A–B forces are stronger than A–A or B–B. Exothermic mixing. Molecules escape less easily. ΔH_mix < 0, volume contracts.

Examples

Water + HNO₃ or HCl
Acetone + chloroform
CHCl₃ + propanone or ethyl ethanoate

Large negative deviations → negative azeotrope (maximum boiling point):

📌 HNO₃–Water Azeotrope

bp = 120.5°C (above pure HNO₃ 86°C and water 100°C)
Composition: 68% HNO₃ by mass (mole fraction ≈ 0.384)

Definition of Azeotrope

📜 Definition

An azeotrope (constant boiling mixture) is a mixture of two or more liquids in such a ratio that its composition cannot be changed by distillation. At the azeotropic composition, the liquid and vapour have the same composition.

Summary Table

Deviation	VP vs ideal	BP vs ideal	Azeotrope type	Example
Small positive	Slightly higher	Slightly lower	None	Ethanol + benzene (small)
Large positive	Maximum (hump)	Minimum (dip)	Min boiling	Ethanol + water (78.2°C)
Small negative	Slightly lower	Slightly higher	None	CHCl₃ + ether
Large negative	Minimum (dip)	Maximum (hump)	Max boiling	HNO₃ + water (120.5°C)

💨

Vacuum Distillation

SO 9.2

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Vacuum distillation is distillation carried out at reduced pressure (less than atmospheric). Boiling occurs when the vapour pressure of the liquid just exceeds the external pressure. Lowering external pressure lowers the boiling point.

💡 Principle

Reduced pressure → lower boiling point → less energy needed → compounds that decompose below their normal bp can now be distilled safely.

Advantages (SO 9.2)

Allows distillation of thermally unstable compounds that decompose before their normal boiling point.
Reduced energy requirements — lower temperatures needed for both heating and cooling.
Can be carried out with or without heating the mixture.
Sometimes preferred over steam distillation as it can produce purer product.

Key Examples

Compound	Normal bp (°C)	Pressure (atm)	Reduced bp (°C)
Phenylamine (aniline)	184	0.02	77
Dimethyl sulfoxide (DMSO)	189	0.1	85
Glycerol	290	0.05	167

⚠️ Apparatus Note

A capillary tube is used in the vacuum distillation apparatus to introduce tiny bubbles into the liquid, preventing 'bumping' — sudden violent boiling that can damage the apparatus or cause injury.

Industrial Application

In petroleum refining, after atmospheric distillation, the heavy residue is processed by vacuum distillation. This extracts lighter, more valuable fractions (e.g. lubricating oils, wax) that would decompose at the temperatures needed to distil them at atmospheric pressure.

♨️

Steam Distillation

SO 9.3

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Steam distillation purifies temperature-sensitive organic compounds that are immiscible with water and would decompose at their normal boiling points.

Principle (SO 9.3)

When an organic compound immiscible with water is mixed with water and heated, each component independently exerts its own vapour pressure as if the other is absent:

P_total = P°_water + P°_organic

This total exceeds atmospheric pressure at a temperature below 100°C — the mixture boils lower than either pure component.

📌 Bromobenzene

Normal bp = 156°C. With water: distils at ~95°C. Decomposition prevented.

📌 Phenylamine

Normal bp = 184°C. With water: distils at 98°C. Industrial purification method.

📌 Nitrobenzene

Normal bp = 211°C. With water (VP_water = 733 mmHg at 99°C): distils at 99°C. P_NB = 760 − 733 = 27 mmHg.

Steam Distillation Calculations (SO 9.3)

Since each component is independent:

P_water / P_A = n_water / n_A P_water / P_A = (m_water / M_water) / (m_A / M_A)

Applications (SO 9.7)

Essential oils from plant materials — eugenol from cloves, eucalyptus oil, citrus oils from lemon/orange peel, oils in perfumes from various plants.
Purification of nitrobenzene (bp 211°C → 99°C with steam)
Purification of phenylamine/aniline (bp 184°C → 98°C)
Fragrance industry (SO 9.7: include perfume, rum, petroleum)

