CLASS 12 CHEMISTRY · CHAPTER 1 · STUDY NOTES
Solutions
A comprehensive study of liquid solutions, covering concentration units, solubility laws, vapour pressure, colligative properties, and abnormal molar masses.
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Introduction to Solutions and Their Types
First identify the final phase, then identify what is dissolved in it.A uniform mixture can be gas, liquid or solid
Read the assumptions
Particle sizes, clock time and vessel drawings are enlarged teaching models. Graphs marked schematic show a relationship, not measurements. Raoult’s-law and colligative calculations use ideal or dilute-solution assumptions; real interactions and dissociation can change the result. Exploration stays on this page and does not record chapter completion.
Original illustrations and calculations · NCERT Chemistry I, Solutions · source figures inform the models
- Solution
- A homogeneous mixture whose composition and properties are uniform on the macroscopic scale.
| Type of Solution | Solute | Solvent | Common Example |
|---|---|---|---|
| Gaseous | Gas | Gas | Mixture of oxygen and nitrogen |
| Gaseous | Liquid | Gas | Chloroform mixed with nitrogen |
| Gaseous | Solid | Gas | Camphor in nitrogen gas |
| Liquid | Gas | Liquid | Oxygen dissolved in water |
| Liquid | Liquid | Liquid | Ethanol dissolved in water |
| Liquid | Solid | Liquid | Glucose dissolved in water |
| Solid | Gas | Solid | Solution of hydrogen in palladium |
| Solid | Liquid | Solid | Amalgam of mercury with sodium |
| Solid | Solid | Solid | Copper dissolved in gold (Alloys) |
A binary solution has two components. The solvent usually contributes the larger amount and determines the solution’s physical state; the other component is the solute. A gas, liquid or solid may supply the solute, so there are nine starting-state combinations in the chapter’s classification.
Air illustrates a gaseous mixture, glucose in water a liquid solution, and copper in gold a solid solution. A dissolved gas is dispersed in the liquid: visible bubbles are a separate gas phase. Likewise, droplets floating in air are not a homogeneous gaseous solution.
Let’s look a little closer.
Go deeper: Uniform does not mean motionless
Molecules in a liquid or gas keep moving even when macroscopic composition is uniform. A homogeneous composition is not a claim that every microscopic volume contains exactly the same particles.
In the alloy picture, orange atoms replace some host atoms. This substitutional lattice is one way a solid solution can form; hydrogen in palladium is a different, interstitial-type example. The picture does not classify all alloys as single-phase solutions.
Take a moment. What do you think?
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Question
A sample contains dissolved oxygen and visible oxygen bubbles. Is the whole sample one homogeneous solution? Identify the solution phase.
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Methods of Expressing Concentration
Choose the denominator before substituting numbers.Same solution, different denominators
Read the assumptions
Particle sizes, clock time and vessel drawings are enlarged teaching models. Graphs marked schematic show a relationship, not measurements. Raoult’s-law and colligative calculations use ideal or dilute-solution assumptions; real interactions and dissociation can change the result. Exploration stays on this page and does not record chapter completion.
Original illustrations and calculations · NCERT Chemistry I, Solutions · source figures inform the models
- Molality m
- Moles of solute per kilogram of solvent; the denominator excludes the solute mass.
Mass percentage divides component mass by total solution mass and multiplies by 100. Volume percentage uses component volume divided by final solution volume; mixing liquid volumes need not be additive. Mass-by-volume percentage means grams of solute per 100 mL final solution. ppm is a fraction multiplied by 10⁶, with its mass or volume basis stated.
Mole fraction xi = ni/Σn is dimensionless, with Σxi = 1. Molarity M = nsolute/Vsolution uses litres of final solution; molality m = nsolute/msolvent uses kilograms of solvent. In a closed non-reacting sample, mass fractions, mole fractions and molality stay constant when temperature changes, while volume expansion can change molarity.
Go deeper: Keep preparation and composition separate
A specified mass of solute and solvent does not by itself tell you the final solution volume. Use a measured final volume or a stated density when converting to molarity. The interactive volume control describes a separately prepared solution, not an invented density relation.
1 mg/kg final solution is 1 ppm by mass. In very dilute water whose density is close to 1 kg/L, 1 mg/L is approximately 1 ppm; this shortcut fails when the density differs appreciably. Units such as mg/mL are not automatically ppm.
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Question
A student dissolves 12 g glucose in 188 g water. Find mass percentage and molality using 180 g/mol. Can you find molarity without more information?
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Solubility of Solids in Liquids
Watch the two directions before changing temperature.Saturation is a moving balance
Read the assumptions
Particle sizes, clock time and vessel drawings are enlarged teaching models. Graphs marked schematic show a relationship, not measurements. Raoult’s-law and colligative calculations use ideal or dilute-solution assumptions; real interactions and dissociation can change the result. Exploration stays on this page and does not record chapter completion.
