Solution:

Given:

• Solvent front distance = 10.0 cm (distance solvent traveled)
• Blue spot distance = 8.0 cm from origin
• Red spot distance = 5.0 cm from origin
• Yellow spot distance = 2.0 cm from origin

Find: (a) Rf values, (b) Most polar pigment, (c) Effect of more polar solvent

Part (a): Calculate Rf values

Rf (Retention factor) formula:

$R_f = \frac{\text{distance traveled by component}}{\text{distance traveled by solvent}}$

For blue pigment:

$R_f(\text{blue}) = \frac{8.0 \text{ cm}}{10.0 \text{ cm}} = 0.80$

For red pigment:

$R_f(\text{red}) = \frac{5.0 \text{ cm}}{10.0 \text{ cm}} = 0.50$

For yellow pigment:

$R_f(\text{yellow}) = \frac{2.0 \text{ cm}}{10.0 \text{ cm}} = 0.20$

Answer (a):

$\boxed{R_f(\text{blue}) = 0.80, \quad R_f(\text{red}) = 0.50, \quad R_f(\text{yellow}) = 0.20}$

Summary table:

Part (b): Which pigment is most polar?

Principle of paper chromatography:

Paper = stationary phase (polar cellulose) Solvent = mobile phase

How polarity affects Rf:

Polar compounds:

• Strong attraction to polar paper (stationary phase)
• Stick to paper more
• Move slowly with solvent
• Low Rf value

Nonpolar compounds:

• Weak attraction to polar paper
• Dissolve better in solvent (mobile phase)
• Move quickly with solvent
• High Rf value

Relationship:

$\text{More polar} \rightarrow \text{Lower } R_f$

$\text{Less polar} \rightarrow \text{Higher } R_f$

Comparing our pigments:

Answer (b):

$\boxed{\text{Yellow pigment is most polar}}$

Explanation:

• Yellow has lowest Rf (0.20)
• Traveled least distance
• Strong attraction to polar paper
• Therefore most polar

Part (c): Effect of more polar solvent

Current situation:

• Paper: Polar (stationary phase)
• Solvent: Some polarity (mobile phase)
• Pigments partition between paper and solvent

If solvent becomes MORE polar:

Effect on pigment behavior:

1. More polar solvent better dissolves polar pigments

   • Polar pigments pulled into mobile phase more
   • Less attraction to stationary phase (relatively)

2. All pigments carried farther up paper

   • Better solvation by solvent
   • Move faster/farther

3. All Rf values INCREASE

Specific predictions:

Yellow (most polar):

• Currently Rf = 0.20
• More polar solvent → much better dissolved
• Rf increases significantly
• Example: Might go to Rf ≈ 0.35-0.40

Red (intermediate):

• Currently Rf = 0.50
• Moderately affected
• Rf increases
• Example: Might go to Rf ≈ 0.60-0.65

Blue (least polar):

• Currently Rf = 0.80
• Already moves well
• Slight increase (already near maximum)
• Example: Might go to Rf ≈ 0.85-0.90

Answer (c):

$\boxed{\text{All } R_f \text{ values would INCREASE (pigments travel farther)}}$

Reasoning:

• More polar solvent better dissolves all pigments
• Pigments spend more time in mobile phase
• Less time stuck to stationary phase
• Travel farther up paper
• Higher Rf values

Visual representation:

Current (less polar solvent):

• Solvent front: 10 cm
• Blue spot: 8 cm (Rf = 0.80)
• Red spot: 5 cm (Rf = 0.50)
• Yellow spot: 2 cm (Rf = 0.20)
• Origin: 0 cm

With more polar solvent:

• Solvent front: 10 cm
• Blue spot: 9 cm (Rf = 0.90)
• Red spot: 6.5 cm (Rf = 0.65)
• Yellow spot: 4 cm (Rf = 0.40)
• Origin: 0 cm

Note: Spots still in same order (blue > red > yellow), but all higher

Additional insights:

Why does order stay the same?