⚠️ Advantages vs Disadvantages

Advantages

Below 100°C → no decomposition
Cheap, readily available steam
No organic solvent needed
Can be combined with vacuum distillation (e.g. bromobenzene at 61°C)

Disadvantages

Only works for immiscible liquids
Distillate forms two layers (needs separating funnel)
Large volumes of water produced
Not suitable for water-miscible compounds

Laboratory Procedure

Generate steam in a separate steam generator flask
Pass steam through the organic mixture in a second flask
Mixed vapour (steam + organic) passes through the condenser
Condensate forms two layers in receiver: water layer + organic layer
Separate layers using a separating funnel
Dry organic layer with anhydrous MgSO₄; filter; distil off solvent

⚗️

Solvent Extraction & Partition Law

SO 9.4 · 9.5

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Solvent extraction separates compounds based on their preferential solubilities in two immiscible liquids (usually water and an organic solvent). The solute transfers from the phase in which it is less soluble to the phase in which it is more soluble.

The Partition Law (SO 9.4)

📜 Partition Law

When a solute distributes itself between two immiscible solvents at equilibrium and constant temperature, the ratio of its concentrations in the two solvents is a constant — the partition coefficient (K_d or K_p).

K_d = [solute]_organic / [solute]_aqueous = constant (at fixed T)

The partition coefficient depends on the nature of both the solute and the two solvents. A large K_d means the solute prefers the organic phase.

💡 Multiple Extractions — More Efficient

Using the same total volume of solvent in multiple small extractions removes more solute than a single large extraction. Each fresh solvent maintains a large concentration gradient, driving more solute across the interface.

Common Organic Solvents

Solvent	Advantage
Diethyl ether	Low bp (34.6°C) — easily distilled off and recycled; most widely used
Trichloromethane	Good for polar organic compounds
Ethyl acetate	Lower toxicity
Dichloromethane	Good for non-polar to moderately polar compounds

Laboratory Procedure — Separating Funnel

Add crude organic product (aqueous) + organic solvent to separating funnel. Shake; periodically open tap to release vapour pressure.
Allow to stand → two distinct layers form → drain layers separately. The organic layer (the extract) is kept.
Repeat extraction with fresh solvent at least twice. Combine all organic extracts.
Dry with anhydrous ionic salt (e.g. MgSO₄, Na₂SO₄) — takes several hours. Filter off the hydrated salt.
Distil off the organic solvent to recover the pure solid/liquid product.

Selecting the Correct Method (SO 9.5)

Situation	Best Method
Miscible liquids, bp difference >25°C	Simple distillation
Miscible liquids, similar bp	Fractional distillation
High bp liquid or thermally sensitive	Vacuum or steam distillation
Organic compound immiscible with water (high bp)	Steam distillation + separating funnel
Organic product in aqueous solution	Solvent extraction (separating funnel)
Two immiscible liquids (density difference)	Separating funnel only

Interactive Simulations

Explore phase separation concepts through interactive charts, animated apparatus, and live calculations.

Raoult's Law — Vapour-Liquid Equilibrium

Partial & total pressures vs composition for binary mixtures

Liquid line (bubble point) Vapour line (dew point) Ideal (Raoult)

🧮 Live Calculator

Mole fraction x_A:0.50

P_A (partial)

—

P_B (partial)

—

P_total

—

y_A (vapour)

—

Recorded Data

x_A	P_A (mmHg)	P_B (mmHg)	P_total (mmHg)	y_A
No data recorded yet.

Simple Distillation Apparatus

Distillation Log

Time (s)	Flask Temp (°C)	Vapour Temp (°C)	Distillate (mL)	EtOH%
Start distillation to record data.

Controls

Mixture

Heat Intensity

50%

Live Readings STOPPED

Flask Temp	25.0°C
Vapour Temp	25.0°C
Distillate	0.0 mL
Ethanol %	—

🔬 Observation

Select a mixture and press Start to begin.

