← Chemistry — Std 12
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Solutions

Ch. 2Std 12

Easy Overview

You are making lemonade and you add sugar — does it all dissolve? How much sugar can you add before it stops dissolving? What happens to the boiling point of water when you dissolve salt in it? Solutions are everywhere — saline IV drips in hospitals, the alloys in your jewelry, the air you breathe — and understanding them means understanding concentration, solubility, and the weird things that happen when you add stuff to solvents. A solution is a homogeneous mixture where one substance (the solute) is uniformly dispersed in another (the solvent). The solvent is the component present in larger amount. Solutions can be gaseous (air), liquid (seawater, ethanol in water), or solid (brass, steel). In this chapter, you will focus mainly on liquid solutions, especially solids dissolved in liquids. Concentration can be expressed in many ways — and you need to know ALL of them. Mass percentage is mass of solute divided by total mass times 100. Volume percentage follows the same idea with volumes. Mole fraction (x) is moles of one component divided by total moles. Molarity (M) is the most common — moles of solute per liter of solution. But molarity changes with temperature because liquids expand when heated. Molality (m) fixes this — it is moles of solute per kilogram of solvent. Since mass does not change with temperature, molality is temperature-independent. Parts per million (ppm) is used for very dilute solutions like pollutants or trace minerals in water. Then there is solubility. The rule 'like dissolves like' is your best friend: polar dissolves polar, nonpolar dissolves nonpolar. That is why oil does not dissolve in water. Temperature affects solubility in opposite ways for solids vs gases. For most solids, solubility increases with temperature. But for gases, solubility DECREASES — that is why a warm soda goes flat faster. Henry's law governs gas solubility under pressure. The crown jewels of this chapter are the colligative properties — properties that depend ONLY on the number of solute particles, not their identity. Adding salt to water lowers the freezing point, raises the boiling point, lowers vapor pressure, and creates osmotic pressure. These four properties make antifreeze work, roads get salted in winter, and your cells stay healthy in IV fluids.

Types of solutions and concentration units — the toolbox

A solution is a homogeneous mixture with components uniformly distributed at the molecular level. The solvent is the major component; the solute is the minor one. Solutions can be gas-gas (air), gas-liquid (soda water), liquid-liquid (ethanol in water), solid-liquid (saltwater), or solid-solid (alloys). You need to be fluent in ALL concentration units. Mass percentage = (mass of solute / mass of solution) × 100. Volume percentage = (volume of solute / volume of solution) × 100. Parts per million (ppm) = (mass of solute / mass of solution) × 10⁶ — used for very dilute solutions like fluoride in water (1 ppm prevents tooth decay). Mole fraction (x): for a binary solution A + B, xₐ = nₐ / (nₐ + n_B) and x_B = n_B / (nₐ + n_B), where xₐ + x_B = 1. Molarity (M) = moles of solute / volume of solution in L. This is the most common unit in lab work — but it changes with temperature because volume expands. Molality (m) = moles of solute / mass of solvent in kg. Since mass is temperature-independent, molality is preferred for colligative property calculations. The key difference: if you need temperature-independent measurements, use molality. For lab titrations, use molarity. A typical problem: 2 g of NaOH (M = 40 g/mol) dissolved in water to make 500 mL solution gives M = (2/40) / 0.5 = 0.1 M. To find molality, you would need the mass of solvent, which requires knowing the solution density.

