7.1
Partition Coefficient & Solvent Extraction
Partition Coefficient KDWhen a solute distributes between two immiscible solvents, at equilibrium the ratio of concentrations is constant at a given temperature:
KD = [solute in solvent A] / [solute in solvent B] = constant (at fixed T)
Solvent Extraction Principle
Used to extract a compound from one solvent (often water) into another (e.g. ether, chloroform).
Multiple small extractions > one large extraction.
Example: 100 mL water contains 10 g solute; Kᵈ = [ether]/[water] = 4.
One extraction with 100 mL ether:
Let x = mass extracted. [ether] = x/100; [water] = (10-x)/100
Kᵈ = (x/100)/((10-x)/100) = x/(10-x) = 4 → x = 8 g extracted (80%)
Two extractions with 50 mL ether each:
1st: x/(10-x) = (50/100)×4... → x = 6.67 g (66.7% of 10)
Remaining 3.33 g, 2nd extraction: extracts 2.22 g more
Total: 6.67+2.22 = 8.89 g (89%) — BETTER than single 100 mL extraction
Distribution Formula
Mass remaining after n extractions with volume Vs of solvent:
mₙ = m₀ × (Vᵧ/(Vᵧ + Kᵈ·V))ⁿ
where Vᵧ = volume of original solvent (water), V = volume of extraction solvent per extraction
Key insight: n small extractions always extract more than 1 large extraction of same total volume.
7.2
Raoult’s Law & Colligative Properties
Raoult’s LawThe partial vapour pressure of a component in an ideal solution is equal to the mole fraction of that component times its pure vapour pressure:
pᶅ = xᶅ · pᶅ° where xᶅ = mole fraction, pᶅ° = vapour pressure of pure A
Colligative Properties
Properties that depend only on the number of solute particles (not their identity). All arise from the lowering of solvent chemical potential by the solute.
1. Vapour pressure lowering: Δp = xᵲ·pᵧ° (xᵲ = mole fraction of solute)
2. Boiling point elevation: ΔTᵬ = Kᵬ·m (Kᵬ = ebullioscopic constant; m = molality mol/kg)
3. Freezing point depression: ΔTᵮ = Kᵮ·m (Kᵮ = cryoscopic constant)
4. Osmotic pressure: π = cRT (c = molar concentration; R = 8.314 J/mol K)
Van’t Hoff factor i: for electrolytes, replace m with i·m
NaCl: i = 2 (Na⁺ + Cl⁻); CaCl₂: i = 3; glucose: i = 1
Key Constants
| Solvent | Kᵬ (°C·kg/mol) | Kᵮ (°C·kg/mol) | Normal BP (°C) | Normal FP (°C) |
|---|---|---|---|---|
| Water | 0.512 | 1.86 | 100 | 0 |
| Benzene | 2.53 | 5.12 | 80.1 | 5.5 |
| Cyclohexane | 2.79 | 20.2 | 80.7 | 6.5 |
Worked Examples
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Example 1: Boiling point elevation
10 g urea CO(NH₂)₂ (M=60) in 500 g water.
m = (10/60)/(0.500) = 0.333 mol/kg
ΔTᵬ = Kᵬ·m = 0.512 × 0.333 = 0.171°C New BP = 100.17°C
Example 2: Freezing point depression
5 g NaCl (M=58.5) in 100 g water. i=2.
m = (5/58.5)/0.100 = 0.855 mol/kg
ΔTᵮ = Kᵮ·i·m = 1.86 × 2 × 0.855 = 3.18°C New FP = −3.18°C
Example 3: Osmotic pressure
0.1 mol/L glucose at 25°C (298 K):
π = cRT = 0.1 × 8.314 × 298 = 247.7 kPa ≈ 2.45 atm
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Exercises
- The partition coefficient of compound X between ether and water is Kᵈ = 5 (ether:water). 20 g of X is dissolved in 200 mL water. Calculate mass extracted by (a) one extraction with 200 mL ether; (b) two extractions with 100 mL ether each.(a) Kᵈ = (x/200)/((20-x)/200) = x/(20-x) = 5 → 5(20-x)=x → 100=6x → x = 16.67 g (83.3%). (b) 1st: x/(20-x) = (100/200)×5 = 2.5 → x = 2.5(20-x) = 50-2.5x → 3.5x=50 → x=14.29 g. Remaining = 5.71 g. 2nd: y/(5.71-y)=2.5 → 3.5y=14.28 → y=4.08 g. Total = 14.29+4.08 = 18.37 g (91.9%). Multiple extractions are more efficient.
