Electrolytes & Colligative Properties — Print · Grade 11 Science

Chapter: Solutions & Dissolved Particles

Electrolytes & Colligative Properties

Learning goals

Readings for this unit

Before this unit

Moles, concentration, phase changes, and ions.

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Start with something to explain

Why can equal mole amounts of two solutes shift freezing differently?

Illustrative evidence case. One selected solute remains molecular while another dissociates into several ions.

Colligative comparisons depend on effective particle amountColligative comparisons depend on effective particle amount1 mol ideal soluteSelected dissociationIdeal particlesGlucoseMolecules remain intact1 molNaClNa⁺ + Cl⁻2 molCaCl₂Ca²⁺ + 2Cl⁻3 molvan’t Hoff factor i: effective particle count ratio in the model
Original ideal particle-count table; it does not assume measured i is exactly an integer at every concentration.

Notice and wonder

Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.

Why can equal mole amounts of two solutes shift freezing differently?

Your first model

Count effective dissolved entities and identify the solvent-mass denominator.

Build the language from the idea

molality
Moles solute per kilogram of solvent.
particle factor i
Effective number of dissolved particles per solute formula unit in a model.
osmosis
Net solvent flow across a selectively permeable membrane.

A solution’s behavior depends on both particle identity and particle count. Ionic solutes may dissociate into ions; molecular solutes such as sugar can remain neutral molecules. A strong electrolyte produces many mobile ions and conducts; a weak electrolyte ionizes partially. Solubility and electrical conductivity answer different questions.

Colligative comparisons depend on effective particle amountColligative comparisons depend on effective particle amount1 mol ideal soluteSelected dissociationIdeal particlesGlucoseMolecules remain intact1 molNaClNa⁺ + Cl⁻2 molCaCl₂Ca²⁺ + 2Cl⁻3 molvan’t Hoff factor i: effective particle count ratio in the model
Original ideal particle-count table; it does not assume measured i is exactly an integer at every concentration.

Molality b is moles of solute per kilogram of solvent, unlike molarity which uses liters of solution. Molality is useful when studying temperature-dependent properties because solvent mass does not expand with temperature. Convert grams of solvent to kilograms before dividing.

A temperature-shift model requires the appropriate concentration conventionA temperature-shift model requires the appropriate concentration conventionΔTf = iKf bb = 0.100 mol solute / 0.500 kg solvent = 0.200 mol/kgi=2, Kf=1.86 K·kg/mol → ΔTf=0.744 KThe denominator is solvent kilograms, not solution liters.
Original selected ideal NaCl-like i=2 calculation with a supplied water Kf approximation; no measured effective dissociation factor is claimed.

In a dilute ideal model, adding nonvolatile solute lowers vapor pressure and changes freezing and boiling points. ΔTf = iKf b and ΔTb = iKb b; i is the effective dissolved-particle factor. Sugar has i about 1, whereas complete ideal dissociation of NaCl gives about 2. Real solutions can depart from ideal values.

A phase-equilibrium explanation is stronger than saying particles block freezingA phase-equilibrium explanation is stronger than saying particles block freezingDepression magnitude ΔTf (K)00.511.522.500.20.40.6Molality b (mol/kg solvent)Selected dilute model: slope = iKf = 3.72 K·kg/mol.
Original calculated ideal-model curve, not observed freezing data or a complete water–salt phase diagram.

The freezing point is the pure-solvent freezing temperature minus ΔTf, while boiling point is pure-solvent boiling temperature plus ΔTb. For water near ordinary pressure, Kf ≈1.86°C·kg/mol and Kb≈0.512°C·kg/mol. State pressure, dilute-solution assumptions, and solute choice.

Osmosis is net solvent movement across a selectively permeable barrier toward the side with lower solvent chemical potential, often the more concentrated solute side in a simple comparison. Osmotic pressure in an ideal dilute solution is π=iMRT. Use consistent pressure units with R and kelvin temperature.

Osmotic flow depends on membrane selectivity and pressure conditionsOsmotic flow depends on membrane selectivity and pressure conditionsLower effective soluteHigher effective soluteSolvent-permeable / solute-restricting membraneOsmotic equilibrium includes pressure as well as composition.
Original qualitative two-side membrane model. The single arrow shows the selected initial net direction under equal pressure, not a perpetual flow guarantee.