Solubility and the 'like dissolves like' principle

Solubility is the maximum amount of solute that can dissolve in a given amount of solvent at a specific temperature. The golden rule: 'like dissolves like.' Polar solvents (water, ethanol) dissolve polar or ionic solutes (salt, sugar, acids). Nonpolar solvents (benzene, CCl₄, hexane) dissolve nonpolar solutes (iodine, grease, oils). Why? Dissolving requires solvent molecules to pull apart solute particles. Polar water molecules can surround and stabilize ions through ion-dipole interactions. Nonpolar solvents cannot do this — try mixing oil and water and you will see two distinct layers. Temperature effects: for most solids in liquids, solubility increases with temperature (more thermal energy helps overcome the lattice energy of the solid). But there are exceptions — Ce₂(SO₄)₃ becomes LESS soluble as temperature rises (its dissolution is exothermic). For gases in liquids, solubility ALWAYS decreases with temperature — gas molecules have more kinetic energy and escape the solution. This is why cold water holds more dissolved oxygen than warm water (important for fish). Pressure has little effect on solid/liquid solubility but a huge effect on gas solubility — that is Henry's law. Common mistake: 'solubility' and 'rate of dissolving' are not the same. Solubility is the equilibrium amount that CAN dissolve; rate is how FAST it dissolves (affected by stirring, particle size, temperature).

Henry's law — gases under pressure

Henry's law: the mass of a gas dissolved in a given volume of liquid at constant temperature is directly proportional to the pressure of the gas above the liquid. Equation: p = K_H × x, where p = partial pressure of the gas, x = mole fraction of the gas in solution, and K_H = Henry's constant (depends on the gas and temperature — increases with temperature, meaning gases become less soluble as temperature rises). If you double the pressure, you double the dissolved gas amount. Applications: (1) Carbonated beverages — CO₂ is dissolved at high pressure (3-4 atm) in the sealed bottle. Open it, pressure drops to 1 atm, CO₂ becomes less soluble and fizzes out. (2) Deep-sea diving — divers breathe compressed air; at depth, N₂ dissolves in their blood. Surface too quickly and the rapid pressure drop causes N₂ to form bubbles in the blood — decompression sickness (the bends). Divers ascend slowly to allow slow N₂ elimination. (3) High-altitude breathing — lower partial pressure of O₂ means less O₂ dissolves in blood, causing altitude sickness. Henry's law also explains why fish survive under ice — cold water dissolves more O₂. A typical numerical: 'The K_H for CO₂ in water is 1.67 × 10³ atm at 25°C. Calculate the mole fraction of CO₂ in soda at 4 atm pressure.' Answer: x = p/K_H = 4/(1.67 × 10³) = 2.4 × 10⁻³.

Raoult's law for volatile liquids

Raoult's law applies to solutions of volatile liquids. It states: the partial vapor pressure of each component is directly proportional to its mole fraction. For a binary mixture of A and B: pₐ = p°ₐ × xₐ and p_B = p°_B × x_B, where p° is the vapor pressure of the pure component. Total vapor pressure P_total = pₐ + p_B = p°ₐxₐ + p°_Bx_B. Since x_B = 1 - xₐ, P_total = p°_B + (p°ₐ - p°_B)xₐ. This is a linear equation — ideal solutions give a straight line plot. The composition of the vapor is NOT the same as the liquid — the vapor is richer in the more volatile component (higher p°). This is the basis of fractional distillation. For example, benzene (p° = 75 mmHg) and toluene (p° = 22 mmHg): the vapor contains more benzene than the liquid does. By repeated condensation and re-evaporation, you separate them — this is how crude oil is refined. Exam tip: Raoult's law for the solvent holds when the solute is non-volatile. When both are volatile, it applies to both components. The law assumes ideal behavior — no special interactions between unlike molecules.

Ideal and non-ideal solutions — deviations

An ideal solution obeys Raoult's law at all compositions. Characteristics: ΔH_mix = 0 (no heat change — A-B interactions equal A-A and B-B), and ΔV_mix = 0 (volume is additive). Examples: benzene + toluene, n-hexane + n-heptane. Non-ideal solutions deviate. Positive deviation: vapor pressure is HIGHER than predicted. A-B interactions are WEAKER than A-A and B-B — molecules prefer their own kind and escape more easily. ΔH_mix > 0 (endothermic), ΔV_mix > 0 (volume expands). Example: ethanol + water. Negative deviation: vapor pressure LOWER than predicted. A-B interactions are STRONGER than A-A and B-B — molecules cling together. ΔH_mix < 0 (exothermic), ΔV_mix < 0 (volume contracts). Example: acetone + chloroform (they form H-bonds). Azeotropes are mixtures that boil at constant composition — the vapor has the same composition as the liquid, so they cannot be separated by distillation. Minimum-boiling azeotropes (ethanol-water: 95.6% ethanol, boils at 78.2°C) show positive deviation. Maximum-boiling azeotropes (HCl-water: 20.2% HCl, boils at 110°C) show negative deviation. Exam favorite: compare ideal vs non-ideal solutions with examples.