Positive Azeotrope

Ethanol–Water · min bp = 78.2°C at x = 0.894

Bubble pointDew pointAzeotrope

Negative Azeotrope

HNO₃–Water · max bp = 120.5°C at x = 0.384

Bubble pointDew pointAzeotrope

Ideal Mixture — Fractional Distillation Path

Benzene–Toluene · trace how fractional distillation enriches the more volatile component

Starting x_benz: 0.30

Steam Distillation Apparatus

Controls

Organic Compound

Steam Pressure

101 kPa

Calculated Results

Distillation Temp	—
P°_water	—
P°_organic	—
Mole ratio (water/org)	—
Mass % organic	—

📌 How It Works

Steam Distillation Data

Compound	P_total (kPa)	T_dist (°C)	P_water (mmHg)	P_org (mmHg)	Mass% org
No data recorded yet.

Worked Examples

Step-by-step solutions to CAPE-style questions. Click to expand each example.

Raoult's Law — Partial & Total Vapour Pressure

Raoult's Law · Ideal Solution

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At 25°C, the vapour pressure of pure benzene is 95.1 mmHg and that of pure toluene is 28.4 mmHg. Calculate the partial vapour pressures, total vapour pressure, and mole fraction of benzene in the vapour above a mixture containing a mole fraction of benzene x = 0.60.

Step 1 — Identify known quantities

P°_benzene = 95.1 mmHg, P°_toluene = 28.4 mmHg, x_benzene = 0.60
∴ x_toluene = 1 − 0.60 = 0.40

Step 2 — Apply Raoult's Law to each component

P_benz = P°_benz × x_benz = 95.1 × 0.60 = 57.1 mmHg P_tol = P°_tol × x_tol = 28.4 × 0.40 = 11.4 mmHg

Step 3 — Total vapour pressure (Dalton's Law)

P_total = P_benz + P_tol = 57.1 + 11.4 = 68.5 mmHg

Step 4 — Mole fraction in vapour phase

y_benz = P_benz / P_total = 57.1 / 68.5 = 0.834 Notice: y_benz (0.834) > x_benz (0.60) — the vapour is enriched in benzene (the more volatile component). This is the basis of fractional distillation.

✓ Answers: P_benz = 57.1 mmHg · P_tol = 11.4 mmHg · P_total = 68.5 mmHg · y_benz = 0.834

Steam Distillation — Molar Mass of Nitrobenzene

Steam Distillation · Molar Mass Calculation

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A mixture of nitrobenzene and water distils at 99°C under atmospheric pressure (760 mmHg). The vapour pressure of water at 99°C is 733 mmHg. The distillate contains 40 g of water and 10 g of nitrobenzene. Calculate the molar mass of nitrobenzene.

Step 1 — Find partial pressure of nitrobenzene

P_NB = P_total − P_water = 760 − 733 = 27 mmHg

Step 2 — Set up the molar mass ratio

For steam distillation of immiscible liquids, partial pressures equal molar ratios: P_water / P_NB = n_water / n_NB = (m_water/M_water) / (m_NB/M_NB) 733 / 27 = (40/18) / (10/M_NB)

Step 3 — Solve for M_NB

27.15 = (2.222) / (10/M_NB) 10/M_NB = 2.222 / 27.15 = 0.08185 M_NB = 10 / 0.08185 = 122.2 g mol⁻¹

✓ Molar mass of nitrobenzene ≈ 122 g mol⁻¹ (actual: 123.1 g mol⁻¹ — close agreement)

Steam Distillation — Mass of Phenylamine Obtained

Steam Distillation · SG Textbook Worked Example 2

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A mixture of phenylamine (M = 93 g mol⁻¹) and water distils at 98°C. At 98°C, P°_water = 707 mmHg and P_total = 760 mmHg. The distillate contains 25 g of water. Calculate the mass of phenylamine in the distillate.

Step 1 — Find partial pressure of phenylamine

P_PA = 760 − 707 = 53 mmHg

Step 2 — Apply mole ratio equation

P_water / P_PA = n_water / n_PA 707 / 53 = (25/18) / (m_PA/93) 13.34 = 1.389 / (m_PA/93)

Step 3 — Solve for mass of phenylamine

m_PA/93 = 1.389 / 13.34 = 0.1041 m_PA = 0.1041 × 93 = 9.68 g ≈ 9.7 g

✓ Mass of phenylamine = 9.7 g (per 25 g water in distillate)

Ethanol–Water Azeotrope Analysis

Azeotropes · Fractional Distillation · CE Q1(c)(d)

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(c) What is an azeotrope? (d) Explain why the fractional distillation of a mixture of ethanol and water can only produce a maximum of approximately 96% ethanol by mass and not pure ethanol.