Road salt works only within practical temperature and concentration limits and can damage soils and waterways. Do not infer that twice the nominal salt always doubles the real effect. Safe paper calculations explain particle-count models; measurements should compare conditions and include limitations.

Build your understanding

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Part 1 · Count independent dissolved particles

An electrolyte produces mobile ions in a suitable solution. Introductory colligative properties depend on effective dissolved-particle amount under an approximate solution model rather than only the solute’s formula mass. The van’t Hoff factor i expresses an effective particle ratio: ideal complete NaCl dissociation suggests two, and CaCl₂ three. Ion association and nonideal interactions can make measured effective values differ, so formula dissociation counts are a starting approximation, not exact experimental constants.

Colligative comparisons depend on effective particle amountColligative comparisons depend on effective particle amount1 mol ideal soluteSelected dissociationIdeal particlesGlucoseMolecules remain intact1 molNaClNa⁺ + Cl⁻2 molCaCl₂Ca²⁺ + 2Cl⁻3 molvan’t Hoff factor i: effective particle count ratio in the model
Original ideal particle-count table; it does not assume measured i is exactly an integer at every concentration.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Why can one mole of CaCl₂ have a larger ideal particle effect than one mole of glucose?

Visual reasoning — after your attempt

The selected complete dissociation produces three moles of ions per mole of CaCl₂, while glucose stays as one mole of intact molecules.

Colligative effects depend on dissolved-particle amount in suitable dilute ideal models. Electrolyte dissociation can increase particle count; association and nonideality can make the effective van ’t Hoff factor differ from an integer. Specify whether formula units or effective dissolved particles are being counted.

Ideal NaCl gives approximately two ions per formula unit; glucose remains one molecule. This does not mean every real NaCl solution has exactly i=2.

Pause and explain

Why can a measured electrolyte factor differ from the ideal count?

Check your reasoning after answering

Ion interactions, association, and nonideal behavior can change the effective particle contribution relative to complete independent dissociation.

Part 2 · Use molality for temperature-shift models

For a dilute idealized solution with the specified nonvolatile solute behavior, freezing-point depression is approximated by ΔTf=iKf b, where b is molality and Kf is a solvent-specific constant. Boiling-point elevation uses a corresponding Kb under its own conditions. A positive depression magnitude means the freezing temperature is lower than the pure solvent’s reference. The model does not mean every solute produces the same effect at equal grams or that concentrated mixtures follow the dilute equation exactly.

A temperature-shift model requires the appropriate concentration conventionA temperature-shift model requires the appropriate concentration conventionΔTf = iKf bb = 0.100 mol solute / 0.500 kg solvent = 0.200 mol/kgi=2, Kf=1.86 K·kg/mol → ΔTf=0.744 KThe denominator is solvent kilograms, not solution liters.
Original selected ideal NaCl-like i=2 calculation with a supplied water Kf approximation; no measured effective dissociation factor is claimed.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Why would substituting 0.200 mol/L directly for b be unjustified without more information?

Visual reasoning — after your attempt

The required b is mol/kg solvent. Converting from molarity needs density/composition information rather than a changed unit label.

Molality is moles solute per kilogram solvent. Introductory freezing depression and boiling elevation use ΔTf=iKfm and ΔTb=iKbm with supplied solvent constants. These differ from molarity because volume is not the denominator. State dilute, nonvolatile-solute assumptions where required.

0.10 mol solute in 0.50 kg water gives 0.20 mol/kg, not 0.10 M unless a final solution volume is supplied.

Pause and explain

Why cannot molarity be substituted automatically for molality?

Check your reasoning after answering

Their denominators differ: liters of solution versus kilograms of solvent. They are only approximately related under suitable conditions.

Part 3 · Explain effects through phase equilibrium

A dissolved solute changes the solvent’s chemical potential and therefore the conditions for phase equilibrium. In the selected dilute model, more effective solute particles give a larger freezing-temperature shift. This is not a literal wall of solute particles physically blocking every water molecule from crystallizing. Real phase diagrams, eutectic limits and concentration changes during freezing add detail beyond the straight-line approximation. Interpret the graph within its stated range and avoid extending it indefinitely to impossible temperatures.