Colligative property 1 — relative lowering of vapor pressure

Adding a non-volatile solute lowers the vapor pressure of the solvent. Solute particles occupy some surface area, so fewer solvent molecules escape into the vapor. For a solution with a non-volatile solute, Raoult's law for the solvent: P = p° × x₁ (x₁ = mole fraction of solvent). Since x₁ + x₂ = 1: P = p°(1 - x₂). Rearranging: (p° - P)/p° = x₂. The left side (ΔP/p°) is the relative lowering of vapor pressure — and it equals the mole fraction of the solute. This is colligative — it depends only on the number (mole fraction) of solute particles, not their identity. Extended for molar mass determination: ΔP/P° = (w₂ × M₁)/(w₁ × M₂), where w₂ = mass of solute, w₁ = mass of solvent, M₂ = molar mass of solute, M₁ = molar mass of solvent. By measuring vapor pressure of pure solvent and solution, you can calculate M₂. However, vapor pressure measurements are experimentally tricky, so other colligative properties are more commonly used. Important: this only works for non-volatile solutes. If the solute is volatile, both partial pressures change and the formula does not apply.

Colligative property 2 — boiling point elevation

A liquid boils when its vapor pressure equals the external pressure (usually 1 atm). Adding a non-volatile solute lowers the vapor pressure, so the solution needs a HIGHER temperature to reach the same external pressure — the boiling point increases. ΔT_b = T_b(solution) - T_b(pure solvent) = K_b × m. K_b is the ebullioscopic constant (specific to the solvent), and m is molality. For water, K_b = 0.512 K kg mol⁻¹. This means a 1 m aqueous solution of a non-electrolyte boils at 100.512°C. In practical form: ΔT_b = (K_b × w₂ × 1000)/(w₁ × M₂), where w₂ = mass of solute (g), w₁ = mass of solvent (g), M₂ = molar mass (g/mol). Example: 10 g of glucose (M = 180 g/mol) in 200 g water. Molality = (10/180)/(200/1000) = 0.2778 m. ΔT_b = 0.512 × 0.2778 = 0.142°C. So the solution boils at 100.142°C. Real-world: adding salt to pasta water does elevate the boiling point, but the effect is tiny — 1 tablespoon in 4 L changes it by only about 0.01°C. The real reason is flavor. A more important application: antifreeze (ethylene glycol) in car radiators raises the boiling point of coolant to prevent summer boil-over (and lowers freezing point in winter).

Colligative property 3 — freezing point depression

Adding a non-volatile solute lowers the freezing point. ΔT_f = T_f(pure solvent) - T_f(solution) = K_f × m. For water, K_f = 1.86 K kg mol⁻¹. So a 1 m aqueous solution freezes at -1.86°C. Practical form: ΔT_f = (K_f × w₂ × 1000)/(w₁ × M₂). Applications: (1) Salting roads in winter — NaCl or CaCl₂ spread on icy roads dissolves in the thin water layer on ice, lowering its freezing point below ambient temperature, so the ice melts. NaCl works to about -10°C; CaCl₂ works to about -30°C. Environmental cost: salt runoff damages vegetation and corrodes cars. (2) Antifreeze — ethylene glycol in car radiators prevents freezing in winter. A 50:50 mixture with water freezes at about -37°C. (3) Making ice cream — a mixture of ice and salt around the container gets cold enough (-21°C) to freeze the cream mixture. The freezing point depression method is the most common experimental technique for molar mass determination because temperature measurements are more accurate than vapor pressure measurements. Lab tip: use a Beckmann thermometer for precise ΔT_f readings. Common mistake: confusing the subtraction — ΔT_f = T_f(°pure) - T_f(solution) while ΔT_b = T_b(solution) - T_b(°pure). The order is different!