- Calculate the boiling point of a solution of 18 g glucose (M=180) in 500 g water. Kᵬ(water) = 0.512 °C·kg/mol.mol glucose = 18/180 = 0.10 mol. m = 0.10/0.500 = 0.20 mol/kg. ΔTᵬ = 0.512 × 0.20 = 0.1024°C. BP = 100 + 0.102 = 100.10°C.
- What is the freezing point of a solution of 5.85 g NaCl (M=58.5) in 250 g water? Kᵮ = 1.86, assume full dissociation.mol NaCl = 5.85/58.5 = 0.100 mol. m = 0.100/0.250 = 0.400 mol/kg. i = 2 (NaCl → Na⁺ + Cl⁻). ΔTᵮ = 1.86 × 2 × 0.400 = 1.488°C. FP = 0 − 1.488 = −1.49°C.
- A 0.5 mol/L sucrose solution at 37°C has an osmotic pressure of: π = cRT. Calculate π in kPa.π = cRT = 0.5 × 8.314 × (37+273) = 0.5 × 8.314 × 310 = 1288.7 kPa ≈ 1289 kPa ≈ 12.7 atm. This is why red blood cells (internal ≈0.3 mol/L solutes) burst in pure water (hypotonic) and shrink in concentrated salt solution (hypertonic).
- State and explain two practical applications of colligative properties.1. Anti-freeze (freezing point depression): ethylene glycol added to car radiator water prevents freezing in winter. At 50% v/v ethylene glycol, FP is lowered to about −37°C. Also raises BP slightly, preventing summer boil-over. 2. Saline drip (osmotic pressure): hospital IV saline (0.9% NaCl, isotonic) has the same osmotic pressure as blood plasma — no net water movement across red blood cell membranes. Pure water IV would cause haemolysis (red cells burst by osmosis); concentrated saline would cause crenation (cells shrink). 3. (Also valid): Determination of molar mass from colligative properties; road salting for ice melting (FP depression).
- Explain why electrolytes show greater colligative effects than non-electrolytes of the same molality.Colligative properties depend on the NUMBER of solute particles. Electrolytes dissociate into ions in solution, increasing the total number of particles. Van’t Hoff factor i > 1 for electrolytes. Example: 0.1 mol NaCl → 0.1 mol Na⁺ + 0.1 mol Cl⁻ = 0.2 mol particles (i=2). 0.1 mol glucose (non-electrolyte) = 0.1 mol particles (i=1). Therefore NaCl causes twice the freezing point depression of glucose at the same molality. CaCl₂ (i=3) causes three times the effect.
🧠
Quiz — 25 Questions
Unit 7: Solvent Extraction & Colligative Properties
25 QsQ1
Raoult's Law states that:
pᶅ = xᶅ · pᶅ°. In an ideal solution, each component's partial vapour pressure is proportional to its mole fraction. Non-ideal solutions deviate positively (weaker interactions than pure) or negatively (stronger interactions, e.g., H-bonds between unlike molecules).
Q2
The partition coefficient Kᵈ = [ether]/[water] = 3 means:
Kᵈ = [solute in ether]/[solute in water] = 3 means at equilibrium, the concentration in ether is 3 times that in water. High Kᵈ → compound prefers ether → easily extracted from water. Low Kᵈ → compound prefers water → needs many extractions or different solvent.
Q3
Multiple small extractions are more efficient than one large extraction because:
Mathematical proof: mₙ = m₀(Vᵧ/(Vᵧ+KᵈV))ⁿ. Smaller V per extraction raises (Vᵧ/(Vᵧ+KᵈV)) but more extractions (higher n) lowers mₙ more. Geometric series argument: each extraction takes a constant fraction, so n fractions always give more than 1 fraction of same total volume.
Q4
A colligative property depends on:
Colligative = Latin 'collected together'. The property depends only on the number of particles (moles). Same freezing point depression for 0.1 mol glucose, 0.1 mol ethanol, 0.1 mol sucrose — all give ΔTᵮ = 1.86 × 0.1 = 0.186°C regardless of identity.
Q5
ΔTᵬ = Kᵬ · m. If Kᵬ for water = 0.512 and m = 1.0 mol/kg, the new BP is:
ΔTᵬ = 0.512 × 1.0 = 0.512°C elevation. New BP = 100 + 0.512 = 100.512°C. Boiling point is always elevated by solute. Dissolving anything in water raises its boiling point.