A phase-equilibrium explanation is stronger than saying particles block freezingA phase-equilibrium explanation is stronger than saying particles block freezingDepression magnitude ΔTf (K)00.511.522.500.20.40.6Molality b (mol/kg solvent)Selected dilute model: slope = iKf = 3.72 K·kg/mol.
Original calculated ideal-model curve, not observed freezing data or a complete water–salt phase diagram.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

What does the slope describe in the selected linear model, and why should it not be extrapolated without limit?

Visual reasoning — after your attempt

It is iKf. At larger concentration, nonideality and other phase behavior can invalidate the simple dilute relation.

A suitable nonvolatile solute lowers solvent vapor pressure in an ideal mixture and changes phase-equilibrium temperatures. The molecular explanation concerns solvent chemical potential and particle mixing, not heavier water molecules or a permanent change in the solvent identity. Exact behavior needs a suitable solution model.

A freezing-point depression is a change in when solvent-rich solid can equilibrate with solution. Solute concentration can change as freezing proceeds, so one initial shift is not an entire freezing history.

Pause and explain

Does solute make each water molecule heavier?

Check your reasoning after answering

No. It changes the solution’s thermodynamic and particle environment, not the mass or identity of each solvent molecule.

Part 4 · Keep osmotic pressure and membrane conditions explicit

Osmosis is net solvent movement across a selectively permeable membrane in response to chemical-potential differences. Under equal-pressure simple conditions with effectively retained solute, solvent tends toward the higher effective solute side. An opposing pressure can balance that tendency; dilute ideal osmotic pressure is approximated by π=i cRT with appropriate concentration and temperature units. A membrane permeable to the solute changes the long-term account, so “water always moves toward more solute” is incomplete without pressure and permeability conditions.

Osmotic flow depends on membrane selectivity and pressure conditionsOsmotic flow depends on membrane selectivity and pressure conditionsLower effective soluteHigher effective soluteSolvent-permeable / solute-restricting membraneOsmotic equilibrium includes pressure as well as composition.
Original qualitative two-side membrane model. The single arrow shows the selected initial net direction under equal pressure, not a perpetual flow guarantee.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

What can stop net solvent flow even when the two sides have different retained-solute concentrations?

Visual reasoning — after your attempt

A suitable opposing pressure difference can balance the solvent chemical-potential difference; membrane permeability and conditions must be stated.

For a dilute ideal solution, osmotic pressure can be modeled as π=iCRT with consistent units. Osmosis involves a selectively permeable membrane and solvent transfer. Without a suitable membrane or concentration difference, the simple osmotic setup is not present.

Use a supplied membrane that passes solvent but not the solute. A membrane allowing both freely requires a different account.

Pause and explain

Why does membrane selectivity matter?

Check your reasoning after answering

The osmotic model requires different permeability for solvent and relevant solute. Free passage of both changes the conditions for sustained osmotic behavior.

Apply the whole unit · Compare ideal freezing-point depressions

Illustrative solutions each have molality 0.100 mol/kg in water. A is glucose with i=1; B is NaCl with supplied ideal i=2. Use Kf=1.86°C·kg/mol.

Your task

Calculate depression for each, predict initial freezing temperatures relative to pure water at 0°C, and explain why the comparison is an ideal particle model rather than an exact guarantee for all concentrations.

What your response should show

  • Choose the appropriate chemical model or relationship and state its conditions.
  • Show substitutions, units, particle or charge counts, and intermediate reasoning as relevant.
  • Use supplied evidence to justify the conclusion; distinguish observation from inference.
  • Identify a meaningful limitation, uncertainty, or next measurement.
Response guidance — after your attempt

  1. A: ΔTf=1×1.86×0.100=0.186°C.

  2. B: ΔTf=2×1.86×0.100=0.372°C.

  3. Initial predicted freezing temperatures are −0.186°C and −0.372°C.

The ideal NaCl prediction has twice the depression because its effective particle factor is supplied as two. Real association and concentration effects can alter that factor, and continuing freezing changes the liquid concentration. State molality and solvent conditions; do not compare molarity records as though their denominators were identical.

Scientific references and source-model foundations

Parent / teacher teaching notes

Ask students to count particles and identify the denominator before substituting. Separate a temperature shift from the direction of heat transfer. Ask the learner to explain each particle symbol, axis, unit and arrow before calculating. Distinguish hypothetical models and calculated curves from measured records. Public OpenSciEd excerpts here are selected foundation prompts, not complete high-school teacher editions or an endorsement. All required advanced explanations and practice are provided locally.