Original illustrations and calculations · NCERT Chemistry I, Solutions · source figures inform the models
- Saturated solution
- A solution at its equilibrium solubility under specified conditions; in contact with excess solid, dissolution and crystallisation have equal average rates.
Solubility is the maximum equilibrium amount dissolved in a specified amount of solvent at a specified temperature and pressure. An unsaturated solution can dissolve more solute under those conditions. At saturation in contact with excess solid, particles still leave and rejoin the crystal; constant concentration does not mean that exchange has stopped.
Solute–solvent interactions must compete with those in the separate substances. Polar or ionic solutes often dissolve in polar solvents, and non-polar solutes in non-polar solvents. For the simple temperature comparison, endothermic dissolution is favoured by heating and exothermic dissolution by cooling. Ordinary pressure changes have little effect on solid–liquid solubility because both phases are relatively incompressible.
Go deeper: Rate is not the same as capacity
Stirring and breaking a crystal can speed the approach to equilibrium without necessarily changing equilibrium solubility. Adding more solid to an already saturated solution leaves excess solid rather than increasing the equilibrium dissolved concentration.
The temperature curves are qualitative, not universal straight lines or measurements for a named salt. Actual phase transitions, hydration and solution non-ideality can complicate the simple trend. “Like dissolves like” is a tendency, not a substitute for evidence.
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Question
Two saturated solid–liquid systems have endothermic and exothermic dissolution respectively. Predict the simple equilibrium response to heating. Does stirring necessarily give the same change?
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Solubility of Gases in Liquids: Henry's Law
Use the pressure of this gas, not automatically the total gas pressure.Use the gas’s partial pressure
Read the assumptions
Particle sizes, clock time and vessel drawings are enlarged teaching models. Graphs marked schematic show a relationship, not measurements. Raoult’s-law and colligative calculations use ideal or dilute-solution assumptions; real interactions and dissociation can change the result. Exploration stays on this page and does not record chapter completion.
Original illustrations and calculations · NCERT Chemistry I, Solutions · source figures inform the models
- Henry’s law
- At fixed temperature and in the applicable dilute gas-solution regime, gas partial pressure p is proportional to dissolved gas mole fraction x: p = KH x.
With the chapter’s convention, x = p/KH. At the same temperature and gas partial pressure, larger KH gives smaller dissolved mole fraction. Pressure and KH must be in matching units. Different gases, solvents and temperatures have different constants; keep those conditions attached to the value.
Increasing gas partial pressure raises equilibrium dissolved concentration. Opening a carbonated bottle lowers CO₂ partial pressure, allowing dissolved gas to leave. The chapter’s O₂ and N₂ constants increase between 293 and 303 K, so at fixed pressure these gases are less soluble in warmer water. Gas dissolution commonly releases heat, explaining this trend through equilibrium.
Go deeper: A useful law with a specified regime
The bottle animation illustrates a transient, whereas Henry’s law specifies equilibrium at fixed temperature. Bubble production and equilibration time depend on nucleation and transport; the law alone does not calculate the movie.
For a gas mixture use the species’ partial pressure. At altitude reduced oxygen partial pressure contributes to reduced availability; blood transport also involves haemoglobin and physiology, so water-solubility calculations are not a complete medical model. Diving-gas choice and ascent procedures need professional guidance, not a fixed textbook recipe.
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Question
For a hypothetical dilute gas solution KH = 5.0 × 10⁴ bar at a fixed temperature. Find x at gas partial pressures 0.40 and 0.80 bar. What happens if KH doubles at the same pressure?
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Vapour Pressure and Raoult's Law
Calculate liquid fractions first, then pressure, then vapour fractions.The vapour can have a different composition
Read the assumptions
Particle sizes, clock time and vessel drawings are enlarged teaching models. Graphs marked schematic show a relationship, not measurements. Raoult’s-law and colligative calculations use ideal or dilute-solution assumptions; real interactions and dissociation can change the result. Exploration stays on this page and does not record chapter completion.
Original illustrations and calculations · NCERT Chemistry I, Solutions · source figures inform the models
- Raoult’s law
- For an ideal liquid solution, the partial vapour pressure of component i is pi = xi pi°, with pure-liquid pressure pi° at the same temperature.
For two volatile ideal components A and B, ptotal = xA pA° + xB pB° with xB = 1−xA. The total is linear in liquid composition and joins the two non-zero pure-component pressures. Dalton’s relation gives vapour yi = pi/ptotal; it does not generally equal liquid xi.
