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Solutions · TOPIC LESSON

Colligative Properties

Count independently behaving dissolved particles and match the concentration unit to the property.

JEE Main · JEE Advanced · NEET UG38 min study guide9 indexed sections

BY THE END, YOU CAN

  • Explain the solvent-based origin of colligative effects under dilute ideal assumptions.
  • Calculate vapor-pressure lowering, boiling elevation and freezing depression.
  • Apply osmotic pressure to concentration and molar-mass problems.
  • Use van ’t Hoff factors for idealized dissociation or association without confusing them with universal exact counts.
  • Handle mixtures, apparent molar masses and limitations of linear formulas.

BEFORE YOU START

  • Moles, molarity, molality and mole fraction.
  • Phase equilibrium and vapor pressure.
  • Ideal-gas equation and basic solution composition.

Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.

CONNECT THE CONCEPT TO THE EXAM

Selected exam questions & worked solutions

These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.

Available in this lesson: JEE Advanced 2021. This is a limited lesson sample, not a complete exam-question bank.

1 sourced sample question available across the listed exams and years.

Matched topics: Colligative Properties.

JEE Advanced · 2021 · Paper 1 · Chemistry · Q9

Paraphrased exam question

Independently derived and checked solution. Archived key finality is unverified; this is not presented as an official final answer.

Find the boiling temperature of 0.1 molal aqueous AgNO₃, assuming complete dissociation. Use Kb=0.5 K kg mol⁻¹ and pure-water boiling temperature 100 °C.

Reveal answer & worked solution

Answer: 100.10 °C (independently derived).

  1. Each AgNO₃ unit gives Ag⁺ and NO₃⁻, so the van’t Hoff factor is i=2.
  2. The elevation is ΔTb=iKb m=2×0.5×0.1=0.10 K.
  3. Add this elevation to 100 °C: the solution boils at 100.10 °C.

Source: JEE (Advanced) 2021, Paper 1, Chemistry Q9; independently derived solution; key finality unverified. Solution written for this lesson.

01 / UNDERSTAND & APPLY

1. What “colligative” holds constant

Colligative properties are solution effects that, in the dilute ideal limit, depend mainly on the number of independently behaving dissolved solute particles relative to solvent rather than on the solute’s detailed chemical identity. Common examples are solvent vapor-pressure lowering, boiling-point elevation, freezing-point depression and osmotic pressure.

The qualification matters. Solute identity can affect association, dissociation, chemical reaction and nonideal interactions, all of which alter how many effective particles the solution contains or how they influence solvent chemical potential. Two equal formula-unit concentrations do not necessarily give equal effects if one solute dissociates and the other does not.

For a nonvolatile solute in an ideal solution, mixing lowers the solvent’s chemical potential relative to the pure solvent at the same temperature and pressure. This changes the conditions for vaporization, freezing and solvent transfer through a membrane. The equations are different expressions of that solvent-equilibrium change, not four unrelated empirical tricks.

Before calculating, identify the solvent, whether the solute is volatile, whether it reacts or forms ions, and which concentration unit the formula requires. Molality uses kilograms of solvent; molarity uses litres of final solution; mole fraction uses total moles. Interchanging these denominators can produce a large error while leaving the numerical arithmetic looking tidy.

02 / UNDERSTAND & APPLY

2. A nonvolatile solute reduces the solvent vapor pressure

For an ideal solution, the solvent partial pressure follows Raoult’s law p_A=x_Ap_A°. Here p_A° is the pure solvent vapor pressure at the same temperature and x_A is its liquid-phase mole fraction. If the solute is nonvolatile, the solution’s vapor pressure is just this solvent contribution under the simple model.

The relative lowering is (p_A°−p_A)/p_A°=1−x_A=x_B for a binary solution. In a dilute solution, x_B=n_B/(n_A+n_B)≈n_B/n_A if n_B is much smaller than n_A. The equality to x_B is exact within ideal Raoult behavior; the further replacement by n_B/n_A is a dilute approximation.

For a solute that dissociates or associates, an ideal particle-count model replaces formula-unit moles with effective dissolved-particle moles. In real electrolyte solutions, activities make this more subtle. A volatile solute contributes its own vapor pressure, so total vapor pressure is not obtained simply by lowering the solvent contribution and ignoring the solute.