The key idea

Colligative changes depend approximately on the number of dissolved particles, not merely solute grams.

Worked examples

Example 1

Dissolve .10 mol sugar in .50 kg water. Find ideal freezing depression.
  1. b=.10/.50=.20 mol/kg.
  2. Sugar i≈1.
  3. ΔTf=1×1.86×.20.
Answer: 0.372°C; freezing near −0.372°C

Example 2 · Worked application

Compare ideal dilute 0.10 m glucose and 0.10 m NaCl solutions with i=1 and i=2.
  1. Colligative change scales with particle molality i×m.
  2. Glucose gives 0.10 particle-molality units; NaCl gives 0.20.
  3. The ideal NaCl effect is twice as large, with real nonideality possible.
Answer: NaCl has twice the ideal freezing-point depression under the supplied model.
Check the reasoning

Use molality, not molarity, for the standard freezing-point relation.

Example 3

Compare equal molality NaCl and sugar ideally.
  1. Sugar gives one particle per molecule.
  2. NaCl gives about two ions per formula unit.
  3. Use ΔTf proportional to i.
Answer: NaCl gives roughly twice the depression in the ideal dilute model.

Example 4 · Find and fix the mistake

Context: Compare ideal dilute 0.10 m glucose and 0.10 m NaCl solutions with i=1 and i=2. A student claims: “Equal formula-unit amounts always give equal colligative effects.” Evaluate the claim and write a corrected explanation.
  1. Test the claim against the stated conditions: Use molality, not molarity, for the standard freezing-point relation.
  2. Dissociation changes the number of dissolved particles; use the supplied van't Hoff factor and model limits.
Answer: Dissociation changes the number of dissolved particles; use the supplied van't Hoff factor and model limits.
Check the reasoning

Use molality, not molarity, for the standard freezing-point relation.

Example 5 · Reason from the visual model

A phase-equilibrium explanation is stronger than saying particles block freezingA phase-equilibrium explanation is stronger than saying particles block freezingDepression magnitude ΔTf (K)00.511.522.500.20.40.6Molality b (mol/kg solvent)Selected dilute model: slope = iKf = 3.72 K·kg/mol.
Original calculated ideal-model curve, not observed freezing data or a complete water–salt phase diagram.
What does the slope describe in the selected linear model, and why should it not be extrapolated without limit?
  1. A dissolved solute changes the solvent’s chemical potential and therefore the conditions for phase equilibrium. In the selected dilute model, more effective solute particles give a larger freezing-temperature shift. This is not a literal wall of solute particles physically blocking every water molecule from crystallizing. Real phase diagrams, eutectic limits and concentration changes during freezing add detail beyond the straight-line approximation. Interpret the graph within its stated range and avoid extending it indefinitely to impossible temperatures.
  2. It is iKf. At larger concentration, nonideality and other phase behavior can invalidate the simple dilute relation.
Answer: It is iKf. At larger concentration, nonideality and other phase behavior can invalidate the simple dilute relation.
Check the reasoning

Original calculated ideal-model curve, not observed freezing data or a complete water–salt phase diagram.

Avoid these mistakes

Check your understanding

Why compare equal mole amounts rather than equal grams for colligative effects?

Discuss the answer

Different molar masses and dissociation factors produce different dissolved-particle counts per gram.

Return to the phenomenon

Return to the initial case. Why can equal mole amounts of two solutes shift freezing differently? Revise your drawing or explanation using two named ideas and one stated limitation.

Compare explanations after your attempt

The dilute colligative model uses effective particle amount, not just formula moles or grams. Dissociation, association and phase-equilibrium conditions determine its useful scope.

What should we investigate next?

Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.

Investigate, calculate and revise

Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.

Activity 1 · Return to the opening case

Assigned reading and preparation

Use the opening phenomenon, unit definitions and explained Parts 1–4.

Your task

Return to the initial case. Why can equal mole amounts of two solutes shift freezing differently? Revise your drawing or explanation using two named ideas and one stated limitation.

Success criteria

  • Use the stated observations and appropriate units.
  • Connect a labeled model with the chemical mechanism.
  • State assumptions and an evidence limitation.
  • Explain a substantive revision or next test.
Hint

Keep species identity, material transfer and energy account distinct.

