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State Board · Class 12 · Chemistry

Lesson 2: Solutions

7 concepts · 56 practice questions with answers · Free to read

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Types of solutions and concentration

From daily life

Tea turns sweeter with more sugar — sweetness is concentration.

Simple definition

A solution is a homogeneous mixture; concentration is expressed as molarity, molality, mole fraction, mass % or ppm.

ConcentrationExpressing concentrationQuantityFormulaUnitMolarity Mn(solute) / V(L)mol L⁻¹Molality mn(solute) / kg solventmol kg⁻¹Mole fraction xn₂ / (n₁ + n₂)no unitMass %w₂ / w(soln) × 100%ppm10⁶ × w₂ / w(soln)ppmEx. 18 g glucose (M = 180) + 100 g waterm = 0.1 mol / 0.100 kg = 1.0 mol kg⁻¹x₂ = 0.1 / (0.1 + 5.556) = 0.0177M depends on temperature; m and x do not

In the figure : Concentration terms.

Rule 1: Types of solutions — Nine types by the physical state of solvent and solute: gas–gas (air), liquid–liquid (water–ethanol), solid–solid (brass), gas–liquid (soda).

  • Easy : Brass = Cu + Zn.
  • Medium : Soda = CO₂ + water.
  • Exam-type : Air is a gaseous mixture.

Rule 2: Molarity and molality — M = mol per litre of solution (depends on T); m = mol per kg of solvent (independent of T).

  • Easy : 18 g glucose + 100 g water ⇒ m = 1.0.
  • Medium : 2 mol in 4 L ⇒ M = 0.5.
  • Exam-type : m does not change with T.

Rule 3: Mole fraction and ppm — x = n/N with Σx = 1; ppm = 10⁶ × solute mass / solution mass.

  • Easy : x = 0.1/5.656 = 0.0177.
  • Medium : 1 mg/kg = 1 ppm.
  • Exam-type : Water pollutants are given in ppm.

Calculation: 18 g glucose (M = 180) + 100 g water

  • Step 1: n(glucose) = 18/180 = 0.1 mol; n(water) = 100/18 = 5.556 mol.
  • Step 2: m = 0.1/0.100 = 1.0 mol kg⁻¹.
  • Step 3: x = 0.1/(0.1 + 5.556) = 0.0177.

CET level (MCQ)

  • CET 1) Independent of temperature? 1) molality 2) molarity 3) volume % 4) none — Answer: 1).
  • CET 2) Sum of all mole fractions? 1) 1 2) 0 3) 100 4) 10 — Answer: 1).

Real-life use: Drug doses, saline, water quality, beverages.

Remember

M = n/V(L); m = n/kg solvent; x = n/N; ppm = 10⁶ w₂/w.

Common mistake

Using kg of solution in molality.

CET trap

Molality uses kg of solvent, not solution.

Shortcut

Memory clue: "molarity per litre, molality per kilo".

Spot the difference

QuantityUnit
Mmol L⁻¹
mmol kg⁻¹
xnone
ppmmg kg⁻¹

Practice questions

Easy
  1. 1What is a solution?
  2. 2Molarity vs molality.
  3. 3What is mole fraction?
Medium
  1. 4What is ppm?
  2. 518 g glucose + 100 g water ⇒ m.
  3. 6Its x.
Challenging
  1. 7What type is brass?
  2. 8Why does M change with T?
Show answers (try them yourself first)
  1. 1Homogeneous mixture.
  2. 2L solution vs kg solvent.
  3. 3n/N.
  4. 4One part per million.
  5. 51.0.
  6. 60.0177.
  7. 7Solid–solid.
  8. 8Volume changes.

Solubility and Henry's law

From daily life

Fizz appears when a soda bottle is opened — pressure drops and CO₂ solubility falls.

Simple definition

Solubility is the amount in a saturated solution at a given T. Henry's law: p = K_H x; gas solubility is proportional to pressure.

HenryHenry's law: p = KH · xpressure psolubilityKH ↓KH ↑Large KH ⇒less solubleT↑ ⇒ KH↑solubility ↓Gas solubility ∝ pressure; falls as temperature rises1 atm ⇒ 0.03 mol/L; 3 atm ⇒ 0.09 mol/L (example)Uses: soda bottles, diver's bends, high-altitude hypoxia

In the figure : Henry's law.