Vapor pressure is an equilibrium property at a given temperature. The rate at which a liquid evaporates in an open room also depends on air flow, humidity and mass transfer. Lower vapor pressure should not be converted into an exact evaporation-time prediction without those additional factors.

p_A=x_Ap_A°; (p_A°−p_A)/p_A°=x_B

Ideal binary solution with nonvolatile solute for the total-pressure interpretation.

ORIGINAL WORKED EXAMPLE

Count solvent and solute moles

An ideal solution contains 0.200 mol nonvolatile nonelectrolyte and 9.80 mol solvent. The pure solvent vapor pressure is 50.0 kPa at the stated temperature. Find the solution vapor pressure.

  1. Total moles=10.00, so x_A=9.80/10.00=0.980.
  2. p_A=0.980×50.0=49.0 kPa.
  3. The lowering is 1.00 kPa and relative lowering 0.0200, equal to solute mole fraction.
  4. The solute is assumed not to contribute appreciably to vapor pressure.

Answer p=49.0 kPa.

03 / UNDERSTAND & APPLY

3. Boiling elevation and freezing depression use molality

A liquid boils when its vapor pressure equals the imposed external pressure. A nonvolatile solute lowers solvent vapor pressure at a given temperature, so the solution must generally be heated to a higher temperature to reach the same external pressure. In the dilute limit, ΔT_b=K_bm for a nonelectrolyte, with K_b the solvent’s molal boiling-point-elevation constant.

Freezing equilibrium compares the chemical potentials of solvent in the liquid and solid phases. If the solid that forms is essentially pure solvent and the solute remains in the liquid, lowering liquid-solvent chemical potential lowers the freezing temperature. The dilute relation is ΔT_f=K_fm, where ΔT_f is defined as the positive amount by which freezing temperature is lowered.

Use T_b(solution)=T_b(pure)+ΔT_b and T_f(solution)=T_f(pure)−ΔT_f. The positive constants K_b and K_f belong to the solvent, not the solute. The molality m is moles of dissolved solute particles per kilogram of solvent under the nonelectrolyte ideal model.

The formulas are first-order dilute approximations. At higher concentration, nonideality and temperature-dependent composition can matter. If solute enters the solid phase, forms a separate compound, or produces a eutectic complication, the simple freezing-depression picture may fail. A phase diagram is then more informative than a linear extrapolation.

ΔT_b=iK_bm; ΔT_f=iK_fm

Dilute model; m is formula-unit molality when a van ’t Hoff factor i is included.

ORIGINAL WORKED EXAMPLE

Use kilograms of solvent

Dissolve 0.0100 mol nonelectrolyte in 500.0 g water. Using K_f=1.86 K kg mol⁻¹, estimate the freezing depression and freezing point relative to pure water at 0.00 °C.

  1. Solvent mass=0.5000 kg, so m=0.0100/0.5000=0.0200 mol kg⁻¹.
  2. ΔT_f=1.86×0.0200=0.0372 K.
  3. The solution freezing point is lowered to approximately−0.0372 °C.
  4. The solute mass is not included in the denominator because molality uses solvent mass.

Answer ΔT_f=0.0372 K; estimated freezing point−0.0372 °C.

04 / UNDERSTAND & APPLY

4. A selective membrane reveals a solvent chemical-potential difference

A semipermeable membrane permits solvent transfer but blocks the specified solute on the observation timescale. When pure solvent and solution at the same temperature and initial pressure are separated by such a membrane, solvent tends to enter the solution. The resulting hydrostatic pressure difference can oppose that transfer. Osmotic pressure is the excess pressure required on the solution side to stop net solvent flow under the stated conditions.

For an ideal dilute solution, π=CRT=nRT/V for a nonelectrolyte, or π=iCRT using formula-unit molarity C and an effective particle factor. Unlike freezing and boiling formulas, this expression uses solution volume and molarity. Use absolute temperature and choose R in units compatible with the requested pressure and volume.

The direction is determined by solvent chemical potential, not by solute particles physically pulling water as tiny suction devices. A pressure greater than the osmotic pressure can drive reverse osmosis under suitable membrane conditions. The model assumes membrane selectivity and integrity; a solute that crosses freely does not sustain the same osmotic pressure difference.