Part (c) — Definition of Azeotrope

An azeotrope (constant boiling mixture) is a mixture of two or more liquids in such a ratio that its composition cannot be changed by distillation. At the azeotropic composition, the liquid and vapour phases have the same composition.

Part (d) — Why 96% is the maximum

Ethanol–water shows a positive deviation from Raoult's Law: the actual vapour pressure exceeds the ideal. This creates a minimum boiling azeotrope at 78.2°C with a composition of 95.6% ethanol by mass (mole fraction ≈ 0.894).

On the T-x,y diagram, as distillation proceeds from a dilute ethanol mixture, the vapour successively becomes richer in ethanol. However, once the azeotrope composition is reached, the liquid and vapour lines meet. The vapour at this point has the same composition as the liquid — no further enrichment can occur by distillation alone. Any distillate beyond this point would carry over both ethanol and water in the fixed azeotrope ratio.

Additional Detail — How to Get Anhydrous Ethanol

To obtain 100% ethanol, the water must be removed by a chemical drying agent:
• Calcium oxide (CaO/quicklime) — reacts with water: CaO + H₂O → Ca(OH)₂
• Anhydrous copper(II) sulfate, silica gel, or molecular sieves
Then the anhydrous ethanol is distilled off from the drying agent.

✓ The ethanol–water system forms a minimum-boiling azeotrope (78.2°C, 95.6% EtOH). At this composition the liquid and vapour are identical, so distillation cannot produce further enrichment.

Partition Coefficient — Single vs Multiple Extractions

Solvent Extraction · Partition Law · CE Q

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A compound X has K_d = 5.0 (ether/water). 0.60 g of X is dissolved in 60 cm³ of water. (a) Calculate the mass extracted in a single extraction using 60 cm³ of ether. (b) Calculate the total mass extracted using TWO extractions each with 30 cm³ ether. (c) State which procedure is more efficient.

Part (a) — Single extraction: 60 cm³ ether

Let mass extracted = m grams. Mass remaining in water = (0.60 − m) g. K_d = [X]_ether / [X]_water = (m/60) / ((0.60−m)/60) = 5.0 m / (0.60 − m) = 5.0 m = 3.0 − 5m → 6m = 3.0 → m = 0.50 g Mass extracted = 0.50 g (83.3%)

Part (b) — First extraction: 30 cm³ ether

Let mass extracted in 1st extraction = m₁. K_d = (m₁/30) / ((0.60−m₁)/60) = 5.0 m₁ × 60 / (30 × (0.60−m₁)) = 5.0 2m₁ / (0.60−m₁) = 5.0 2m₁ = 3.0 − 5m₁ → 7m₁ = 3.0 → m₁ = 0.4286 g Remaining in water = 0.60 − 0.4286 = 0.1714 g

Part (b) — Second extraction: 30 cm³ ether on remaining 0.1714 g

2m₂ / (0.1714−m₂) = 5.0 2m₂ = 0.857 − 5m₂ → 7m₂ = 0.857 → m₂ = 0.1224 g Total extracted = 0.4286 + 0.1224 = 0.5510 g (91.8%)

✓ (a) 0.50 g (83.3%) vs (b) 0.55 g (91.8%) — multiple small extractions are more efficient. Each fresh solvent restores a maximum concentration gradient, pulling more solute across the interface.

Reading a BP–Composition Diagram

Fractional Distillation · T-x,y Diagram · CE Q2(b)(c)

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A boiling point–composition diagram for benzene–toluene at 1 atm shows: pure benzene bp = 80.1°C, pure toluene bp = 110.6°C. (b) At 95°C, read the liquid composition (x_benz) and vapour composition (y_benz). (c) Explain how fractional distillation enriches the vapour in benzene.