Rule 1: Factors — "Like dissolves like"; solids: usually ↑ with T; gases: ↓ with T.

  • Easy : Salt in water, naphthalene in benzene.
  • Medium : Less O₂ in warm water.
  • Exam-type : Pressure hardly affects solids.

Rule 2: Henry's law — p = K_H x; larger K_H ⇒ lower solubility; T↑ ⇒ K_H↑.

  • Easy : 1 atm ⇒ 0.03 mol/L ⇒ 3 atm ⇒ 0.09.
  • Medium : N₂ has a large K_H.
  • Exam-type : Double pressure ⇒ double solubility.

Rule 3: Applications — Soda bottles, diver's bends (N₂ in blood), high-altitude hypoxia (less O₂).

  • Easy : Divers use helium mixtures.
  • Medium : p(O₂) falls at altitude.
  • Exam-type : Bubbles = pressure drop.

Calculation: 0.03 mol/L at 1 atm; at 3 atm?

  • Step 1: solubility ∝ p.
  • Step 2: 0.03 × 3 = 0.09.
  • Step 3: 0.09 mol/L.

CET level (MCQ)

  • CET 1) Gas solubility as T rises? 1) decreases 2) increases 3) same 4) zero — Answer: 1).
  • CET 2) Large K_H means? 1) lower solubility 2) higher 3) same 4) infinite — Answer: 1).

Real-life use: Soft drinks, diving, mountaineering.

Remember

p = K_H x; solubility ∝ p; T↑ ⇒ gas solubility ↓.

Common mistake

Thinking larger K_H means more soluble.

CET trap

Larger K_H ⇒ lower solubility.

Shortcut

Memory clue: "more pressure, more dissolved".

Spot the difference

FactorEffect
p↑gas solubility ↑
T↑ (gas)solubility ↓
K_H↑solubility ↓

Practice questions

Easy
  1. 1State Henry's law.
  2. 2Meaning of K_H.
  3. 3How does bends occur?
Medium
  1. 4Why does soda fizz?
  2. 5Effect of T on gas solubility.
  3. 61 atm ⇒ 3 atm calculation.
Challenging
  1. 7Explain 'like dissolves like'.
  2. 8Hypoxia at altitude.
Show answers (try them yourself first)
  1. 1p = K_H x.
  2. 2Large K_H ⇒ low solubility.
  3. 3N₂ bubbles in blood.
  4. 4Pressure drops.
  5. 5Decreases.
  6. 60.09.
  7. 7Similar polarity.
  8. 8Low p(O₂).

Raoult's law: ideal and non-ideal solutions

From daily life

Two similar friends mixing change nothing — an ideal solution.

Simple definition

Raoult's law: p_A = x_A p°A. An ideal solution obeys it for all components with ΔH_mix = 0 and ΔV_mix = 0.

RaoultRaoult's law: ideal solutionxBpp°A = 80p°B = 40p (total)pA = xA p°ApB = xB p°Bp = pA + pBEx. xA = 0.4: p = 0.4×80 + 0.6×40 = 56 kPaVapour yA = 32/56 = 0.571 (richer in volatile A)Positive deviation: p above Raoult, ΔH > 0 (ethanol + hexane)Negative deviation: p below Raoult, ΔH < 0 (acetone + chloroform)

In the figure : Raoult's law and deviations.

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Raoult's law: ideal and non-ideal solutions

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Rule 1: Ideal solution — p = x_A p°A + x_B p°B; A–B interactions equal A–A and B–B; e.g. benzene + toluene, hexane + heptane.

  • Easy : p°A = 80, p°B = 40, x_A = 0.4 ⇒ p = 56 kPa.
  • Medium : y_A = 32/56 = 0.571.
  • Exam-type : Vapour is richer in the more volatile component.

Rule 2: Positive deviation — A–B attraction weaker ⇒ p higher, ΔH > 0, ΔV > 0; ethanol + hexane, CCl₄ + CHCl₃.

  • Easy : Solution cools.
  • Medium : Minimum-boiling azeotrope.
  • Exam-type : Ethanol + water (95.6 %).

