Solutions · Lesson 11 of 11
Chapter Summary and Practice
“Connect solution composition, molecular interactions, phase behaviour and particle counting through an integrated review and fully answered problems.”
• Choose an appropriate concentration measure and reference amount. • Connect molecular interactions with solubility and vapour behaviour. • Select and justify a colligative equation with its assumptions. • Integrate composition, phase-equilibrium and particle-count calculations.
Follow a dissolved particle through the chapter
A solution begins with components and interactions. Composition tells us how much of each component is present; equilibrium tells us how those particles distribute between phases. Once particles are dispersed, their number relative to the solvent changes the solvent’s escaping tendency. That single connection explains why ideal non-volatile solutes lower vapour pressure, shift boiling and freezing conditions, and cause osmotic transfer across a selective membrane.
The order of reasoning matters. First identify what has been measured: mass, solvent mass, solution volume, pressure, temperature difference or composition. Then identify the chemical entities present. Finally choose the relationship that connects those quantities under a defensible model. A formula chosen only because it contains the available numbers can give a plausible-looking answer with the wrong physical meaning.
Composition is the starting point
| Measure | Definition or calculation | Reference amount and caution |
|---|---|---|
| Mass percent | 100 × solute mass / solution mass | Solution includes both solvent and solute. |
| Volume percent | 100 × component volume / final solution volume | State temperature; component volumes may not add. |
| Mass/volume percent | g solute per 100 mL final solution | Do not reinterpret as mass percent. |
| Mass-based ppm | 10⁶ × solute mass / solution mass | Exactly mg kg⁻¹; mg L⁻¹ needs density. |
| Mole fraction | nᵢ / sum of all component amounts | Dimensionless; fractions sum to one. |
| Molarity C | Solute mol / solution L | Temperature-dependent through volume. |
| Molality m | Solute mol / solvent kg | Mass-based; distinguish analytical amount from effective particles. |
For a percentage conversion, choosing 100 g solution usually fixes the bookkeeping immediately. For molarity, convert total solution mass to volume through density. For molality, remove solute mass from the total to identify solvent mass. For a hydrate, include the complete hydrate molar mass when converting its weighed mass to moles. Balanced reaction coefficients must be applied before converting a required reactant amount into a solution volume.
Interactions control both dissolution and volatility
Solute–solvent contacts compete with the contacts broken in the original materials. Similar interactions often support mixing, while an ionic lattice can resist dissolution in a non-polar liquid. A saturated solution represents a dynamic balance between dissolution and return to the solid. Temperature changes that balance according to the enthalpy of dissolution; pressure especially affects gas solubility. The nature of the dissolved species and possible chemical reactions determine whether a simple physical-solubility model is adequate.
| Relationship | Symbols and units | Conditions and interpretation |
|---|---|---|
| p_g=K_Hx_g | Gas partial pressure and K_H in the same pressure unit; x_g dimensionless | Dilute gas solubility; check constant convention and reactions. |
| pᵢ=xᵢpᵢ° | Liquid xᵢ; pressures at the same temperature | Ideal solution component or suitable solvent limit. |
| p_total=Σpᵢ; yᵢ=pᵢ/p_total | Vapour yᵢ is dimensionless | Ideal-gas vapour mixture; x and y need not match. |
| ΔmixH=0; ΔmixV=0 | Mixing enthalpy and volume changes | Ideal solution benchmark, not absence of interactions. |
Real mixtures can deviate above or below Raoult’s reference pressure. Weaker average unlike contacts commonly give positive deviation, while more favourable unlike contacts commonly give negative deviation. Sufficient deviation may produce an azeotrope with y=x at its boiling composition. Ordinary distillation enriches only when vapour and liquid compositions differ. An azeotrope is therefore a phase-equilibrium limitation, not a special chemical compound.
Choose the colligative relationship and its denominator
| Property | Relation | Inputs, units and validity |
|---|---|---|
| Relative pressure lowering | r=(p°−p)/p°=n₂/(n₁+n₂) | Finite-composition ideal binary non-electrolyte; non-volatile solute. |
| Dilute pressure lowering | r≈i n₂/n₁ | Analytical solute amount; small solute/solvent ratio. |
| Boiling elevation | ΔT_b=iK_bm; T_b=T_b°+ΔT_b | K_b in K kg mol⁻¹; same external pressure; non-volatile dilute solute. |
| Freezing depression | ΔT_f=iK_fm; T_f=T_f°−ΔT_f | K_f in K kg mol⁻¹; pure solvent separates as solid; dilute model. |
| Osmotic pressure | Π=iCRT | T in K; final solution volume; compatible R and pressure units; retained solute. |
| Apparent molar mass | M_apparent=M_normal/i | The model used to infer mass assumed independent single solute units. |
The thermal shifts are positive magnitudes with opposite effects on the final temperature: add a boiling rise and subtract a freezing depression. Osmosis is a pressure-difference problem, not a thermal problem. To obtain pascals with R=8.314, use cubic metres; to use mol L⁻¹, choose an L-based gas constant. Reverse osmosis requires enough opposing pressure to exceed the osmotic difference and, in a working membrane, overcome resistance.
Count effective particles once
Analytical molality counts weighed formula units or monomers before the correction. Multiplying by i converts it to an effective particle measure once. Do not use an already multiplied particle concentration and multiply by i again. A low apparent molar mass can signal dissociation; a high one can signal association. Measured electrolyte factors also reflect ionic interactions, so a simple particle equation must not be used to assert a unique microscopic dissociation fraction without justification.