Two dilute ideal solutions at the same temperature are isotonic with respect to a particular perfectly solute-blocking membrane when their effective particle concentrations give equal osmotic pressures. Equal molarities are insufficient if their i values differ. In biological systems, membrane permeability and active processes add complexity, so this thermodynamic comparison is not a medical formulation rule.

Osmotic pressure can be sensitive enough to estimate large molar masses at low solute concentrations, where boiling or freezing changes may be very small. An appropriate membrane must retain the macromolecule while allowing solvent equilibration. Nonideality and aggregation require extrapolation or a more detailed model for precise measurements.

π=iCRT=i(n/V)RT

Ideal dilute solution; C is formula-unit molarity and V is final solution volume.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Osmosis and stopping pressure across a selective membraneWith pure solvent and dilute solution at equal temperature and pressure, a membrane that passes solvent but retains solute permits net solvent flow toward the solution. Applying additional pressure equal to the osmotic pressure on the solution side stops the net flow, while molecular exchange continues in both directions. For an ideal dilute nonelectrolyte osmotic pressure is CRT.Osmosis and its stopping pressureDilute solution; solute cannot cross the membranePure solventSolutionSemipermeable membranenet solventflow initially● = retained soluteEqual pressure: net solvent flow toward solution.At osmotic equilibriumApply extra pressure to the solution sideSolventSolutionΔP = πNo net solvent flow at the stopping pressure.Molecular exchange continues in both directions.Ideal dilute nonelectrolyte: π = CRTWith particle-factor model: π ≈ iCRTSchematic chambers; hydrostatic head neglected.
Isothermal schematic, not a particle trajectory simulation. A perfectly solute-retaining membrane and negligible hydrostatic head are assumed; C is analytical concentration when using the stated particle-factor model.

ORIGINAL WORKED EXAMPLE

An osmotic-pressure calculation

A 3.00 g sample of a nonelectrolyte with molar mass 180 g mol⁻¹ is dissolved to make 250.0 mL solution at 300 K. Estimate π with R=0.082057 L atm mol⁻¹ K⁻¹.

  1. n=3.00/180=0.01667 mol.
  2. C=0.01667/0.2500=0.06667 M.
  3. π=0.06667×0.082057×300≈1.641 atm.
  4. The volume is final solution volume; no solvent density is needed for this molarity calculation.

Answer π≈1.64 atm.

05 / UNDERSTAND & APPLY

5. Dissociation and association modify the effective particle count

The van ’t Hoff factor i is the observed colligative effect divided by the effect predicted for the same formula-unit concentration of an ideal nonelectrolyte. Under an ideal dissociation model, if a fraction α of formula units yields ν separate particles, i=1+(ν−1)α. Complete dissociation gives i=ν and no dissociation gives i=1.

For MX₂→M²⁺+2X⁻, ν=3. An ideal i=2.50 corresponds to α=(2.50−1)/(3−1)=0.750. For a 1:1 electrolyte ν=2, so the same algebra gives i=1+α. Do not use ν equal to the sum of ion charge magnitudes; count particles from the balanced dissociation.

For association into n-mers, if α is the fraction of monomer units associated, particle counting gives i=1−α(1−1/n). Dimerization gives i=1−α/2. Complete dimerization halves the particle count. The fraction of monomer units associated is not necessarily the mole fraction of dimers in the final mixture.

Real strong-electrolyte solutions can have i below the ideal integer because interionic interactions change colligative behavior, even when the substance is conventionally regarded as extensively dissociated. Thus a measured i is not always a literal dissociation percentage. Use the α relation when the problem explicitly adopts ideal independent-particle counting.

ORIGINAL WORKED EXAMPLE

Infer a model dissociation fraction

A 0.100 mol kg⁻¹ solution of hypothetical MX₂ depresses a solvent’s freezing point by 0.465 K. K_f=1.86 K kg mol⁻¹. Find i and the idealized α.

  1. i=ΔT_f/(K_fm)=0.465/(1.86×0.100)=2.50.
  2. MX₂ yields three particles per fully dissociated formula unit.
  3. α=(i−1)/(3−1)=1.50/2=0.750.
  4. This is an idealized dissociation interpretation, not a general exact conclusion for real concentrated electrolytes.

Answer i=2.50; ideal-model α=0.750.