Rule 3: Negative deviation — Stronger A–B attraction ⇒ p lower, ΔH < 0, ΔV < 0; acetone + chloroform, HNO₃ + water.

  • Easy : Solution warms.
  • Medium : Maximum-boiling azeotrope.
  • Exam-type : HNO₃ + water (68 %).

Calculation: p°A = 80, p°B = 40 kPa; x_A = 0.4

  • Step 1: p_A = 0.4 × 80 = 32; p_B = 0.6 × 40 = 24.
  • Step 2: p = 32 + 24 = 56 kPa.
  • Step 3: y_A = 32/56 = 0.571.

CET level (MCQ)

  • CET 1) ΔH_mix of an ideal solution? 1) 0 2) + 3) – 4) infinite — Answer: 1).
  • CET 2) Negative deviation ⇒? 1) ΔH < 0 2) ΔH > 0 3) ΔH = 0 4) ΔV > 0 — Answer: 1).

Real-life use: Distillation, alcohol production, pharmaceutical solvents.

Remember

p = x_A p°A + x_B p°B; ideal ΔH = 0; positive ⇒ ΔH > 0; negative ⇒ ΔH < 0.

Common mistake

Writing negative ΔH for positive deviation.

CET trap

Positive deviation ⇒ ΔH +; negative ⇒ –.

Shortcut

Memory clue: "positive = apart, negative = closer".

Spot the difference

TypeΔHp
Ideal0Raoult
Positive+higher
Negative–lower

Practice questions

Easy
  1. 1State Raoult's law.
  2. 2Properties of ideal solutions.
  3. 3Find p and y_A.
Medium
  1. 4Why positive deviation?
  2. 5Why negative deviation?
  3. 6What is an azeotrope?
Challenging
  1. 7Examples of ideal solutions.
  2. 8Sign of ΔV.
Show answers (try them yourself first)
  1. 1p_A = x_A p°A.
  2. 2ΔH = 0, ΔV = 0.
  3. 356 kPa, 0.571.
  4. 4Weaker A–B attraction.
  5. 5Stronger A–B attraction.
  6. 6Constant-boiling mixture.
  7. 7Benzene + toluene.
  8. 8Positive and negative.

Colligative properties and lowering of vapour pressure

From daily life

Salt makes water boil later — the effect depends on the number of solute particles.

Simple definition

Colligative properties depend only on the number of solute particles: vapour-pressure lowering, boiling-point elevation, freezing-point depression and osmotic pressure.

LoweringLowering of vapour pressure↑↑↑↑↑↑↑↑Pure solvent↑↑↑↑↑Solution (non-volatile solute)p°p < p°(p° – p)/p° = x₂ = n₂/(n₁ + n₂)0.1 mol glucose + 90 g water (5 mol): x₂ = 0.0196p° = 23.76 mmHg ⇒ Δp = 0.47 mmHgSolute particles occupy the surface ⇒ less evaporation

In the figure : Vapour-pressure lowering.

Rule 1: Relative lowering — (p° – p)/p° = x₂ for a non-volatile solute.

  • Easy : x₂ = 0.1/5.1 = 0.0196.
  • Medium : p° = 23.76 ⇒ Δp = 0.47 mmHg.
  • Exam-type : Raise x₂ ⇒ larger Δp.

Rule 2: Finding M₂ — For dilute solutions (p° – p)/p ≈ w₂M₁/(M₂w₁), hence M₂ = w₂M₁p/((p° – p)w₁).

  • Easy : Solvent water M₁ = 18.
  • Medium : Measure p° and p.
  • Exam-type : Accuracy is limited.

Rule 3: Explanation — Solute particles occupy the surface ⇒ less evaporation ⇒ p falls.

  • Easy : Pure liquid has highest p.
  • Medium : Solute must be non-volatile.
  • Exam-type : Depends on number of particles.

Calculation: 0.1 mol glucose + 90 g water, p° = 23.76 mmHg

  • Step 1: n(water) = 90/18 = 5 mol.
  • Step 2: x₂ = 0.1/5.1 = 0.0196.
  • Step 3: Δp = 0.0196 × 23.76 = 0.47 mmHg.

