Solutions

A solution is a homogeneous mixture of two or more pure substances whose composition and properties are the same throughout. A solution with two components is a binary solution. The component present in the largest amount is the solvent; it decides the physical state of the solution. The other component is the solute.

Since each component can be a solid, a liquid or a gas, there are nine types of solutions. The examiner wants the NCERT examples, so learn them as a table in your head: gas in gas (a mixture of oxygen and nitrogen), liquid in gas (chloroform vapour mixed with nitrogen), solid in gas (camphor in nitrogen), gas in liquid (oxygen dissolved in water), liquid in liquid (ethanol in water), solid in liquid (glucose in water), gas in solid (hydrogen in palladium), liquid in solid (amalgam of mercury with sodium) and solid in solid (copper dissolved in gold).

Two things are asked repeatedly. First, 'give an example of a solid solution in which the solute is a gas'; the answer is hydrogen in palladium, and the solvent is the solid. Second, in a solid solution both components are solids, as in alloys such as brass; the solvent is again the component present in the larger amount. After this the chapter spends nearly all its time on liquid solutions, mostly aqueous ones, which is where the numericals come from.

Concentration can be stated qualitatively (dilute, concentrated) but a numerical needs a number. Learn all seven ways and, more importantly, learn to convert between them.

Mass percentage (w/w): mass of the component × 100 ÷ total mass of the solution. Commercial bleaching solution contains 3.62% (w/w) sodium hypochlorite in water. Volume percentage (v/v): volume of the component × 100 ÷ total volume of the solution; used for liquid in liquid, as in a 35% (v/v) ethylene glycol antifreeze. Mass by volume (w/V): mass of solute in grams per 100 mL of solution, common in medicine and pharmacy. Parts per million: parts of the component per million parts of the solution, used for traces; 5.8 × 10⁻³ g of dissolved oxygen per kg of sea water is 5.8 ppm. Mole fraction x_A = n_A/(n_A + n_B); the mole fractions of all components add up to 1. Molarity M is moles of solute per litre of solution. Molality m is moles of solute per kilogram of solvent.

The point examiners test: molarity depends on temperature because the volume of the solution changes on heating, while molality, mole fraction and mass percentage do not, since they involve masses only. That is why molality is used in the colligative-property formulas.

The standard conversion trick: whenever a solution is given as a mass percentage, assume 100 g of solution. For a 10% (w/w) glucose solution that gives 10 g glucose and 90 g water, so molality = (10/180)/0.090 = 0.617 mol kg⁻¹. If the density is also given, volume of solution = 100 g ÷ density, and molarity = moles ÷ that volume in litres. Write every step with units; the marking scheme gives a mark for the correct set-up even when the arithmetic slips.

Solubility is the maximum amount of a substance that can dissolve in a fixed amount of solvent at a given temperature and pressure. It depends on the nature of the solute and the solvent, on temperature and on pressure.

Nature: like dissolves like. Polar and ionic solutes such as NaCl dissolve in polar solvents like water; non-polar solutes such as naphthalene dissolve in non-polar solvents like benzene. Ionic compounds do not dissolve in benzene because the ion–dipole attraction that stabilises the ions in water is absent there.

Dynamic equilibrium: when a solid is added to a solvent, dissolution starts; as the concentration rises, some dissolved particles collide with the solid and crystallise. When the rates of dissolution and crystallisation become equal, no more solid dissolves: Solute + Solvent ⇌ Solution. Such a solution is saturated; one that can dissolve more solute at that temperature is unsaturated.

Temperature: since this is an equilibrium, Le Chatelier's principle applies. If dissolution is endothermic (Δ_solH > 0) solubility rises with temperature; if it is exothermic (Δ_solH < 0) solubility falls. Pressure: solids and liquids are almost incompressible, so pressure has no significant effect on the solubility of a solid in a liquid. Contrast this with gases in the next section, where pressure is the whole story.

Gases dissolve in liquids to different extents; oxygen dissolves only a little in water yet keeps aquatic life alive, while HCl is highly soluble. Solubility of a gas rises with pressure. Henry's law gives the quantitative form: at a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas. In the form the examiner wants: the partial pressure of the gas in the vapour phase is proportional to the mole fraction of the gas in the solution, p = K_H x, where K_H is Henry's law constant.