• Blue always least polar → always highest Rf
• Yellow always most polar → always lowest Rf
• Changing solvent polarity shifts ALL Rf values but maintains order

Practical implications:

Too polar solvent:

• All Rf values very high (0.8-1.0)
• Poor separation
• Spots close together near top

Too nonpolar solvent:

• All Rf values very low (0.0-0.2)
• Poor separation
• Spots close together near bottom

Optimal solvent:

• Rf values spread out (0.2-0.8)
• Good separation
• Easy to distinguish components

Summary of key concepts:

1. Rf = distance(component) / distance(solvent)

   • Always between 0 and 1
   • Characteristic for each compound (under same conditions)

2. Lower Rf = more polar (for polar stationary phase)

   • Polar compounds stick to polar paper

3. More polar solvent → higher Rf values

   • Better dissolves polar compounds
   • Carries them farther

4. Chromatography separates by polarity

   • Different affinities for stationary vs mobile phases

Solution:

Given:

• Mixture: Ethanol (C₂H₅OH) and water (H₂O)
• BP(ethanol) = 78°C
• BP(water) = 100°C
• Two miscible liquids (form homogeneous solution)

Find: (a) Best separation technique, (b) Which distills first, (c) Why not 100% pure, (d) Better technique

Part (a): Most appropriate separation technique

Analysis of mixture:

Type: Two miscible liquids (form solution)

• Cannot use filtration (both dissolved)
• Cannot use chromatography easily (not good for large amounts)
• Cannot use crystallization (both liquids)

Key property difference: Boiling points

• BP difference = 100°C - 78°C = 22°C
• Moderate difference

Best technique: DISTILLATION

Why distillation?

1. Liquids have different BPs

   • Component with lower BP vaporizes first
   • Can be selectively evaporated and condensed

2. Components are miscible

   • Cannot separate by density/filtration
   • Need to use volatility difference

3. Want to purify/collect both components

   • Distillation allows collection of volatile component
   • Leaves less volatile component in flask

Answer (a):

$\boxed{\text{Distillation (specifically simple or fractional distillation)}}$

Reason: Separates miscible liquids based on boiling point differences. Ethanol (lower BP) vaporizes first, travels to condenser, and is collected. Water (higher BP) remains in flask.

Part (b): Which component distills first?

Principle: Component with lower boiling point vaporizes first

Comparison:

• Ethanol BP: 78°C (lower)
• Water BP: 100°C (higher)

During distillation:

Step 1: Heat mixture

• Temperature rises toward lowest BP
• At ~78°C, ethanol begins to boil

Step 2: Vaporization

• Ethanol vaporizes preferentially
• Vapor is enriched in ethanol (more ethanol than water in vapor)

Step 3: Condensation

• Ethanol vapor enters condenser
• Cools back to liquid
• Collected as distillate

Answer (b):

$\boxed{\text{Ethanol distills over first}}$

Reason: Lower boiling point (78°C < 100°C). Component with lower BP is more volatile and evaporates first.

Part (c): Why can't 100% pure ethanol be obtained?

Key concept: Ethanol-water forms an AZEOTROPE

Azeotrope: Mixture that boils at constant temperature with constant composition

• Vapor has same composition as liquid
• Cannot be separated further by simple distillation

Ethanol-water azeotrope:

• Composition: 95.6% ethanol, 4.4% water (by volume)
• Boiling point: 78.1°C
• This is as pure as you can get by normal distillation

Why azeotrope forms:

Intermolecular forces:

• Ethanol and water form hydrogen bonds with each other
• Ethanol: O-H can H-bond
• Water: O-H can H-bond

H-bonding between ethanol and water:

$C2H5OH ··· H-O-H$

These interactions:

• Stabilize the mixture
• Change vapor pressure behavior
• Create composition where liquid and vapor have same ratio

Distillation behavior:

Starting with dilute ethanol (e.g., 10% ethanol):

1. Heat to ~78°C
2. Vapor enriched in ethanol (more volatile)
3. Collect distillate (higher % ethanol)
4. Repeat: gradually increase ethanol concentration

But at 95.6% ethanol:

• Reach azeotrope
• Vapor composition = liquid composition
• Cannot increase ethanol % further
• No matter how many times you distill!

Answer (c):

\boxed{\text{Ethanol-water forms an azeotrope at 95.6% ethanol}}

Explanation: At this composition, liquid and vapor have identical composition. Further distillation cannot increase purity beyond 95.6%. Strong hydrogen bonding between ethanol and water creates this constant-boiling mixture.