CET level (MCQ)

  • CET 1) Colligative properties depend on? 1) particle number 2) kind 3) colour 4) size — Answer: 1).
  • CET 2) Relative lowering = ? 1) x₂ 2) x₁ 3) m 4) M — Answer: 1).

Real-life use: Cold storage, molar mass, mixture studies.

Remember

(p° – p)/p° = x₂; number of particles; non-volatile solute.

Common mistake

Using x₁ instead of x₂.

CET trap

Relative lowering equals the solute mole fraction x₂.

Shortcut

Memory clue: "lowering equals solute fraction".

Spot the difference

PropertyDirection
Vapour pressure↓
T_b↑
T_f↓
π↑

Practice questions

Easy
  1. 1What are colligative properties?
  2. 2The four properties.
  3. 3Formula of relative lowering.
Medium
  1. 4Find Δp.
  2. 5Why non-volatile?
  3. 6Why does p fall?
Challenging
  1. 7How is M₂ found?
  2. 8Find x₂.
Show answers (try them yourself first)
  1. 1Depend on particle number.
  2. 2p, T_b, T_f, π.
  3. 3(p° – p)/p° = x₂.
  4. 40.47 mmHg.
  5. 5Solute must not evaporate.
  6. 6Surface occupied.
  7. 7w₂M₁p/((p°–p)w₁).
  8. 80.0196.

Elevation of boiling point and depression of freezing point

From daily life

Salt on ice makes it melt — the freezing point drops.

Simple definition

ΔT_b = K_b m and ΔT_f = K_f m; for water K_b = 0.52 and K_f = 1.86 K kg mol⁻¹.

ΔTΔTb and ΔTfTpΔTfΔTbΔTb = Kb mΔTf = Kf mKb water 0.52Kf water 1.861.0 m glucose: Tb = 100.52 °C; Tf = –1.86 °CM₂ = Kf × w₂ × 1000 / (ΔTf × w₁)1.5 g solute, 50 g water, ΔTf = 0.62 ⇒ M₂ = 90Antifreeze, road salt, molar mass determination

In the figure : ΔT_b and ΔT_f.

Live figure

Elevation of boiling point and depression of freezing point

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Live figure

Elevation of boiling point and depression of freezing point

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Open in the app

Rule 1: Boiling-point elevation — ΔT_b = K_b m; 1.0 m glucose ⇒ 0.52 K ⇒ T_b = 100.52 °C.

  • Easy : 0.5 m ⇒ 0.26 K.
  • Medium : K_b depends on the solvent.
  • Exam-type : Lower p ⇒ higher T needed.

Rule 2: Freezing-point depression — ΔT_f = K_f m; 1.0 m ⇒ 1.86 K ⇒ T_f = –1.86 °C.

  • Easy : Antifreeze (ethylene glycol).
  • Medium : Road salt.
  • Exam-type : Seawater freezes near –1.9 °C.

Rule 3: Finding M₂ — M₂ = K_f w₂ × 1000/(ΔT_f × w₁); 1.5 g in 50 g water with ΔT_f = 0.62 ⇒ 90.

  • Easy : w₁ in grams.
  • Medium : Factor of 1000.
  • Exam-type : Small ΔT_f ⇒ large M.

Calculation: 1.5 g solute, 50 g water, ΔT_f = 0.62 K

  • Step 1: M₂ = K_f w₂ × 1000/(ΔT_f w₁).
  • Step 2: = 1.86 × 1.5 × 1000/(0.62 × 50).
  • Step 3: M₂ = 90 g mol⁻¹.

CET level (MCQ)

  • CET 1) K_f of water? 1) 1.86 2) 0.52 3) 5.12 4) 3.9 — Answer: 1).
  • CET 2) ΔT_b of 0.5 m glucose? 1) 0.26 2) 0.52 3) 1.04 4) 0.13 — Answer: 1).

Real-life use: Antifreeze, ice-cream making, roads.

Remember

ΔT_b = K_b m; ΔT_f = K_f m; M₂ = K_f w₂×1000/(ΔT_f w₁).

Common mistake

Taking w₁ in kg and still multiplying by 1000.

CET trap

w₁ in grams ⇒ use 1000; in kg ⇒ do not.

Shortcut

Memory clue: "water: boil 0.52, freeze 1.86".