Read K_H the right way round. For a given pressure, a higher K_H means a smaller mole fraction, so lower solubility. At 293 K, K_H for oxygen (34.86 kbar) is lower than for nitrogen (76.48 kbar), so oxygen is the more soluble of the two. K_H depends on the gas, the solvent and the temperature.

Temperature: dissolving a gas is like condensation and is exothermic, so by Le Chatelier's principle solubility falls as the temperature rises; K_H increases with temperature. This is why boiling water drives off dissolved air and why aquatic species prefer cold water.

Applications (three are enough for full marks): (1) soda bottles are sealed under high CO₂ pressure to increase its solubility; (2) scuba divers breathe air at high pressure, so more nitrogen dissolves in the blood; when they surface, the pressure drops, nitrogen bubbles out and blocks capillaries, causing the painful 'bends'; their tanks are therefore filled with air diluted with helium (about 11.7% He, 56.2% N₂, 32.1% O₂), helium being much less soluble; (3) at high altitude the partial pressure of oxygen is low, so less oxygen dissolves in blood and tissues, giving climbers anoxia: weakness and inability to think clearly.

Consider two volatile liquids 1 and 2 in a closed vessel. Both evaporate and at equilibrium the total vapour pressure is the sum of their partial pressures. Raoult's law (1886): for a solution of volatile liquids, the partial vapour pressure of each component is directly proportional to its mole fraction in the solution: p₁ = p₁⁰ x₁ and p₂ = p₂⁰ x₂, where p⁰ is the vapour pressure of the pure component at the same temperature.

By Dalton's law the total pressure is p_total = p₁ + p₂ = p₁⁰ x₁ + p₂⁰ x₂. Substituting x₁ = 1 − x₂ gives p_total = p₁⁰ + (p₂⁰ − p₁⁰) x₂. Three conclusions the examiner looks for: the total vapour pressure varies linearly with the mole fraction of either component; it lies between the vapour pressures of the two pure components; a plot of p₁, p₂ and p_total against x₂ gives three straight lines, with p_total running from p₁⁰ at x₂ = 0 to p₂⁰ at x₂ = 1.

Composition of the vapour phase: if y₁ and y₂ are the mole fractions in the vapour, then p₁ = y₁ p_total and p₂ = y₂ p_total, so y₁ = p₁/p_total. The vapour is always richer in the more volatile component, the one with the higher p⁰. This is the basis of fractional distillation.

Standard numerical (NCERT intext 1.8): p_A⁰ = 450 mm Hg, p_B⁰ = 700 mm Hg, p_total = 600 mm Hg. Then 600 = 450 x_A + 700 (1 − x_A) = 700 − 250 x_A, so x_A = 0.40 and x_B = 0.60. Vapour: y_A = 450 × 0.40/600 = 0.30, y_B = 0.70. The vapour is richer in B, the more volatile liquid. Always state Raoult's law in words before writing the equation; the statement carries its own mark.

This short section is a favourite 2-mark question: 'how is Raoult's law a special case of Henry's law?' Compare the two. Henry's law for a gas dissolved in a liquid: p = K_H x. Raoult's law for a volatile component of a solution: p = p⁰ x. In both, the partial pressure of the volatile component in the vapour phase is directly proportional to its mole fraction in the solution; only the proportionality constant differs. Henry's constant K_H depends on the gas–solvent pair, whereas in Raoult's law the constant is the vapour pressure of the pure component, p⁰. So Raoult's law becomes a special case of Henry's law in which K_H equals p⁰.

The physical picture: in a solution of a gas in a liquid, one component is so volatile that it exists as a gas, and its solubility is described by Henry's law; a volatile liquid component of a solution is described by Raoult's law. Both laws say the same thing, that the escaping tendency of a component is proportional to how much of it is present at the surface, which is what the mole fraction measures.

For full marks write both equations, say what each constant is, and finish with the sentence 'Raoult's law is a special case of Henry's law where K_H = p⁰'. Do not write that the two laws are the same; they coincide only when the constants are equal, which is the ideal case.