Q6
The van’t Hoff factor i for MgCl₂ (assuming complete dissociation) is:
MgCl₂ → Mg²⁺ + 2Cl⁻ → 3 ions. i = 3. The colligative effect is 3× that of a non-electrolyte of the same molality. ΔTᵮ(MgCl₂) = Kᵮ · 3 · m.
Q7
Osmotic pressure π = cRT. A 0.2 mol/L solution at 300 K has π =:
π = cRT = 0.2 mol/L × 8.314 J/(mol·K) × 300 K = 498.8 J/L = 498.8 kPa ≈ 4.92 atm. (1 J/L = 1 kPa). Check: 0.2 mol/L × 0.08206 L·atm/mol·K × 300 K = 4.92 atm ≈ 499 kPa. Consistent.
Q8
Road salting in winter works because:
NaCl (i=2) or CaCl₂ (i=3) dissolved in thin water film on road surface increases solute particle count → freezing point depression ΔTᵮ = Kᵮ·i·m. Water remains liquid at temperatures below 0°C. CaCl₂ is more effective (i=3) and works to lower temperatures.
Q9
Vapour pressure lowering by a non-volatile solute is:
Δp = xᵲ·pᵧ° (Raoult's Law for non-volatile solute). xᵲ = mole fraction of solute = nᵲ/(nᵲ+nᵧ). Solute molecules occupy surface sites, reducing the fraction of solvent molecules at the surface → fewer escape to vapour phase → lower vapour pressure.
Q10
Which property is NOT a colligative property?
Colour depends on the identity of the solute (chromophore, electronic structure), not the number of particles. Osmotic pressure, BP elevation, FP depression, and vapour pressure lowering all depend only on particle count — colligative.
Q11
The ebullioscopic constant Kᵬ for water (0.512 °C·kg/mol) is smaller than for benzene (2.53). This means:
Larger Kᵬ for benzene: 1 mol/kg solute in benzene raises BP by 2.53°C vs only 0.512°C in water. Kᵬ = RTᵬ²Mᵧ/(1000ΔHᵗᵧᵎᵖ) — proportional to BP squared and inversely proportional to enthalpy of vaporisation. Water has very high ΔHᵗᵧᵎᵖ (40.7 kJ/mol) → small Kᵬ.
Q12
Molar mass of an unknown compound can be determined from colligative properties because:
Rearranging: Mᵣ = Kᵬ · wᵲ / (ΔTᵬ · wᵧ) where w = mass (g) and wᵧ in kg. By measuring ΔTᵬ precisely (to 0.001°C) and knowing Kᵬ, Mᵣ of any solute can be found. Used historically to determine molar masses of polymers and biomolecules.
Q13
At 25°C, Kᵈ(ether/water) for compound Y = 2. To extract 90% of Y from 100 mL water, you need:
mₙ/m₀ = (Vᵧ/(Vᵧ+KᵈV))ⁿ. For 90% extraction, mₙ/m₀ = 0.10. With one 100 mL ether: (100/(100+2×100))¹ = 100/300 = 0.333 → 66.7% extracted. Need multiple small extractions. Three 50 mL ether: (100/(100+2×50))³ = (100/200)³ = 0.125 → 87.5% extracted (close to 90%).
Q14
If Kᵮ for cyclohexane = 20.2 °C·kg/mol, what is the FP depression for 1 mol/kg naphthalene in cyclohexane?
ΔTᵮ = Kᵮ·m = 20.2 × 1.0 = 20.2°C. Cyclohexane has large Kᵮ (20.2) — useful for molar mass determination because a small amount of solute gives a large, easily measurable FP depression. Normal FP of cyclohexane = 6.5°C; new FP = 6.5 − 20.2 = −13.7°C.
Q15
An isotonic saline solution has the same osmotic pressure as blood because:
Isotonic (same osmotic pressure, ~0.9% NaCl = ~308 mOsm/L) — matches blood plasma osmolarity. No net water movement across cell membranes (no osmosis). Hypotonic solution → water enters cells → haemolysis. Hypertonic → water leaves → crenation.
Q16
The boiling point of a solvent is elevated by a non-volatile solute because:
BP elevation and VP lowering are linked: solute lowers VP (Raoult's Law). At normal BP (100°C for water), VP is now below 101.3 kPa. Higher temperature is needed until VP equals atmospheric pressure again. ΔTᵬ = Kᵬ·m accounts for this.
Q17
For a 0.1 mol/kg solution of CaCl₂ in water (i=3, Kᵮ=1.86):
ΔTᵮ = Kᵮ · i · m = 1.86 × 3 × 0.1 = 0.558°C. Ca²⁺ + 2Cl⁻ = 3 ions (i=3). FP = 0 − 0.558 = −0.558°C.