Spot the difference

ConstantWater
K_b0.52
K_f1.86

Practice questions

Easy
  1. 1Formula of ΔT_b.
  2. 2Formula of ΔT_f.
  3. 3T_b of 1.0 m glucose.
Medium
  1. 4T_f of 1.0 m glucose.
  2. 5Find M₂ (1.5 g, 50 g, 0.62).
  3. 6Why salt on roads?
Challenging
  1. 7Unit of K_f.
  2. 8Why do T_b and T_f change?
Show answers (try them yourself first)
  1. 1K_b m.
  2. 2K_f m.
  3. 3100.52 °C.
  4. 4–1.86 °C.
  5. 590.
  6. 6Lowers freezing point.
  7. 7K kg mol⁻¹.
  8. 8Because p falls.

Osmosis and osmotic pressure

From daily life

A raisin swells in water — water moves through its skin.

Simple definition

Osmosis is flow of solvent through a semipermeable membrane from dilute to concentrated; osmotic pressure π = CRT.

OsmosisOsmosis and osmotic pressuremembranesolventsolution→π = C R Tπ = (n/V) R TR = 0.0821 L atmK⁻¹ mol⁻¹0.1 M glucose, 300 K: π = 0.1×0.0821×300 = 2.46 atm2 g polymer/L, π = 0.0050 atm, 300 K ⇒ M = 9852Isotonic: equal π; hypertonic: higher; hypotonic: lowerBest colligative property for large molecules (proteins, polymers)

In the figure : Osmotic pressure.

Live figure

Osmosis and osmotic pressure

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Rule 1: π = CRT — π = (n/V)RT with R = 0.0821 L atm K⁻¹ mol⁻¹; 0.1 M at 300 K ⇒ 2.46 atm.

  • Easy : T↑ ⇒ π↑.
  • Medium : Double C ⇒ double π.
  • Exam-type : This is van't Hoff's law.

Rule 2: Finding M₂ — M₂ = w₂RT/(πV); 2 g/L, 0.0050 atm, 300 K ⇒ 9852.

  • Easy : Large molecules: proteins, polymers.
  • Medium : Measured at fixed T.
  • Exam-type : Small but measurable π.

Rule 3: Isotonic solutions — Equal π ⇒ isotonic (0.9 % saline); hypertonic: cells shrink; hypotonic: cells swell; reverse osmosis uses pressure > π for desalination.

  • Easy : Pickle preservation.
  • Medium : Desalinating seawater.
  • Exam-type : Cell wall.

Calculation: 0.1 M glucose at 300 K

  • Step 1: π = CRT.
  • Step 2: = 0.1 × 0.0821 × 300.
  • Step 3: π = 2.46 atm.

CET level (MCQ)

  • CET 1) Pressure needed for reverse osmosis? 1) greater than π 2) less than π 3) equal to π 4) zero — Answer: 1).
  • CET 2) Isotonic solutions have equal? 1) π 2) M 3) T 4) p — Answer: 1).

Real-life use: Desalination, saline, pickles, cell biology.

Remember

π = CRT; isotonic = equal π; reverse osmosis uses P > π.

Common mistake

Using the wrong R.

CET trap

R = 0.0821 for atm; 8.314 for kPa.

Shortcut

Memory clue: "water flows from less to more concentrated".

Spot the difference

SolutionEffect on cell
Hypertonicshrinks
Hypotonicswells
Isotonicno change

Practice questions

Easy
  1. 1What is osmosis?
  2. 2Explain π = CRT.
  3. 30.1 M, 300 K ⇒ π.
Medium
  1. 4Find M₂ (2 g/L, 0.005 atm).
  2. 5Reverse osmosis.
  3. 6Isotonic solution.
Challenging
  1. 7Why is π used for large molecules?
  2. 8Why is saline 0.9 %?
Show answers (try them yourself first)
  1. 1Solvent flow.
  2. 2n/V × RT.
  3. 32.46 atm.
  4. 49852.
  5. 5P > π gives pure water.
  6. 6Equal π.
  7. 7Gives a measurable π.
  8. 8Isotonic with blood.

van't Hoff factor: dissociation and association

From daily life

One NaCl unit gives two ions in water, so the effect doubles: i = 2.