Part (d): Techniques for better separation

To get > 95.6% ethanol (absolute ethanol), use:

Method 1: Add drying agent

• Add anhydrous calcium oxide (CaO) or molecular sieves
• Drying agent chemically reacts with or absorbs water
• Removes water from azeotrope
• Then distill to get pure ethanol

Chemical reaction:

$CaO + H2O \rightarrow Ca(OH)2$

Method 2: Azeotropic distillation

• Add third component (e.g., benzene, cyclohexane)
• Forms new azeotrope with water
• This azeotrope boils at different temperature
• Can distill off water with third component
• Leaves pure ethanol

Method 3: Fractional distillation

• Won't break azeotrope, but...
• Better separation before reaching azeotrope
• Uses fractionating column
• Multiple vaporization-condensation cycles
• Gets to 95.6% more efficiently than simple distillation

For initial question (a) - should specify:

If BP difference is moderate (20-25°C):

• Fractional distillation preferred over simple distillation
• Better separation
• Higher purity

Answer (d):

$\boxed{\text{Fractional distillation (for better separation to azeotrope)}}$ \boxed{\text{+ Drying agent (to break azeotrope and get 100% ethanol)}}

Fractional distillation advantages:

• Fractionating column provides multiple theoretical distillations
• Better separation than simple distillation
• More efficient path to 95.6% purity

To get absolute ethanol (100%):

• Must use chemical method (drying agent)
• Or azeotropic distillation with third component
• Physical distillation alone cannot break azeotrope

Summary comparison:

Key concepts:

1. Choose separation by property difference

   • Different BPs → distillation
   • Different polarities → chromatography/extraction
   • Different sizes → filtration

2. Lower BP → distills first

   • More volatile
   • Evaporates at lower temperature

3. Azeotropes limit distillation

   • Constant-boiling mixtures
   • Same composition in liquid and vapor
   • Cannot be separated further by distillation alone

4. Fractional > Simple distillation

   • Better separation
   • Multiple stages
   • Use when BP difference < ~25°C

5. Break azeotrope chemically

   • Add drying agent
   • Add third component
   • Change IMFs in mixture

Solution:

Given mixture components:

1. Sand (SiO₂): Insoluble in water, nonpolar, high melting point, does not sublime
2. Salt (NaCl): Ionic, very soluble in water, high melting point
3. Naphthalene (C₁₀H₈): Nonpolar organic solid, insoluble in water, sublimes at 80°C

Objective: Separate and isolate ALL THREE components in pure form

Key properties to exploit:

SEPARATION SCHEME:

STEP 1: SUBLIMATION

(i) Technique: Sublimation

(ii) Property exploited: Naphthalene sublimes (solid → gas directly) at 80°C; sand and salt do not sublime

(iii) What is separated: Naphthalene (collected) from sand + salt (remain)

Procedure:

1. Place mixture in evaporating dish
2. Cover with inverted funnel with cold water flask on top
3. Heat gently to ~80°C
4. Naphthalene sublimes (becomes vapor)
5. Vapor contacts cold surface, deposits as solid crystals
6. Collect pure naphthalene from cold surface

Result after Step 1:

• ✓ Pure naphthalene (collected on cold surface)
• Sand + salt mixture (remains in dish)

STEP 2: DISSOLUTION

(i) Technique: Dissolution / Extraction with water

(ii) Property exploited: Salt is soluble in water; sand is not

(iii) What is separated: Salt (dissolves) from sand (does not dissolve)

Procedure:

1. Add distilled water to sand + salt mixture
2. Stir thoroughly to dissolve all salt
3. Result: Aqueous solution of NaCl + undissolved sand

Result after Step 2:

• Aqueous solution containing dissolved NaCl
• Solid sand (undissolved)

STEP 3: FILTRATION

(i) Technique: Filtration

(ii) Property exploited: Particle size - sand particles too large to pass through filter

(iii) What is separated: Sand (trapped) from salt solution (passes through)

Procedure:

1. Pour mixture through filter paper in funnel
2. Sand trapped on filter paper (residue)
3. Salt solution passes through (filtrate)

Result after Step 3:

• Wet sand on filter paper
• Salt solution collected as filtrate

STEP 4: DRYING SAND

(i) Technique: Drying / Evaporation

(ii) Property exploited: Water evaporates; sand does not

(iii) What is separated: Water (evaporates) from sand (remains)

Procedure:

1. Leave sand on filter paper in warm place
2. Or place in drying oven at low temperature (100-110°C)
3. Water evaporates

Result after Step 4:

• ✓ Pure dry sand

STEP 5: EVAPORATION / CRYSTALLIZATION

(i) Technique: Evaporation (or crystallization for better purity)

(ii) Property exploited: Water evaporates; salt does not (non-volatile)

(iii) What is separated: Water (evaporates) from salt (remains)

Procedure:

Method A - Evaporation (faster):

1. Heat salt solution in evaporating dish
2. Water evaporates
3. Salt crystals remain

Method B - Crystallization (purer):

1. Heat salt solution to near boiling
2. Evaporate some water (concentrated solution)
3. Cool slowly
4. Salt crystals form
5. Filter to collect crystals
6. Dry crystals

Result after Step 5:

• ✓ Pure dry salt (NaCl)

COMPLETE FLOW CHART:

Step 1: SUBLIMATION (80°C)

• Input: Sand + Salt + Naphthalene
• Property: Naphthalene sublimes
• Output: NAPHTHALENE (pure) + remaining (Sand + Salt)

Step 2: ADD WATER

• Input: Sand + Salt
• Property: Salt dissolves, sand doesn't
• Output: Sand + Salt solution

Step 3: FILTRATION

• Input: Sand + Salt solution
• Property: Particle size
• Output: SAND (wet, on filter) + Salt solution (filtrate)

Step 4: DRY SAND

• Input: Wet sand
• Property: Water evaporates
• Output: SAND (pure, dry)

Step 5: EVAPORATION

• Input: Salt solution
• Property: Water evaporates
• Output: SALT (pure, dry)

ALTERNATIVE SCHEME (Different order):

Could also start with dissolution:

1. MIXTURE + ADD WATER
2. FILTRATION gives: Salt solution + (Sand + Naphthalene)
3. EVAPORATE salt solution → SALT
4. SUBLIMATION of (Sand + Naphthalene) → NAPHTHALENE

But first scheme is better because:

• Sublimation first removes naphthalene without solvents
• Cleaner separation
• Less contamination risk

DETAILED SUMMARY TABLE:

KEY CONSIDERATIONS:

Order matters:

• Must do sublimation before adding water
• If you add water first, naphthalene might stick to sand (harder to separate)
• Sublimation is cleanest first step

Temperature control:

• Sublimation: ~80°C (not too high or salt might decompose)
• Drying sand: 100-110°C (above water BP)
• Evaporating salt solution: Can heat to boiling

Purity checks:

• Naphthalene: Check melting point (should be 80°C)
• Sand: Rinse with water again to ensure no salt
• Salt: Could recrystallize for higher purity

Common mistakes to avoid: ❌ Filtering before dissolving salt (would trap salt + sand together) ❌ Adding water before sublimation (naphthalene hard to remove when wet) ❌ Not drying sand properly (would have salt contamination from residual water)

ALTERNATIVE TECHNIQUES (if available):

Instead of sublimation, could use:

• Solvent extraction with nonpolar solvent
  • Add hexane or toluene
  • Naphthalene dissolves (nonpolar)
  • Sand and salt don't dissolve
  • Filter, evaporate hexane → naphthalene

Advantage of sublimation over extraction:

• No organic solvents needed
• Environmentally friendlier
• Simpler procedure
• Direct collection of pure solid

FINAL ANSWER:

Complete separation scheme:

1. Sublimation (80°C) → isolate naphthalene
2. Add water → dissolve salt
3. Filtration → separate sand
4. Dry sand (100°C) → pure sand
5. Evaporate water → pure salt

All three components isolated in pure form!

$\boxed{\text{Naphthalene (sublimation) → Sand (filtration + drying) → Salt (evaporation)}}$

Key concepts applied:

• Multiple techniques needed for complex mixture
• Order of operations critical
• Choose technique based on property differences
• Physical properties enable separation (sublimation, solubility, size)