Part (b) — Reading the diagram at 95°C

Draw a horizontal tie-line at T = 95°C across the two-phase region:
• The intersection with the liquid (bubble point) curve gives: x_benz ≈ 0.41 • The intersection with the vapour (dew point) curve gives: y_benz ≈ 0.63 Note: y_benz > x_benz — confirming the vapour is richer in benzene (more volatile).

Part (c) — Mechanism of enrichment

Starting from a mixture with x_benz = 0.41:
1. Heat to 95°C → first vapour formed has y_benz ≈ 0.63 (richer in benzene).
2. This vapour rises up the fractionating column and condenses. The condensate (x ≈ 0.63) is now richer in benzene.
3. This new liquid is re-heated → vapour with even higher y_benz forms. A new tie-line at a lower temperature can be drawn.
4. The process repeats at each 'theoretical plate' up the column.
5. After many such equilibria, vapour of almost pure benzene (x → 1, bp → 80.1°C) reaches the top and exits.

The fractionating column acts as many sequential simple distillations — each step enriches the vapour further in the more volatile component.

✓ At 95°C: x_benz ≈ 0.41 (liquid), y_benz ≈ 0.63 (vapour). The column provides repeated liquid–vapour equilibria, each enriching the vapour in benzene.

Selecting the Correct Distillation Method

Method Selection · SO 9.5 · CE Q3(a)

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For each scenario, identify and justify the most appropriate separation technique:
W: Separating water from a dissolved salt.
X: Purifying nitrobenzene (bp 211°C, immiscible with water).
Y: Separating ethanol (bp 78.5°C) from methanol (bp 64.7°C).

Scenario W — Water from dissolved salt

Technique: Simple distillation
The salt is involatile (no vapour pressure). On heating, only water evaporates and is condensed and collected as pure water. The bp difference is effectively infinite. A fractionating column is unnecessary.

Scenario X — Nitrobenzene (bp 211°C)

Technique: Steam distillation
Nitrobenzene is immiscible with water and has a very high boiling point (211°C). Direct distillation at 211°C would risk decomposition. Steam distillation lowers the distillation temperature to ~99°C (well below 211°C), preventing thermal decomposition. The two-layer distillate is separated using a separating funnel.

Scenario Y — Ethanol (78.5°C) from methanol (64.7°C)

Technique: Fractional distillation
The boiling point difference is only 78.5 − 64.7 = 13.8°C — too small for effective simple distillation. A fractionating column with glass beads provides many liquid–vapour equilibria. Methanol (more volatile) distils first, followed by ethanol. Note: since methanol bp < ethanol bp, methanol is more volatile and will be preferentially enriched in the distillate.

✓ W: Simple distillation · X: Steam distillation · Y: Fractional distillation

Steam Distillation — Vapour Pressure from % Composition

Steam Distillation · Back-calculation · CE Q4(b)

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A sample of aniline (phenylamine, M = 93) is steam distilled at 98°C. The distillate contains 25.5% aniline by mass. At 98°C, P_total = 760 mmHg. Calculate the vapour pressure of aniline at 98°C.

Step 1 — Find the mole ratio from mass percentages

In 100 g distillate: mass of aniline = 25.5 g, mass of water = 74.5 g. n_aniline = 25.5 / 93 = 0.2742 mol n_water = 74.5 / 18 = 4.139 mol Mole ratio n_water / n_aniline = 4.139 / 0.2742 = 15.09

Step 2 — Relate mole ratio to vapour pressures

For an immiscible mixture: n_water / n_aniline = P_water / P_aniline P_water / P_aniline = 15.09 Also: P_water + P_aniline = 760 mmHg P_water = 760 − P_aniline (760 − P_aniline) / P_aniline = 15.09

Step 3 — Solve for P_aniline

760 − P_aniline = 15.09 × P_aniline 760 = 16.09 × P_aniline P_aniline = 760 / 16.09 = 47.2 mmHg P_water = 760 − 47.2 = 712.8 mmHg

✓ Vapour pressure of aniline at 98°C ≈ 47 mmHg · P_water ≈ 713 mmHg

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