Colligative property 4 — osmotic pressure

Osmosis is the spontaneous net movement of solvent molecules from a region of lower solute concentration to higher solute concentration through a semipermeable membrane. Think of it as nature trying to equalize concentrations. Osmotic pressure (π) is the excess pressure that must be applied to the solution side to PREVENT osmosis. van't Hoff equation: π = CRT, where C = molar concentration (mol/L), R = 0.0821 L atm mol⁻¹ K⁻¹, T = temperature in Kelvin. Also: π = (n₂/V)RT = (w₂RT)/(M₂V). Osmotic pressure is the MOST SENSITIVE colligative property — even very dilute solutions give measurable π. For example, 0.001 M solution at 25°C gives π = 0.001 × 0.0821 × 298 = 0.0245 atm ≈ 18.6 mmHg — easily measurable. This makes it ideal for determining molar masses of macromolecules (proteins, polymers). A 1% protein solution gives a small but measurable osmotic pressure — while ΔT_f for the same solution would be fractions of a degree. Reverse osmosis: if you apply pressure GREATER than π on the solution side, solvent flows FROM solution to pure solvent — this is how desalination plants produce fresh water from seawater (30-40 atm needed). This is also how your kidneys work — blood pressure forces water and small solutes through membranes. Common mistake: 'osmotic pressure' vs 'osmosis' — osmosis is the movement; osmotic pressure is the pressure needed to stop it.

Isotonic, hypotonic, and hypertonic solutions

Two solutions with the same osmotic pressure are isotonic. If you place a cell in an isotonic solution, no net water movement occurs — the cell stays healthy. Hypotonic: the solution outside has LOWER osmotic pressure than inside the cell (more dilute). Water flows INTO the cell, causing it to swell and potentially burst (lysis in red blood cells = hemolysis). Hypertonic: the solution outside has HIGHER osmotic pressure. Water flows OUT, causing the cell to shrink (crenation). This is why IV fluids must be isotonic with blood. Normal saline is 0.9% (w/v) NaCl — its osmotic pressure matches blood plasma (about 7.7 atm at body temperature). If you infused pure water, red blood cells would absorb water and burst (hemolysis) — potentially fatal. If you infused concentrated salt solution, cells would shrivel. This is also why you should not drink seawater when dehydrated — seawater is about 3.5% salt (hypertonic), and drinking it would actually DEHYDRATE your cells further. For plant cells, a hypotonic solution makes them turgid (firm, healthy) — the cell wall prevents bursting. A hypertonic solution causes plasmolysis — the cell membrane pulls away from the wall. This is why plants wilt in salty soil.

The van't Hoff factor (i) — correcting for dissociation

Colligative properties depend on the NUMBER of particles. When you dissolve NaCl, it dissociates into Na⁺ and Cl⁻ — two particles. So the actual freezing point depression should be twice what you would calculate assuming molecular NaCl. The van't Hoff factor (i) corrects for this. i = (observed colligative property)/(theoretical property assuming no dissociation). Also i = M_theoretical / M_observed. For non-electrolytes (glucose, urea): i = 1. For strong electrolytes that dissociate completely: i should equal the number of ions. NaCl (2 ions): i = 2. K₂SO₄ (3 ions): i = 3. AlCl₃ (4 ions): i = 4. But in reality, ion pairing at higher concentrations reduces i — for 0.1 m NaCl, i ≈ 1.87. For very dilute solutions, i approaches the theoretical value. For solutes that ASSOCIATE (benzoic acid in benzene forming dimers via H-bonds), i < 1 — two molecules become one particle. Modified equations: ΔT_f = i × K_f × m, ΔT_b = i × K_b × m, π = i × C × R × T, ΔP/P° = i × x₂. Exam trap: they give you a solution of known molality and measured ΔT_f, then ask you to find i and determine whether the result matches complete dissociation.