06 / UNDERSTAND & APPLY

6. Apparent molar mass changes when particle count changes

For a nonelectrolyte sample mass w in solvent mass W_kg, molality is m=(w/M)/W_kg. Substituting into freezing depression gives M=K_fw/(ΔT_fW_kg) when i=1. If W is supplied in grams, an explicit factor 1000 is needed; deriving from kilograms prevents confusion over where that factor belongs.

For osmotic pressure, M=wRT/(πV) for a nonelectrolyte. Here w is in grams if M is required in g mol⁻¹, V uses units compatible with R, and π is the excess osmotic pressure. The formula assumes all of the sample contributes as the intended dissolved species and is retained by the membrane.

If a colligative experiment is interpreted with i=1 when the true effective factor is i, the apparent molar mass is M_apparent=M_formula/i. Dissociation with i>1 makes apparent mass smaller; association with i<1 makes it larger. The atom masses themselves are unchanged; the inference responds to the number of dissolved entities.

Unknown impurities can also distort a molar-mass estimate, especially low-molar-mass contaminants that contribute many particles per gram. A single colligative measurement does not uniquely distinguish association, dissociation and contamination. State the chemical assumptions that make the inference possible.

ORIGINAL WORKED EXAMPLE

Estimate a macromolecular molar mass

A 0.250 g nonelectrolyte sample in 0.1000 L solution has osmotic pressure 0.0100 atm at 300 K. Use R=0.082057 L atm mol⁻¹ K⁻¹ and assume ideal behavior and membrane retention.

  1. M=wRT/(πV).
  2. Numerator=0.250×0.082057×300=6.154275 g L atm mol⁻¹.
  3. Denominator=0.0100×0.1000=0.00100 atm L.
  4. M≈6154 g mol⁻¹, rounded to approximately 6.15×10³ g mol⁻¹.

Answer M≈6.15×10³ g mol⁻¹.

07 / UNDERSTAND & APPLY

7. Add effective particles only after any reaction

For an ideal dilute mixture of nonreacting solutes, their contributions add: ΔT_f≈K_fΣ(i_jm_j) and π≈RTΣ(i_jC_j). Each term must use a consistent solvent-mass or solution-volume basis. Adding separate molalities calculated using different solvent masses is not valid until the solutions are combined and a common basis is established.

If solutes react on mixing, perform stoichiometry first. Neutralization, precipitation, complexation or association changes the set of dissolved particles. Counting all starting ions independently after they have formed a precipitate or a weakly dissociated product overestimates the final colligative effect.

For comparisons in the same solvent and dilute regime, freezing depression ranks by effective molality im, while osmotic pressure at the same temperature ranks by iC. Equal masses of two nonelectrolytes do not imply equal effects: the lower-molar-mass solute gives more moles and generally a larger effect on the same solvent basis.

Equal freezing depressions in two different solvents do not necessarily imply equal particle molalities because K_f differs. Equal osmotic pressures at different temperatures do not imply equal particle molarities because π depends on T. State what is held fixed in a ranking problem.

ORIGINAL WORKED EXAMPLE

Two nonreacting solutes

A solution contains 0.0100 mol solute X and 0.0200 mol solute Y in 0.500 kg solvent. Both are nonelectrolytes and do not react. With K_f=2.00 K kg mol⁻¹, estimate ΔT_f.

  1. Total effective solute amount=0.0100+0.0200=0.0300 mol.
  2. Effective molality=0.0300/0.500=0.0600 mol kg⁻¹.
  3. ΔT_f=2.00×0.0600=0.120 K.
  4. The amounts were combined on one common solvent-mass basis.

Answer ΔT_f=0.120 K.

08 / UNDERSTAND & APPLY

8. Know when the simple formulas stop being reliable

All four elementary relations are models with a dilute or ideal basis. At larger concentration, solvent activity is no longer captured accurately by a simple mole fraction or linear particle count. Ion pairing, specific solvation and interactions can make an effective i depend on concentration and on the measured property.

Molality is independent of thermal expansion because its denominator is solvent mass, while molarity can change with temperature because solution volume changes. This is one reason freezing and boiling relations conventionally use molality. It does not mean every property is temperature independent just because molality is unchanged.

During partial freezing, pure solvent may leave the liquid, increasing the remaining solution’s concentration and lowering its subsequent freezing temperature further. The initial freezing point is therefore different from a claim that the entire sample freezes at one fixed temperature. Similarly, boiling away solvent concentrates a nonvolatile solute.