Q18
Reverse osmosis (water purification) works by:
Reverse osmosis: apply P > π (osmotic pressure of seawater ≈2.7 MPa/27 atm). This forces solvent (water) through the membrane from high solute side to low solute side — against natural osmotic direction. Widely used for seawater desalination (Middle East, 10 million m³/day globally).
Q19
If Δp/p° = 0.02 for a solution of 10 g solute in 90 g water (M=18), what is the molar mass of the solute?
xᵲ = Δp/p° = 0.02. Moles water = 90/18 = 5. xᵲ = nᵲ/(nᵲ+5) = 0.02 → nᵲ = 0.02(nᵲ+5) → 0.98nᵲ = 0.1 → nᵲ = 0.102 mol. Mᵣ = 10/0.102 = 98 g/mol ≈ 100 g/mol.
Q20
An anti-freeze mixture must work down to −20°C. What molality of ethylene glycol is needed? (Kᵮ=1.86)
ΔTᵮ = 20°C = Kᵮ·m (i=1 for ethylene glycol). m = 20/1.86 = 10.75 mol/kg. At 1 kg water: 10.75 mol × 62 g/mol = 667 g ethylene glycol. That’s ~40% by mass. Typical car anti-freeze is 30–50% ethylene glycol/water.
Q21
The osmotic pressure of blood is ~7.7 atm at 37°C. The effective molar concentration of blood solutes is:
c = π/RT = (7.7 atm)/(0.08206 L·atm/mol·K × 310 K) = 7.7/25.44 = 0.303 mol/L ≈ 0.3 mol/L. This is why isotonic saline is 0.9% NaCl (NaCl i=2, so 0.154 mol/L NaCl × 2 = 0.308 mol/L particles ≈ 0.303 mol/L blood).
Q22
Positive deviation from Raoult’s Law occurs when:
Positive deviation: VP > Raoult prediction. Occurs when unlike (solute-solvent) interactions are weaker than like (solvent-solvent) interactions → solvent molecules escape more easily from mixture than from pure solvent. Example: ethanol-water mixtures (minimum boiling azeotrope at 95.6% ethanol).
Q23
The van’t Hoff factor i is less than expected for strong electrolytes in concentrated solution because:
In concentrated strong electrolyte solutions, cations and anions associate into ion pairs (electrostatic attraction). An ion pair acts as ONE particle, not two. So effective i < theoretical. Measured: NaCl at 0.1 mol/kg gives i ≈ 1.87 (not 2.0) due to partial ion pairing. At infinite dilution i→2.
Q24
Which has the lowest freezing point? (all at 0.1 mol/kg)
ΔTᵮ = Kᵮ·i·m. Largest i gives largest ΔTᵮ and lowest FP. CaCl₂ i=3 → ΔTᵮ = 1.86×3×0.1 = 0.558°C → FP = −0.558°C. NaCl: −0.372°C. Glucose/urea i=1: −0.186°C. Ranking lowest FP: CaCl₂ > NaCl > glucose = urea.
Q25
Osmosis is:
Osmosis: solvent moves through a semi-permeable membrane (passes water but not solute) from the more dilute side (higher water chemical potential) to the more concentrated side (lower water potential). Driven by thermodynamics (equalising chemical potential). Continues until osmotic pressure difference is reached.
Unit 7 Quiz — Solvents & Colligative Properties (25 Questions)
Select one answer eachQ1
The partition coefficient (Kpc) is defined as:
Kpc = [solute]_A / [solute]_B at equilibrium, at constant temperature. It is dimensionless and constant.
Q2
For efficient liquid-liquid extraction, it is better to use:
Multiple extractions with small volumes remove more solute than one large volume — a consequence of the partition equilibrium.
Q3
Raoult's Law states that the vapour pressure of a solvent above a solution is:
pA = xA × p°A. Adding solute reduces the mole fraction of solvent (xA < 1), lowering its vapour pressure.
Q4
A colligative property depends on:
Colligative properties (bp elevation, fp depression, osmotic pressure) depend only on particle number, not what the particles are.
Q5
Boiling point elevation is expressed as:
ΔTb = Kb × m × i, where Kb = ebullioscopic constant, m = molality, i = van't Hoff factor (for electrolytes).
Q6
Freezing point depression is used practically in:
Salt on roads lowers the fp of water below 0°C. Antifreeze (ethylene glycol) in radiators prevents freezing in winter.