Degree of dissociation from van't Hoff factor

For an electrolyte A_xB_y that dissociates into x ions of A^(y+) and y ions of B^(x-), total ions n = x + y. If the degree of dissociation is α (fraction dissociated): undissociated = 1 - α, ions produced = nα. Total particles = (1 - α) + nα = 1 + α(n - 1). Therefore: i = 1 + α(n - 1) and α = (i - 1)/(n - 1). For NaCl (n = 2): if observed i = 1.8, then α = (1.8 - 1)/(2 - 1) = 0.8 (80% dissociated). For K₂SO₄ (n = 3): if i = 2.4, α = (2.4 - 1)/(3 - 1) = 0.7 (70% dissociated). For association (dimerization, n = 2): if β = degree of association, total particles = 1 - β + β/2 = 1 - β/2. So i = 1 - β/2, and β = 2(1 - i). For benzoic acid in benzene with i = 0.5: β = 2(1 - 0.5) = 1.0 (100% dimerized). For strong electrolytes at moderate concentrations, i is less than the theoretical number of ions due to ion pairing. The degree of dissociation approaches 1 only at infinite dilution. This relationship is a favorite board exam numerical — practice converting between i and α for different electrolyte types.

Abnormal molar masses — why numbers go wrong

When you use colligative properties to calculate molar mass without accounting for dissociation or association, you get 'abnormal' values. Dissociation: more particles than expected → colligative effect LARGER → calculated molar mass appears SMALLER. Example: NaCl in water — expected ΔT_f for 0.1 m solution = 0.186°C. But NaCl dissociates into 2 ions, so actual ΔT_f ≈ 0.372°C. Using M₂ = (K_f × w₂ × 1000)/(w₁ × ΔT_f) gives a calculated M about half of 58.5 g/mol (about 29 g/mol). Association: fewer particles → colligative effect SMALLER → calculated molar mass appears LARGER. Benzoic acid in benzene dimerizes — a 0.1 m solution behaves like 0.05 m. The observed ΔT_f is half of expected, and calculated M appears twice the real value. The pattern: dissociation → i > 1 → M_obs < M_actual; association → i < 1 → M_obs > M_actual. If a board exam problem gives you the observed ΔT_f and expected ΔT_f, you can immediately tell whether dissociation (observed > expected) or association (observed < expected) occurs. This pattern helps you quickly spot the correct answer in MCQs.

Determination of molar mass — practical methods

Each colligative property can determine molar mass, but the method depends on the solute. (1) Freezing point depression (cryoscopy) — most common in lab practicals. Dissolve known mass of solute in known mass of solvent, measure ΔT_f using a Beckmann thermometer (0.01°C precision), then M₂ = (K_f × w₂ × 1000)/(w₁ × ΔT_f). Best for moderate molar masses. (2) Boiling point elevation (ebullioscopy) — less common because K_b < K_f for most solvents, giving smaller ΔT. (3) Osmotic pressure — best for macromolecules (proteins, polymers). Even dilute solutions give measurable π. M₂ = (w₂RT)/(πV). For proteins with M in tens of thousands to millions g/mol, this is the method of choice. (4) Relative lowering of vapor pressure — measured with a manometer, less common experimentally. The choice of solvent matters: K_f depends on the solvent. Water: 1.86, benzene: 5.12, camphor: 39.7 K kg mol⁻¹. Camphor's enormous K_f makes it useful for the Rast method — micro-scale determinations using just a few mg of solute. Exam tip: always check whether the solute is an electrolyte before deciding whether to use i.