If a solute is non-volatile, only the solvent contributes appreciably to vapour pressure. In the ideal model p1 = x1 p1°, so relative lowering (p1°−p1)/p1° = x2 = n2/(n1+n2). Replacing the denominator by n1 is a dilute-solution approximation.
Go deeper: Connect two pressure laws carefully
Raoult and Henry relations both connect partial pressure to liquid mole fraction but use different proportionality constants and reference limits. In an ideal-solution case KH coincides with the pure-component vapour pressure. Do not interchange constants for a real gas–liquid system without specifying the regime.
Non-volatile solute lowers solvent activity in the ideal model. A surface particle cartoon is an introductory picture, not a universal explanation that solute simply plugs escape holes. Volatile solutes also contribute their own vapour, so the total-pressure conclusion differs.
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Question
Original ideal mixture: pA° = 90 kPa, pB° = 30 kPa and liquid xA = 0.25. Calculate total pressure and vapour yA. Which phase is richer in A?
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Ideal and Non-Ideal Solutions
Compare attractions and pressure curves at the same temperature.Attractions change the escaping tendency
Read the assumptions
Particle sizes, clock time and vessel drawings are enlarged teaching models. Graphs marked schematic show a relationship, not measurements. Raoult’s-law and colligative calculations use ideal or dilute-solution assumptions; real interactions and dissociation can change the result. Exploration stays on this page and does not record chapter completion.
Original illustrations and calculations · NCERT Chemistry I, Solutions · source figures inform the models
- Azeotrope
- A boiling mixture at a specified pressure whose equilibrium liquid and vapour compositions are the same.
| Comparison | Ideal model | Positive deviation | Negative deviation |
|---|---|---|---|
| Vapour pressure at fixed T | Raoult prediction | Above prediction | Below prediction |
| Unlike attractions | Similar to like–like | Typically weaker | Typically stronger |
| Mixing enthalpy | Zero | Typically endothermic in this model | Typically exothermic in this model |
| Mixing volume | Zero | Often expansion; not a universal diagnostic | Often contraction; not a universal diagnostic |
Ideal solutions obey Raoult’s law throughout composition, with ΔmixH = 0 and ΔmixV = 0. Similar unlike and like–like interactions support near-ideal behaviour, as in the chapter’s benzene–toluene example. Real solutions may deviate above or below the ideal partial-pressure relations.
Weaker unlike attractions can give positive deviation, as illustrated by ethanol–acetone. Stronger unlike attractions can give negative deviation, as in chloroform–acetone, where a hydrogen-bond interaction is possible. Large deviations can produce a minimum-boiling or maximum-boiling azeotrope respectively; not every deviation produces one.
Go deeper: Keep a curve’s conditions attached
The pressure-versus-composition graph holds temperature fixed. A boiling-temperature envelope instead holds pressure fixed; liquid and vapour boundaries meet at an azeotrope. A pressure maximum corresponds qualitatively to a boiling minimum, not a second rising-temperature pressure axis.
At an azeotropic composition ordinary fractional distillation cannot enrich the components further by repeated equilibrium evaporation and condensation. The diagrams use teaching curves rather than measured values for a named mixture. Dashed attraction lines do not denote covalent reactions on mixing.
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Question
A binary mixture has weaker unlike attractions and measured total vapour pressure above its ideal prediction. What deviation does it show? Must it have an azeotrope?
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Colligative Properties: Boiling and Freezing
Use particle amount to predict a shift, then use the shift to infer molar mass.More particles shift the phase boundaries
Read the assumptions
Particle sizes, clock time and vessel drawings are enlarged teaching models. Graphs marked schematic show a relationship, not measurements. Raoult’s-law and colligative calculations use ideal or dilute-solution assumptions; real interactions and dissociation can change the result. Exploration stays on this page and does not record chapter completion.
Original illustrations and calculations · NCERT Chemistry I, Solutions · source figures inform the models
- Colligative property
- A solution property whose ideal dilute behaviour depends on dissolved particle amount rather than solute identity, with solvent and conditions specified.
For a dilute non-volatile non-electrolyte, ΔTb = Kb m and ΔTf = Kf m. Boiling-point elevation is Tb−Tb°; freezing-point depression is Tf°−Tf, so both reported shifts are positive magnitudes. Kb and Kf depend on the solvent and have units K kg/mol. At fixed external pressure, lowered solvent vapour pressure requires more heating to boil.
Lowered solvent chemical potential also shifts the solid–liquid equilibrium to lower temperature. The other colligative properties are relative lowering of solvent vapour pressure and osmotic pressure. To infer molar mass from a thermal shift, first find m = ΔT/K, then solute moles from solvent kilograms, then mass per mole. Correct for association or dissociation when necessary.