Q7
The van't Hoff factor (i) for NaCl dissolving in water is approximately:
NaCl → Na⁺ + Cl⁻: 2 ions per formula unit. i ≈ 2 (slightly less due to ion pairing at higher concentrations).
Q8
Osmotic pressure (π) is given by:
π = MRT (van't Hoff equation for dilute solutions). M = molarity, R = gas constant, T = temperature in K.
Q9
Osmosis is the movement of solvent molecules:
Solvent moves through a semipermeable membrane from dilute solution (high water conc.) to concentrated solution.
Q10
Reverse osmosis is used to:
Applied pressure > π forces water from seawater through a membrane, leaving salt behind — desalination.
Q11
The molar mass of an unknown solute can be determined using freezing point depression by:
ΔTf = Kf × m = Kf × (moles solute / kg solvent). Rearranging: moles = ΔTf × kg solvent / Kf, then M = mass/moles.
Q12
An ideal solution follows Raoult's Law perfectly when:
Ideal: ΔHmix = 0, ΔVmix = 0. All intermolecular forces are equal — pure Raoult's Law behaviour throughout.
Q13
A positive deviation from Raoult's Law occurs when:
If A–B forces < A–A and B–B, molecules escape more easily → higher vapour pressure than predicted by Raoult's Law.
Q14
Which pair shows a negative deviation from Raoult's Law?
Acetone C=O accepts H-bond from CHCl₃ — stronger interactions than in pure components → lower vapour pressure.
Q15
The ebullioscopic constant Kb depends on:
Kb is a property of the solvent: Kb = RT²M/(1000 × ΔHvap). Water: Kb = 0.512 K kg mol⁻¹.
Q16
Isotonic solutions have:
Isotonic solutions have equal π. IV saline (0.9% NaCl) is isotonic with blood — prevents cell shrinking or swelling.
Q17
When red blood cells are placed in a hypotonic solution they:
Hypotonic = lower solute than cell interior. Water enters cells by osmosis → cells swell → lysis.
Q18
The distribution ratio in solvent extraction depends on:
Kpc is an equilibrium constant — depends on solute–solvent interactions and temperature, not volumes.
Q19
Steam distillation is used to extract plant oils because:
Steam + oil mixture: P_total = P_steam + P_oil. Mixture boils below 100°C — avoids thermal decomposition of heat-sensitive compounds.
Q20
Cryoscopic constant Kf for water is 1.86 K kg mol⁻¹. A 1 mol/kg glucose solution depresses freezing point by:
ΔTf = Kf × m × i = 1.86 × 1 × 1 = 1.86°C (glucose is non-electrolyte, i = 1).
Q21
Which has the greater osmotic pressure: 0.1 mol/L NaCl or 0.1 mol/L glucose?
NaCl gives ~2× particles (Na⁺ + Cl⁻) vs 1× for glucose. π = MRTi: NaCl has π ≈ 2× that of glucose at same concentration.
Q22
The term 'molality' (m) is defined as:
Molality = n(solute) / mass(solvent) in kg. Unlike molarity, it does not change with temperature (no volume term).
Q23
Henry's Law states that the solubility of a gas in a liquid is:
c = kH × p. Higher gas pressure → more gas dissolved. Explains why carbonated drinks fizz when opened (pressure drops).
Q24
Dialysis uses a semipermeable membrane to separate:
Small ions/molecules pass through pores in the membrane; large molecules (proteins, polymers) cannot — separation by size.
Q25
Antifreeze (ethylene glycol) works by:
Ethylene glycol is a non-volatile solute — lowers fp of water (ΔTf = Kf × m × i). Does not react with water.
📝
Unit Test — 50 marks
Section A
30 marksQ1 [5]
State Raoult's Law and derive expressions for: (a) vapour pressure lowering; (b) boiling point elevation; (c) freezing point depression. Define all symbols. [5]
Raoult's Law: pᵧ = xᵧpᵧ° (partial VP of solvent = mole fraction × pure VP).
(a) VP lowering: pᵧ° − pᵧ = pᵧ°(1-xᵧ) = xᵲpᵧ°. So Δp = xᵲpᵧ° (xᵲ = mole fraction of solute).
(b) BP elevation: ΔTᵬ = Kᵬ·m. Kᵬ = ebullioscopic constant (°C·kg/mol); m = molality (mol/kg solvent). For electrolytes: ΔTᵬ = i·Kᵬ·m.