Numerical problem-solving strategies for solutions

Board exam questions on solutions are among the most numerical-heavy. Here is a systematic approach. Step 1: Write all given data with symbols — mass or moles of solute, mass or volume of solvent, temperature, K_f/K_b values. Step 2: Determine if the solute is an electrolyte. Non-electrolytes: i = 1. Electrolytes: find i (given or calculate from dissociation information). Step 3: Choose the right formula. For ΔT_f/ΔT_b: ΔT = K × m, where m = moles solute / kg solvent. For π: π = i × C × R × T. For vapor pressure: ΔP/P° = i × x₂. Step 4: Convert to consistent units. Mass in kg for molality. Temperature in Kelvin for π. Volume in L for molarity. Step 5: Solve. Watch for common mistakes: (a) Forgetting i for electrolytes. (b) Mixing up ΔT_f = T_f°- T_f vs ΔT_b = T_b - T_b° — the subtraction order is REVERSED. (c) Using molarity instead of molality — they are only equal when density = 1 g/mL. (d) Forgetting that ΔT_f and ΔT_b values in K and °C are numerically equal for differences. Practice the conversion between concentration units — many problems require converting molarity to molality using solution density. The more conversions you practice, the faster you get.

Key Points

  • •Solution = homogeneous mixture; solvent major, solute minor
  • •Molarity (M) = moles/volume(L); changes with temperature (volume expands)
  • •Molality (m) = moles/mass(kg); temperature-independent — use for colligative properties
  • •Mole fraction (x) = moles of component / total moles; dimensionless, sum = 1
  • •Henry's law: p = K_H × x; gas solubility ∝ pressure, ∝ 1/temperature
  • •Raoult's law: p = p° × x; applies to both volatile components in ideal solutions
  • •Ideal: ΔH_mix = 0, ΔV_mix = 0 (benzene + toluene); Positive deviation: weaker A-B interactions
  • •Negative deviation: stronger A-B interactions (acetone + chloroform); Azeotrope = constant-boiling mixture
  • •ΔP/P° = x₂ for non-volatile solute — relative lowering of vapor pressure
  • •ΔT_b = K_b × m; water K_b = 0.512 K kg mol⁻¹ — boiling point elevation
  • •ΔT_f = K_f × m; water K_f = 1.86 K kg mol⁻¹ — freezing point depression
  • •π = CRT (osmotic pressure); most sensitive colligative property — best for macromolecules
  • •Isotonic: same π; Hypotonic: lower π → cell swells (hemolysis); Hypertonic: higher π → cell shrinks
  • •van't Hoff factor i = observed/theoretical property; i > 1 for dissociation, i < 1 for association
  • •α = (i - 1)/(n - 1) where n = number of ions — degree of dissociation
  • •Abnormal M: dissociation → M_obs < M_actual; association → M_obs > M_actual
  • •Reverse osmosis: applied pressure > π purifies water (desalination, kidney function)
  • •Osmotic pressure method is best for protein molar masses — measurable at low concentrations

Practice Questions

  • Calculate the boiling point elevation when 10 g of glucose (C₆H₁₂O₆) is dissolved in 200 g of water. (K_b = 0.512 K kg mol⁻¹) What changes if you use NaCl instead?
  • Distinguish between ideal and non-ideal solutions. Explain positive and negative deviations with examples.
  • A 0.5 M solution of KCl shows freezing point depression twice that of a 0.5 M urea solution. Explain. Calculate the van't Hoff factor for KCl assuming 100% dissociation.
  • Derive the relationship between degree of dissociation and van't Hoff factor for a binary electrolyte. A 0.1 M solution of AB has i = 1.6. Find α.
  • What is osmotic pressure? Derive π = CRT. How would you use osmotic pressure to determine the molar mass of a protein?
  • Explain Henry's law. Why must deep-sea divers surface slowly? Calculate the mole fraction of CO₂ in a soda bottle at 4 atm if K_H = 1.67 × 10³ atm.
  • Calculate the mole fraction of ethanol in a solution of 46 g ethanol and 54 g water. Also calculate molality and molarity if density = 0.9 g/mL.
  • Define colligative properties and list them. Why are they called 'colligative'? Which is most sensitive and why?