Go deeper: State what is being compared
Equal masses of different solutes usually give different numbers of moles. Compare equal particle amounts in equal solvent amounts, under dilute ideal conditions, to isolate a colligative effect. A non-volatile assumption matters: adding a volatile component can alter boiling behaviour differently.
Using masses in grams, M2 = K w2 × 1000/(ΔT w1) for i = 1. With an effective particle factor, replace K by iK. Keep temperature differences in K or °C consistently; a difference of 1 K equals a difference of 1 °C, but absolute temperature in osmotic equations must be kelvin.
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Question
Original example: 4.0 g of a non-electrolyte in 0.250 kg water gives ΔTf = 0.372 K. Using Kf = 1.86 K kg/mol and i = 1, determine its molar mass.
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Osmosis and Osmotic Pressure
Follow water, the retained solute and the opposing pressure separately.Follow water through a selective membrane
Read the assumptions
Particle sizes, clock time and vessel drawings are enlarged teaching models. Graphs marked schematic show a relationship, not measurements. Raoult’s-law and colligative calculations use ideal or dilute-solution assumptions; real interactions and dissociation can change the result. Exploration stays on this page and does not record chapter completion.
Original illustrations and calculations · NCERT Chemistry I, Solutions · source figures inform the models
- Osmotic pressure Π
- The extra solution-side pressure required to prevent net solvent entry through an ideal semipermeable membrane at specified conditions.
An ideal semipermeable membrane passes solvent and rejects the solute. With pure solvent on one side and solution on the other at initially equal pressures, the net solvent movement is into the solution. Molecules can cross both ways; osmosis describes the net movement. A rising solution column develops an opposing pressure difference.
For a dilute ideal non-electrolyte Π = CRT = nRT/V, with absolute T and consistent volume/pressure units. Extra pressure equal to Π prevents net osmosis; greater solution-side pressure can drive reverse osmosis. An actual membrane’s selectivity and operating conditions matter.
Go deeper: Concentration alone does not determine the direction in every setup
Net flow depends on both composition and pressure. It is therefore misleading to say solvent always flows towards a more concentrated solution: sufficient pressure can stop or reverse it. The moving-level scene is qualitative; a quantitative endpoint would need changing concentration, vessel geometry and hydrostatic pressure.
Isotonic, hypotonic and hypertonic compare the effective osmotic conditions outside a cell with those inside. Water can swell or shrink the cell; the drawing is a conceptual comparison, not a medical recommendation. For macromolecular molar mass, osmotic measurements can be useful near room temperature when heating may damage the sample.
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Question
A non-electrolyte has C = 0.020 mol/L at 300 K. Use R = 0.08314 L bar mol⁻¹ K⁻¹ to find Π. Predict net flow for imposed excess pressure 0.30 bar and 0.60 bar.
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Abnormal Molar Masses and van't Hoff Factor
Conserve solute units while recounting independently moving particles.Count independent particles after mixing
Read the assumptions
Particle sizes, clock time and vessel drawings are enlarged teaching models. Graphs marked schematic show a relationship, not measurements. Raoult’s-law and colligative calculations use ideal or dilute-solution assumptions; real interactions and dissociation can change the result. Exploration stays on this page and does not record chapter completion.
Original illustrations and calculations · NCERT Chemistry I, Solutions · source figures inform the models
- van’t Hoff factor i
- Observed colligative effect divided by the ideal effect calculated for the initial formula amount without association or dissociation.
Dissociation into more independent species increases particle amount: for AB → A⁺ + B⁻ with fraction α, i = 1+α in the ideal particle-count model. Dimerisation reduces it: for 2A → A₂ with fraction α, i = 1−α/2. Full dissociation of this AB gives i = 2; complete dimerisation gives i = 0.5.
Thermal shifts become ΔT = iKm and osmotic pressure becomes Π = iCRT. If one instead assumes i = 1, the apparent molar mass inferred from a measured effect is Mtrue/i. It is lower for dissociation and higher for association; the compound’s formula mass itself has not changed.
Go deeper: Limiting counts are not exact at every concentration
For dissociation into ν particles, the ideal fraction model gives i = 1+(ν−1)α. For association into r-mers it gives i = 1−α+α/r. Specify what fraction means: α refers to initial solute units involved, not the fraction of final particles that are dimers.
Real electrolyte solutions can have ion–ion interactions and non-ideal activities. Thus i lower than the ideal stoichiometric limit does not automatically prove incomplete dissociation. Values tend towards ideal limiting counts on sufficient dilution; a particle cartoon cannot quantify all interactions.
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Question
Half the initial A units dimerise. Find i. If true formula molar mass is 120 g/mol, what would an i=1 colligative calculation infer?
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Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 23, 24, 25, 26.