Sugar in water, urea in water: the solute is non-volatile, so it contributes nothing to the vapour. The vapour pressure of the solution comes only from the solvent and is lower than that of the pure solvent at the same temperature. Raoult's law in its general form says that for any solution the partial vapour pressure of each volatile component is proportional to its mole fraction. With the solvent as component 1, p₁ = p₁⁰ x₁, and since only the solvent is volatile, p_total = p₁⁰ x₁. A plot of p₁ against x₁ is a straight line through the origin, ending at p₁⁰ when x₁ = 1.

Why does the vapour pressure fall? In the pure solvent the whole surface is solvent molecules. In the solution part of the surface is occupied by solute particles, so fewer solvent molecules can escape per unit time and the equilibrium vapour pressure is lower. Note what the reasoning does not depend on: the nature of the solute. The lowering depends only on the number of solute particles per given number of solvent molecules. That observation is the doorway to colligative properties.

A quick calculation the examiner likes (NCERT 1.17): the vapour pressure of water at 300 K is 12.3 kPa; a 1 molal solution has 1 mol of solute in 1000/18 = 55.56 mol of water, so x₁ = 55.56/56.56 = 0.9823 and p = 12.3 × 0.9823 = 12.08 kPa. Two lines of working, one mark each.

Ideal solutions obey Raoult's law over the entire range of concentration. Two conditions come with it: the enthalpy of mixing is zero (Δ_mixH = 0, no heat absorbed or evolved) and the volume of mixing is zero (Δ_mixV = 0, the volume of the solution equals the sum of the volumes of the components). At the molecular level this happens when the solute–solvent (A–B) interactions are nearly the same as the solute–solute (A–A) and solvent–solvent (B–B) interactions. No solution is perfectly ideal, but n-hexane with n-heptane, bromoethane with chloroethane, and benzene with toluene come very close.

Non-ideal solutions deviate from Raoult's law, positively or negatively. In positive deviation the A–B interactions are weaker than A–A and B–B, so molecules escape more easily and the vapour pressure is higher than Raoult's law predicts. Mixing absorbs heat (Δ_mixH > 0) and the volume expands (Δ_mixV > 0). Ethanol and acetone is the standard example: acetone molecules get between ethanol molecules and break some of their hydrogen bonds. Carbon disulphide with acetone also shows it. In negative deviation the A–B interactions are stronger than A–A and B–B, the vapour pressure is lower than predicted, heat is evolved (Δ_mixH < 0) and the volume contracts (Δ_mixV < 0). Phenol with aniline (hydrogen bonding between the phenolic proton and the lone pair on nitrogen) and chloroform with acetone (a hydrogen bond between the chloroform hydrogen and the acetone oxygen) are the examples to quote.

Azeotropes are binary mixtures having the same composition in the liquid and vapour phases that boil at a constant temperature. Because the vapour has the same composition as the liquid, the components cannot be separated by fractional distillation. Solutions with a large positive deviation form a minimum boiling azeotrope at a particular composition; ethanol–water is one, with about 95% ethanol by volume, which is why fermentation and distillation cannot give ethanol purer than this. Solutions with a large negative deviation form a maximum boiling azeotrope; nitric acid–water is one, with about 68% nitric acid by mass, boiling at 393.5 K.

For the board, be ready to draw the vapour pressure–composition plots for ideal, positive and negative deviations, and to state the sign of Δ_mixH with the type of deviation.

Colligative properties are properties of a solution that depend on the number of solute particles relative to the total number of particles present, not on the nature of the solute. The name comes from the Latin colligare, to bind together. There are four: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure. Each gives a route to the molar mass of a non-volatile solute, which is what most numericals ask.

Relative lowering of vapour pressure. Start from Raoult's law for the solvent: p₁ = x₁ p₁⁰. The lowering is Δp₁ = p₁⁰ − p₁ = p₁⁰ − p₁⁰ x₁ = p₁⁰(1 − x₁) = p₁⁰ x₂. Dividing by p₁⁰: (p₁⁰ − p₁)/p₁⁰ = x₂ = n₂/(n₁ + n₂). The relative lowering equals the mole fraction of the solute, so it depends only on the number of solute particles: it is colligative. For a dilute solution n₂ is much smaller than n₁, so n₂/(n₁ + n₂) ≈ n₂/n₁ = (w₂/M₂) ÷ (w₁/M₁) = w₂M₁/(M₂w₁). Rearranged: M₂ = w₂M₁p₁⁰/(w₁(p₁⁰ − p₁)).