Simple definition

i = observed ÷ calculated colligative effect; dissociation: i > 1; association: i < 1; glucose: i = 1.

ivan't Hoff factor ii = observed / calculated colligative effect; i = 1 + α(n – 1)SoluteiNoteGlucose1no ionisationNaCl2fullCaCl₂3fullK₂SO₄3fullCH₃COOH / benzene≈ 0.5dimerises0.1 m NaCl1.9α = 0.9Dissociation: i > 1, α = (i – 1)/(n – 1)Association (dimer): i < 1, i = 1 – α/2; i = 0.52 ⇒ α = 0.96ΔTf = i Kf m = 1.86 × 0.1 × 1.9 = 0.353 K

In the figure : Values of i.

Rule 1: Modified formulas — ΔT_f = i K_f m; ΔT_b = i K_b m; π = i C R T; (p°–p)/p° = i x₂.

  • Easy : 0.1 m NaCl, i = 1.9 ⇒ ΔT_f = 0.353 K.
  • Medium : CaCl₂ i = 3.
  • Exam-type : 0.01 M CaCl₂ π = 0.73 atm (298 K).

Rule 2: Dissociation — i = 1 + α(n – 1); α = (i – 1)/(n – 1); NaCl i = 1.9 ⇒ α = 0.9.

  • Easy : n = ions per formula.
  • Medium : Full dissociation α = 1.
  • Exam-type : Weak electrolytes have small α.

Rule 3: Association — CH₃COOH dimerises in benzene ⇒ i ≈ 0.5; i = 1 – α/2; i = 0.52 ⇒ α = 0.96.

  • Easy : Fewer particles ⇒ smaller effect.
  • Medium : Observed M₂ is larger.
  • Exam-type : Due to H-bonding.

Calculation: 0.1 m NaCl, i = 1.9

  • Step 1: α = (1.9 – 1)/(2 – 1) = 0.9.
  • Step 2: ΔT_f = i K_f m = 1.9 × 1.86 × 0.1.
  • Step 3: ΔT_f = 0.353 K.

CET level (MCQ)

  • CET 1) i for fully dissociated CaCl₂? 1) 3 2) 2 3) 1 4) 4 — Answer: 1).
  • CET 2) i for a dimerising solute? 1) < 1 2) > 1 3) = 1 4) = 2 — Answer: 1).

Real-life use: Electrolyte saline, antifreeze, cell fluids.

Remember

i = 1 + α(n–1); association i = 1 – α/2; multiply the formulas by i.

Common mistake

Applying i to non-electrolytes.

CET trap

Glucose and sugar: i = 1.

Shortcut

Memory clue: "splits ⇒ big i, joins ⇒ small i".

Spot the difference

Solutei
NaCl2
CaCl₂3
Glucose1
CH₃COOH/C₆H₆0.5

Practice questions

Easy
  1. 1What is i?
  2. 2i of NaCl and CaCl₂.
  3. 3α = (i – 1)/(n – 1).
Medium
  1. 4i in association.
  2. 5ΔT_f of 0.1 m NaCl.
  3. 6i = 0.52 ⇒ α.
Challenging
  1. 7Why does CH₃COOH dimerise in benzene?
  2. 8π = iCRT.
Show answers (try them yourself first)
  1. 1Observed ÷ calculated.
  2. 22 and 3.
  3. 3Same.
  4. 4< 1.
  5. 50.353 K.
  6. 60.96.
  7. 7H-bonding.
  8. 8Multiply by i.

Quick revision of the whole lesson

ConceptKey formula / fact
Molalitym = n / kg solvent
Henryp = K_H x
Raoultp = x_A p°A + x_B p°B
Relative lowering(p° – p)/p° = x₂
Tb and TfΔT = i K m
M₂K_f w₂ × 1000 / (ΔT_f w₁)
Osmotic pressureπ = i C R T
Dissociationα = (i – 1)/(n – 1)
Live figure

🚀 Quick revision of the whole lesson

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🎯 Most important points for the exam

  • Concentration.
  • Henry.
  • Raoult.
  • Vapour pressure.
  • Tb, Tf.
  • Osmosis.
  • van't Hoff i.
Study mantra

understand the idea → remember the formula → solve the example → solve problems yourself.