A semipermeable membrane picture assumes the relevant solute is blocked. If two solutes have different permeabilities, equal total particle concentrations may not produce equal sustained osmotic effects. A robust answer names the membrane assumption rather than equating every “isotonic” situation with a simple chemical formula count.

Finally, the effect is a change in equilibrium condition, not a guarantee of a rate. Freezing can be delayed by supercooling, boiling can depend on nucleation, and solvent transport can be slow through a membrane. These kinetic observations do not change the definition of the equilibrium colligative quantities.

09 / UNDERSTAND & APPLY

9. Original colligative-property practice

These original problems assume ideal dilute behavior, nonvolatile solutes where vapor-pressure or boiling formulas require them, and no unmentioned chemical reactions. Any dissociation/association fraction is interpreted within the explicit ideal particle-count model.

Before substituting, label the denominator: kilograms of solvent, litres of solution or total moles. Then check whether a reported depression or elevation is a positive magnitude and apply the correct sign when converting it into the actual transition temperature.

ORIGINAL PRACTICE · 1

An ideal binary solution has solvent mole fraction 0.970 and pure-solvent vapor pressure 80.0 kPa. Find the solvent partial pressure.

Reveal answer & reasoning

77.6 kPa.

Raoult’s law gives 0.970×80.0=77.6. If the solute is volatile, total pressure would also include its vapor contribution.

ORIGINAL PRACTICE · 2

For a 0.200 m nonelectrolyte with K_b=0.500 K kg mol⁻¹, find boiling elevation.

Reveal answer & reasoning

0.100 K.

ΔT_b=K_bm=0.500×0.200. Add this positive elevation to the pure-solvent boiling point at the same external pressure.

ORIGINAL PRACTICE · 3

Why is 100 g of solution not automatically 0.100 kg of solvent?

Reveal answer & reasoning

The solution mass includes the solute.

Molality requires solvent mass alone. Subtract the solute mass if the total solution mass and solute mass are supplied.

ORIGINAL PRACTICE · 4

At the same temperature, compare ideal osmotic pressures of 0.100 M glucose and 0.100 M fully dissociated NaCl.

Reveal answer & reasoning

The NaCl model gives twice the osmotic pressure.

Glucose has i=1 and ideal NaCl i=2. With equal C and T, π scales with i. Real electrolyte nonideality can change the exact ratio.

ORIGINAL PRACTICE · 5

A dimerizing solute has ideal i=0.800. Find the fraction of monomer units associated.

Reveal answer & reasoning

α=0.400.

For dimers i=1−α/2, so α=2(1−0.800)=0.400.

ORIGINAL PRACTICE · 6

A compound’s formula molar mass is 200 g mol⁻¹ and effective i=2.00. What apparent molar mass results if the experiment is interpreted with i=1?

Reveal answer & reasoning

100 g mol⁻¹.

M_apparent=M_formula/i=200/2. More particles per gram mimic a smaller molar mass.

ORIGINAL PRACTICE · 7

Two ideal solutions have equal effective particle molarity. One is at 300 K and the other at 330 K. What is π_330/π_300?

Reveal answer & reasoning

1.10.

At fixed effective molarity, osmotic pressure is proportional to absolute temperature:330/300=1.10.

ORIGINAL PRACTICE · 8

If pure solvent freezes out of a solution containing a nonvolatile solute that remains dissolved, does the remaining liquid become more or less concentrated?

Reveal answer & reasoning

More concentrated.

Solvent amount decreases while dissolved solute remains, increasing molality and generally increasing the magnitude of further freezing depression within the model.

KEEP THE BIG IDEAS

Quick revision

  • Colligative formulas count effective particles under ideal or dilute assumptions.
  • Vapor-pressure lowering uses solvent mole fraction; freezing/boiling use molality; osmosis uses molarity.
  • A depression is subtracted from the pure-solvent freezing point; an elevation is added to boiling point.
  • The ideal van ’t Hoff relation is conditional and not a universal literal dissociation meter.
  • Apparent molar mass is inversely proportional to the effective particle factor.
  • Combine particles only after accounting for chemical reactions and a common concentration basis.
  • Membrane selectivity, nonideality and changing phase composition limit the simple equations.
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