An integrated worked comparison
Problem
Compare 0.0100 mol kg⁻¹ glucose, i=1, with 0.0100 mol kg⁻¹ CaCl₂ assigned an effective i=2.50 in water. Use K_f=1.86 and K_b=0.52.
- 1.Both start with the same analytical amount per kilogram solvent, but their effective particle molalities are 0.0100 and 0.0250 mol kg⁻¹.
- 2.Glucose: ΔT_f=1×1.86×0.0100=0.0186 K; ΔT_b=1×0.52×0.0100=0.00520 K.
- 3.CaCl₂: ΔT_f=2.50×1.86×0.0100=0.0465 K; ΔT_b=2.50×0.52×0.0100=0.0130 K.
- 4.The calcium chloride effects are 2.50 times as large, not necessarily three times: the problem supplied an effective factor. Both comparisons require that the same factor and dilute assumptions be adequate at the measurement conditions.
Problem
A student says: “A solution is 10% by mass, so its mole fraction is 0.10; heating does not change concentration; all gas solubilities fall on heating; and every non-ideal mixture is an azeotrope.” Identify the missing reasoning.
- 1.Mass fraction must be converted through the two molar masses before obtaining mole fraction.
- 2.Temperature independence depends on the measure: a closed mixture’s mass composition may stay fixed while molarity changes as volume changes.
- 3.Gas solubility often decreases when dissolution is exothermic, but the statement is not universal for every system.
- 4.An azeotrope requires vapour–liquid composition equality at a specific boiling condition; non-ideality alone is insufficient.
- 5.The corrections all follow the same habit: state the definition, identify the model and check its conditions before generalising.
Quick Quiz
Quiz
Which sequence is most reliable for an unfamiliar numerical problem?
Explanation: Definitions and particle identity determine which equation is meaningful; assumptions and units must be checked before substitution.
Which combination is consistent with a dilute independent non-volatile solute?
Explanation: All three changes follow from reduced solvent escaping tendency in the dilute ideal model.
Which expression uses final solution volume rather than solvent mass?
Explanation: C in the osmotic equation is based on final solution volume; molality-based thermal equations use solvent mass.
A solute has i=0.50 in a simple association model. What is M_apparent/M_normal?
Explanation: i=M_normal/M_apparent, so M_apparent/M_normal=1/i=2.
Which observation directly identifies an azeotropic boiling composition?
Explanation: The phase compositions must coincide; neither equal mass nor non-ideality alone establishes an azeotrope.
Practice Problems
- A 100 g solution contains 10 g glucose and has density 1.20 g mL⁻¹. Using glucose 180 and water 18 g mol⁻¹, calculate molarity, molality and ideal relative vapour-pressure lowering.
- An ideal binary liquid has x(B)=0.30, p(A)°=50 kPa and p(B)°=100 kPa. Find partial pressures, total pressure and vapour fractions.
- A gas has K_H=50000 bar for p=K_Hx at a fixed temperature. Its partial pressure rises from 1 to 3 bar over an unchanged amount of solvent. Compare dissolved mole fractions and explain the approximation for dissolved amounts.
- An ideal non-volatile solute reduces solvent pressure to 90% of the pure value. There are 2 mol solvent and the solute molar mass is 100 g mol⁻¹. Find solute mass exactly within this model.
- For 0.100 mol independent solute in 0.500 kg water, calculate boiling and freezing shifts with K_b=0.52 and K_f=1.86. Give final temperatures using 373.15 and 273.15 K.
- A 1.20 g solute in 0.100 kg solvent gives ΔT_f=0.300 K with K_f=5.00. Calculate apparent molar mass. If normal monomer mass is 100 g mol⁻¹ and only dimerisation occurs, find the associated fraction.
- An ideal solute dissociates into three ions with α=0.80. Its analytical molarity is 0.0100 M at 300 K. Find i and Π in bar using R=0.08314.
- A polymer solution has Π=100 Pa at 300 K, final volume 0.00100 m³ and solute mass 0.500 g. Find molar mass with R=8.314.
- A measured liquid mixture pressure lies below both component Raoult partial-pressure predictions at one composition. Explain the likely interaction picture and whether an azeotrope is proved.
- A 0.200 mol kg⁻¹ solution of a monoprotic acid gives ΔT_f=0.4092 K in water, K_f=1.86. Estimate α and K_a using C≈m.
- A retained-solute solution has osmotic-pressure difference 12 bar from a less concentrated solution. Explain solvent flow for zero excess pressure, 12 bar opposing pressure and 15 bar opposing pressure.
- Describe a full numerical audit for a solution problem and give three reasons a precise-looking result may still be unreliable.
Glucose amount=10/180=0.0555556 mol; water amount=90/18=5 mol. Volume=100/1.20=83.3333 mL=0.0833333 L, so C=0.666667 M. m=0.0555556/0.090=0.617284 mol kg⁻¹. Relative lowering equals x(glucose)=0.0555556/5.0555556=0.0109890 under the finite-composition ideal model. Each denominator refers to a different reference amount.
Key Takeaways
• Begin with component identity, composition definitions and the relevant phases. • Convert masses to amounts before calculating mole fractions or reacting amounts. • Use Henry’s law for an appropriately defined dilute gas-solubility constant and partial pressure. • Use liquid fractions in Raoult’s law and vapour fractions in Dalton’s law. • Ideal and non-ideal mixture behaviour reflects changes in intermolecular interactions. • Colligative equations connect effective particle concentration with solvent properties. • Exact ideal relationships and dilute approximations must be distinguished. • Association, dissociation and ionic interactions affect apparent molar mass. • A reliable answer states its assumptions, checks units and physical bounds, and respects measurement precision.
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Association, Dissociation and Abnormal Molar Mass
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