A form worth knowing for exact working: since p₁⁰/p₁ = (n₁ + n₂)/n₁, we get (p₁⁰ − p₁)/p₁ = n₂/n₁ with no approximation. NCERT uses this in Exercise 1.19.

Worked example (NCERT intext 1.9): 50 g of urea in 850 g of water, p⁰ = 23.8 mm Hg. n₂ = 50/60 = 0.833 mol, n₁ = 850/18 = 47.22 mol, x₂ = 0.833/48.05 = 0.0173. The relative lowering is 0.0173, so p = 23.8 × (1 − 0.0173) = 23.39 mm Hg. Two marking points: the relative lowering, and the final pressure with its unit.

A liquid boils when its vapour pressure equals the external pressure; water boils at 373.15 K at 1 atm. Adding a non-volatile solute lowers the vapour pressure at every temperature, so the solution must be heated to a higher temperature before its vapour pressure reaches 1 atm. The boiling point rises. On a vapour pressure–temperature plot the solution curve lies below the solvent curve and meets the 1 atm line further to the right; this diagram is expected with the derivation.

If T_b⁰ is the boiling point of the pure solvent and T_b that of the solution, the elevation is ΔT_b = T_b − T_b⁰. Experiments show that for dilute solutions the elevation is proportional to the molal concentration of the solute: ΔT_b ∝ m, or ΔT_b = K_b m. K_b is the boiling point elevation constant, molal elevation constant or ebullioscopic constant, with unit K kg mol⁻¹. It is a property of the solvent alone (0.52 K kg mol⁻¹ for water, 2.53 for benzene) and does not change if the molality is doubled, which was a 2023 MCQ.

Derivation of the molar-mass relation. If w₂ g of solute of molar mass M₂ is dissolved in w₁ g of solvent, molality m = (w₂/M₂) ÷ (w₁/1000) = 1000 w₂/(M₂ w₁). Substituting: ΔT_b = K_b × 1000 w₂/(M₂ w₁), so M₂ = 1000 K_b w₂/(ΔT_b w₁). The marking scheme rewards: the proportionality statement, the definition of K_b, the substitution for m and the final expression.

Worked example (CBSE 2015): 1.5 g of a solute in 90 g of benzene raises the boiling point from 353.23 K to 353.93 K, K_b = 2.52 K kg mol⁻¹. ΔT_b = 0.70 K. M₂ = 1000 × 2.52 × 1.5/(0.70 × 90) = 3780/63 = 60 g mol⁻¹.

The freezing point of a substance is the temperature at which its solid and liquid phases have the same vapour pressure. A solution has a lower vapour pressure than the pure solvent, so it reaches the vapour pressure of the solid solvent only at a lower temperature: the freezing point is depressed. On the vapour pressure–temperature diagram the solution curve cuts the solid-solvent curve to the left of the point where the pure-liquid curve cuts it.

The depression is ΔT_f = T_f⁰ − T_f. For dilute solutions ΔT_f ∝ m, or ΔT_f = K_f m, where K_f is the freezing point depression constant, molal depression constant or cryoscopic constant, unit K kg mol⁻¹, again a property of the solvent only: 1.86 for water, 5.12 for benzene. Substituting m = 1000 w₂/(M₂ w₁) exactly as for the boiling point gives M₂ = 1000 K_f w₂/(ΔT_f w₁).

Both K_b and K_f can be calculated from the solvent's boiling or freezing point, its molar mass and its enthalpy of vaporisation or fusion: K_f = R M₁ T_f²/(1000 Δ_fusH). This is rarely asked as a derivation, but a 1-mark 'what does K_f depend on' is common: only the solvent.

Everyday uses that get asked as 'give reasons': ethylene glycol is added to car radiators as antifreeze because it lowers the freezing point of water; salt or CaCl₂ is spread on icy roads because the solution formed freezes below 0 °C, so the ice melts.