(c) FP depression: ΔTᵮ = Kᵮ·m. Kᵮ = cryoscopic constant. FP = 0 − ΔTᵮ for water. For electrolytes: ΔTᵮ = i·Kᵮ·m. Both ΔTᵬ and ΔTᵮ arise because the solute lowers solvent chemical potential: the liquid phase becomes more stable relative to vapour (raises BP) and relative to solid (lowers FP).
(a) VP lowering: pᵧ° − pᵧ = pᵧ°(1-xᵧ) = xᵲpᵧ°. So Δp = xᵲpᵧ° (xᵲ = mole fraction of solute).
(b) BP elevation: ΔTᵬ = Kᵬ·m. Kᵬ = ebullioscopic constant (°C·kg/mol); m = molality (mol/kg solvent). For electrolytes: ΔTᵬ = i·Kᵬ·m.
(c) FP depression: ΔTᵮ = Kᵮ·m. Kᵮ = cryoscopic constant. FP = 0 − ΔTᵮ for water. For electrolytes: ΔTᵮ = i·Kᵮ·m. Both ΔTᵬ and ΔTᵮ arise because the solute lowers solvent chemical potential: the liquid phase becomes more stable relative to vapour (raises BP) and relative to solid (lowers FP).
Q2 [5]
2.5 g of a non-volatile solute dissolved in 50.0 g of cyclohexane (Kᵮ=20.2 °C·kg/mol) lowers the FP by 5.05°C. Calculate the molar mass of the solute. [5]
ΔTᵮ = Kᵮ·m → m = ΔTᵮ/Kᵮ = 5.05/20.2 = 0.250 mol/kg. m = moles/kg solvent → moles = 0.250 × 0.0500 kg = 0.01250 mol. Mᵣ = mass/mol = 2.5/0.01250 = 200 g/mol. (Identity check: naphthalene C⁶₀H⁸ has Mᵣ = 128; glucose has 180; could be something like decalin or a lipid.)
Q3 [5]
Kᵈ(DCM/water) = 6 for compound Z. 15 g Z is dissolved in 150 mL water. Calculate mass extracted by: (a) one extraction with 150 mL DCM; (b) three extractions with 50 mL DCM each. Show all working. [5]
(a) Single 150 mL extraction: x/(15-x) = Kᵈ·(V/Vᵧ) if Kᵈ is in same units... Using Kᵈ = cᵈᴨᴯ/cᵧᵉᵗᵗᴰᵉ = (x/150)/((15-x)/150) = x/(15-x) = 6 → x = 6(15-x) = 90-6x → 7x=90 → x=12.86 g (85.7% extracted).
(b) Three 50 mL extractions. m remaining = m₀·(Vᵧ/(Vᵧ+KᵈV))ⁿ = 15·(150/(150+6×50))³ = 15·(150/450)³ = 15·(1/3)³ = 15·(1/27) = 0.556 g. Mass extracted = 15 − 0.556 = 14.44 g (96.3%). Three extractions extract significantly more than one.
(b) Three 50 mL extractions. m remaining = m₀·(Vᵧ/(Vᵧ+KᵈV))ⁿ = 15·(150/(150+6×50))³ = 15·(150/450)³ = 15·(1/3)³ = 15·(1/27) = 0.556 g. Mass extracted = 15 − 0.556 = 14.44 g (96.3%). Three extractions extract significantly more than one.
Q4 [5]
Calculate the osmotic pressure of a 0.15 mol/L NaCl solution at 37°C. Assume i = 2. Is this solution isotonic with blood (osmotic pressure ≈ 785 kPa)? [5]
Effective concentration = i × c = 2 × 0.15 = 0.30 mol/L. π = cRT = 0.30 × 8.314 × (37+273) = 0.30 × 8.314 × 310 = 773 kPa. Blood osmotic pressure ≈ 785 kPa. Our answer 773 kPa is very close (within 2%). The 0.15 mol/L NaCl solution is approximately isotonic with blood — confirming that 0.9% (w/v) NaCl (= 0.154 mol/L NaCl) is indeed the isotonic saline used in IV drips.