Response guidance — discuss after your attempt

The dilute colligative model uses effective particle amount, not just formula moles or grams. Dissociation, association and phase-equilibrium conditions determine its useful scope.

Activity 2 · Transfer to a changed case

Assigned reading and preparation

Cover the worked answer and use the explained models, relationships and notebook.

Your task

Using Kf=1.86 K·kg/mol, compare ideal freezing depressions for 0.100 mol glucose and 0.100 mol CaCl₂, each in 0.500 kg water. Assume i=1 and 3 respectively.

Success criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.
Hint

Name known quantities and conditions, show intermediate steps and check conservation.

Response guidance — discuss after your attempt

Both formula molalities are 0.200 mol/kg. Glucose gives ΔTf=0.372 K and ideal CaCl₂ gives 1.116 K≈1.12 K. Actual effective i and nonideality can change the measured comparison.

Words to know

molality
Moles solute per kilogram of solvent.
particle factor i
Effective number of dissolved particles per solute formula unit in a model.
osmosis
Net solvent flow across a selectively permeable membrane.

In the real world

Use the chapter investigation to connect this unit to measured evidence and a practical decision.

Exit check

Why compare equal mole amounts rather than equal grams for colligative effects?

Check your explanation after answering

Different molar masses and dissociation factors produce different dissolved-particle counts per gram.

Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.

Problem 1

Find molality of .30 mol solute in .60 kg solvent.
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

0.5 mol/kg

.30/.60=.50.

Problem 2

Why can dissolved sugar fail to conduct well?
Answer and explanation

Neutral sugar molecules do not supply many mobile ions.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 3

Find ideal freezing depression for i=1, Kf=1.86, b=.30 mol/kg.
Hints

List the given quantities, select a relationship, and check units before substituting.

Answer and explanation

0.558 °C

1×1.86×.30=.558.

Problem 4

Compare molarity and molality denominators.
Answer and explanation

Molarity uses solution volume; molality uses solvent mass in kilograms.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 5

Find ΔT_f for m=0.20, i=1 and K_f=1.86°C·kg/mol. Show the reasoning, not just the conclusion.
Hints

Use molality, not molarity, for the standard freezing-point relation.

Answer and explanation

ΔT_f=iK_fm=0.372°C.

Problem 6

Why can an ionic solute conduct electricity while glucose solution usually conducts poorly?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

Mobile ions carry charge; dissolved neutral glucose does not supply comparable ionic carriers.

Problem 7

A phase-equilibrium explanation is stronger than saying particles block freezingA phase-equilibrium explanation is stronger than saying particles block freezingDepression magnitude ΔTf (K)00.511.522.500.20.40.6Molality b (mol/kg solvent)Selected dilute model: slope = iKf = 3.72 K·kg/mol.
Original calculated ideal-model curve, not observed freezing data or a complete water–salt phase diagram.
Interpret the supplied visual. What does the slope describe in the selected linear model, and why should it not be extrapolated without limit?
Answer and explanation

It is iKf. At larger concentration, nonideality and other phase behavior can invalidate the simple dilute relation.

A dissolved solute changes the solvent’s chemical potential and therefore the conditions for phase equilibrium. In the selected dilute model, more effective solute particles give a larger freezing-temperature shift. This is not a literal wall of solute particles physically blocking every water molecule from crystallizing. Real phase diagrams, eutectic limits and concentration changes during freezing add detail beyond the straight-line approximation. Interpret the graph within its stated range and avoid extending it indefinitely to impossible temperatures.

It is iKf. At larger concentration, nonideality and other phase behavior can invalidate the simple dilute relation.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the chemistry relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 8

Using Kf=1.86 K·kg/mol, compare ideal freezing depressions for 0.100 mol glucose and 0.100 mol CaCl₂, each in 0.500 kg water. Assume i=1 and 3 respectively.
Answer and explanation

Both formula molalities are 0.200 mol/kg. Glucose gives ΔTf=0.372 K and ideal CaCl₂ gives 1.116 K≈1.12 K. Actual effective i and nonideality can change the measured comparison.

Both formula molalities are 0.200 mol/kg. Glucose gives ΔTf=0.372 K and ideal CaCl₂ gives 1.116 K≈1.12 K. Actual effective i and nonideality can change the measured comparison.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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