Worked example (CBSE 2015): freezing point of a solution of 31 g of ethylene glycol (M = 62 g mol⁻¹) in 500 g of water. n₂ = 31/62 = 0.5 mol, m = 0.5/0.500 = 1.0 mol kg⁻¹, ΔT_f = 1.86 × 1.0 = 1.86 K, so T_f = 273.15 − 1.86 = 271.29 K, that is −1.86 °C. Always end with the freezing point itself, not just the depression, if that is what was asked.

A semipermeable membrane (SPM) lets solvent molecules through but not solute particles: pig's bladder, parchment, cellophane, or the cellulose acetate film used industrially. If a solution and its pure solvent are separated by an SPM, solvent flows into the solution. This spontaneous flow of solvent from lower to higher solute concentration through an SPM is osmosis.

Osmotic pressure (π) is the excess pressure that must be applied to the solution side to just stop osmosis. For dilute solutions it is proportional to the molarity and to the temperature: π = C R T = (n₂/V) R T. Since n₂ = w₂/M₂, π V = w₂ R T/M₂, giving M₂ = w₂ R T/(π V). Osmotic pressure depends only on the number of solute particles, so it is colligative.

Why it is the preferred method for proteins, polymers and other macromolecules (a standard 2-mark reason): it is measured at room temperature, so heat-sensitive biomolecules survive; molarity is used instead of molality; and even a very dilute solution gives a measurable pressure, whereas the boiling or freezing point changes of the same solution would be too small to measure. NCERT intext 1.12: 1.0 g of a polymer of molar mass 185,000 in 450 mL of water at 37 °C gives π = 30.96 Pa, still measurable.

Isotonic solutions have the same osmotic pressure at the same temperature, so no net osmosis occurs between them; 0.9% (mass/volume) NaCl is isotonic with the fluid inside red blood cells and is used for intravenous injections. In a more concentrated (hypertonic) solution water leaves the cells and they shrink; in a more dilute (hypotonic) solution water enters and they swell and may burst. The same effect explains why raw mangoes shrivel in brine, why wilted flowers revive in fresh water and why pickles do not spoil: bacteria lose water by osmosis.

Reverse osmosis: if a pressure greater than the osmotic pressure is applied to the solution side, solvent flows out of the solution into the pure solvent. It is used for the desalination of sea water; the pressure needed is high and the membrane is a film of cellulose acetate on a porous support, permeable to water but not to the ions in sea water.

The colligative formulas assume that every solute molecule stays as one particle. Two things break that. Dissociation: KCl in water gives K⁺ and Cl⁻, two particles per formula unit, so every colligative effect is nearly doubled and the molar mass calculated from it comes out about half the true value. Association: ethanoic acid in benzene dimerises through hydrogen bonding, halving the number of particles, so the calculated molar mass is about twice the true value; NCERT quotes ethanoic acid showing a molar mass of 120 in benzene. Molar masses that differ from the expected value in this way are called abnormal molar masses.

In 1880 van't Hoff introduced the factor i to account for this: i = normal molar mass ÷ abnormal molar mass = observed colligative property ÷ calculated colligative property = total number of moles of particles after association or dissociation ÷ number of moles of particles before. i < 1 for association, i > 1 for dissociation, i = 1 for solutes like glucose and urea. The four equations become (p₁⁰ − p₁)/p₁⁰ = i n₂/n₁, ΔT_b = i K_b m, ΔT_f = i K_f m and π = i n₂RT/V.

Degree of dissociation. For an electrolyte giving n ions per formula unit, start with 1 mol and let a fraction α dissociate: particles = 1 − α + nα, so i = 1 + α(n − 1) and α = (i − 1)/(n − 1). Degree of association. For nA ⇌ A_n, particles = 1 − α + α/n, so i = 1 − α + α/n and α = (1 − i)/(1 − 1/n); for a dimer, i = 1 − α/2.

Read the words in a numerical carefully: 'completely dissociated' means use the full number of ions (NaCl 2, CaCl₂ 3, K₂SO₄ 3, Al₂(SO₄)₃ 5); '92% ionised' means i = 1 + 0.92(n − 1); 'dimerises' means i lies between 0.5 and 1. Real electrolytes have i slightly below the ideal number (NaCl in 0.1 m solution has i = 1.87) because of attraction between ions, and i approaches the ideal value on dilution.