Q5 [5]
Explain why: (a) adding salt to pasta water raises its boiling point slightly; (b) car engine anti-freeze raises the boiling point AND lowers the freezing point; (c) dialysis machines work on the principle of osmosis. [5]
(a) NaCl dissolves: Na⁺ + Cl⁻ ions (i=2). ΔTᵬ = Kᵬ·i·m → BP elevated. For typical pinch of salt (~5 g NaCl per litre water): m = (5/58.5)/1 = 0.085 mol/kg; ΔTᵬ = 0.512×2×0.085 = 0.087°C. Very small — not culinarily significant! (b) Ethylene glycol (non-electrolyte, i=1): depresses FP (ΔTᵮ = Kᵮ·m, prevents freezing in winter) AND elevates BP (ΔTᵬ = Kᵬ·m, prevents boil-over in summer). Both effects are colligative — the same added solute achieves both. (c) Dialysis: patient's blood (high in urea, creatinine, excess electrolytes) flows on one side of a semi-permeable membrane; dialysis fluid (controlled composition) flows on other side. Small waste molecules (urea, K⁺) diffuse through membrane down concentration gradient. Large molecules (proteins) cannot pass. Blood returned purified — same principle as kidneys (filtration then reabsorption).
Q6 [5]
A solution of 1.8 g of an unknown polymer dissolved in 100 mL water has an osmotic pressure of 0.24 kPa at 25°C. Calculate the molar mass. Why is osmometry preferred over cryoscopy for polymer molar mass determination? [5]
π = cRT → c = π/(RT) = 240 Pa/(8.314 J/mol·K × 298 K) = 240/2477.6 = 0.09687 mol/m³ = 9.687×10⁻⁵ mol/L. Moles = 9.687×10⁻⁵ × 0.100 L = 9.687×10⁻⁶ mol. Mᵣ = 1.8 g / 9.687×10⁻⁶ mol = 1.859×10⁵ g/mol ≈ 1.86 × 10⁵ g/mol (186,000 g/mol = 186 kDa). This is a large polymer! Osmometry is preferred for polymers because: ΔTᵮ = Kᵮ·m would be tiny (0.186/186000 = 10⁻⁶ °C/g/L — impossible to measure). Osmotic pressure π = cRT can be measured precisely even for very low concentrations (very sensitive barometer/manometer). Therefore osmometry is the colligative method of choice for high-molar-mass polymers and proteins.
Section B
20 marksQ7 [10]
(a) Derive the formula for molar mass determination using boiling point elevation. [3]
(b) A student dissolves 3.0 g of an organic acid in 200 g of benzene (Kᵬ=2.53) and measures ΔTᵬ = 0.633°C. She then dissolves the same mass in water and measures ΔTᵬ = 0.256°C using Kᵬ(water)=0.512. Calculate the apparent molar mass in each solvent. Account for the difference. [7]
(a) ΔTᵬ = Kᵬ·m = Kᵬ·(wᵲ/Mᵣ)/(wᵧ/1000) where w is mass in grams, wᵧ in grams. Rearranging: Mᵣ = Kᵬ·wᵲ·1000/(ΔTᵬ·wᵧ) [3]
(b) In benzene: Mᵣ = Kᵬ·wᵲ·1000/(ΔTᵬ·wᵧ) = 2.53×3.0×1000/(0.633×200) = 7590/126.6 = 59.95 ≈ 60 g/mol.
In water: Mᵣ = 0.512×3.0×1000/(0.256×200) = 1536/51.2 = 30 g/mol.
Account for difference: Apparent M in benzene (60) is twice that in water (30). In water, the acid dissociates: HA ⇋ H⁺ + A⁻ (i ≈ 2 for strong acid, or partial ionisation for weak acid). More particles per mole of solute in water → smaller apparent M. In benzene (non-polar, non-ionising): organic acids form hydrogen-bonded dimers: 2HA ⇋ (HA)₂. Dimer has M=2×30=60 — consistent! So actual Mᵣ of the acid monomer = 30 g/mol (likely formic acid HCOOH, M=46... actually would be a better fit if M=30 is not a common acid — perhaps the data gives acetic acid M=60 in benzene, M=30 apparent in water due to ionisation, actual M=60 for acetic acid, i≉2 in water). The key principle: dimerisation in non-polar solvents doubles apparent M; ionisation in water halves apparent M.
(b) In benzene: Mᵣ = Kᵬ·wᵲ·1000/(ΔTᵬ·wᵧ) = 2.53×3.0×1000/(0.633×200) = 7590/126.6 = 59.95 ≈ 60 g/mol.
In water: Mᵣ = 0.512×3.0×1000/(0.256×200) = 1536/51.2 = 30 g/mol.
Account for difference: Apparent M in benzene (60) is twice that in water (30). In water, the acid dissociates: HA ⇋ H⁺ + A⁻ (i ≈ 2 for strong acid, or partial ionisation for weak acid). More particles per mole of solute in water → smaller apparent M. In benzene (non-polar, non-ionising): organic acids form hydrogen-bonded dimers: 2HA ⇋ (HA)₂. Dimer has M=2×30=60 — consistent! So actual Mᵣ of the acid monomer = 30 g/mol (likely formic acid HCOOH, M=46... actually would be a better fit if M=30 is not a common acid — perhaps the data gives acetic acid M=60 in benzene, M=30 apparent in water due to ionisation, actual M=60 for acetic acid, i≉2 in water). The key principle: dimerisation in non-polar solvents doubles apparent M; ionisation in water halves apparent M.
Q8 [10]
Discuss the principles of solvent extraction. Include: the definition of partition coefficient, why multiple extractions are preferred, how pH affects extraction of acidic/basic solutes, and two industrial applications. [10]
Partition coefficient Kᵈ = [solute]ᵧᵂᵗᴧᵗᵗᴷᴹ/[solute]ᵧᵂᵗᴧᵗᴯᵉᵗᴰ (ratio of equilibrium concentrations in two immiscible solvents at constant T). High Kᵈ → compound prefers organic phase → efficiently extracted. Low Kᵈ → compound prefers water → multiple extractions or different solvent needed.
Multiple extractions: m remaining = m₀(Vᵧ/(Vᵧ+KᵈV))ⁿ. For n extractions of volume V each, more solute is recovered than a single extraction of n×V. Mathematical reason: each extraction removes a constant fraction f = KᵈV/(Vᵧ+KᵈV); after n extractions, fraction remaining = (1−f)ⁿ → exponential decay, always lower than single large extraction.
pH effects: acidic compounds (e.g. RCOOH, pKa~5): at low pH (acid conditions), mainly in protonated (RCOOH) form — non-polar — partitions into organic solvent (high Kᵈ in organic/water). At high pH (alkaline), deprotonated to RCOO⁻ — ionic — stays in water (Kᵈ very low). To extract an acid from a mixture: acidify first, then extract with ether. Basic compounds (RNH₂, amines, pKa~10): at high pH (base), uncharged form — extracts into organic phase. At low pH, protonated (RNH₃⁺) — ionic — stays in water. This pH control allows selective extraction of acidic, basic, and neutral compounds from the same mixture (a classic separation scheme).
Industrial applications: (1) Liquid-liquid extraction in pharmaceuticals: extraction of antibiotics (e.g., penicillin) from fermentation broth using butyl acetate at controlled pH. Penicillin is extracted as an acid at low pH, then back-extracted into water at high pH to purify and concentrate. (2) Hydrometallurgy (copper extraction): Cu²⁺ is selectively extracted from dilute sulfate leach liquors using organic chelating agents (e.g., LIX reagents dissolved in kerosene). Strip back with concentrated H₂SO₄ → concentrated CuSO₄ → electrolysis → pure copper. SX-EW (solvent extraction-electrowinning) process: produces ~20% of world copper.
Multiple extractions: m remaining = m₀(Vᵧ/(Vᵧ+KᵈV))ⁿ. For n extractions of volume V each, more solute is recovered than a single extraction of n×V. Mathematical reason: each extraction removes a constant fraction f = KᵈV/(Vᵧ+KᵈV); after n extractions, fraction remaining = (1−f)ⁿ → exponential decay, always lower than single large extraction.
pH effects: acidic compounds (e.g. RCOOH, pKa~5): at low pH (acid conditions), mainly in protonated (RCOOH) form — non-polar — partitions into organic solvent (high Kᵈ in organic/water). At high pH (alkaline), deprotonated to RCOO⁻ — ionic — stays in water (Kᵈ very low). To extract an acid from a mixture: acidify first, then extract with ether. Basic compounds (RNH₂, amines, pKa~10): at high pH (base), uncharged form — extracts into organic phase. At low pH, protonated (RNH₃⁺) — ionic — stays in water. This pH control allows selective extraction of acidic, basic, and neutral compounds from the same mixture (a classic separation scheme).
Industrial applications: (1) Liquid-liquid extraction in pharmaceuticals: extraction of antibiotics (e.g., penicillin) from fermentation broth using butyl acetate at controlled pH. Penicillin is extracted as an acid at low pH, then back-extracted into water at high pH to purify and concentrate. (2) Hydrometallurgy (copper extraction): Cu²⁺ is selectively extracted from dilute sulfate leach liquors using organic chelating agents (e.g., LIX reagents dissolved in kerosene). Strip back with concentrated H₂SO₄ → concentrated CuSO₄ → electrolysis → pure copper. SX-EW (solvent extraction-electrowinning) process: produces ~20% of world copper.