From salt water to a sugar solution — the chemistry of mixing, dissolving, and how dissolved particles change the properties of the whole. With interactive calculators, NCERT exercises, flashcards, and 25+ real-life examples.
Before diving in, here's the entire chapter on one page. This is your mental map — how concentration terms connect to vapour pressure, Raoult's Law, and finally to colligative properties.
A solution is a homogeneous mixture. Concentration terms (molarity, molality) express composition. Solubility depends on nature of solute/solvent + temperature + pressure. Colligative properties (VP↓, BP↑, FP↓, osmosis) depend only on number of particles — corrected by Van't Hoff factor i for electrolytes.
Solution: A homogeneous mixture of two or more substances, where particles are uniformly distributed at the molecular or ionic level (size < 1 nm).
Solute: Component in smaller amount (dissolved).
Solvent: Component in larger amount (dissolving medium).
Binary solution: Contains only two components (one solute + one solvent).
Solutions can exist in all three physical states. The table below shows the nine possible types:
| State of Solute | State of Solvent | Example | Common Name |
|---|---|---|---|
| Gas | Gas | Air (O₂ in N₂) | Gas mixture |
| Gas | Liquid | CO₂ in water | Soda water |
| Gas | Solid | H₂ in Pd | — |
| Liquid | Gas | Moist air | Humidity |
| Liquid | Liquid | Ethanol in water | Alcoholic drinks |
| Liquid | Solid | Hg in Na (amalgam) | Amalgam |
| Solid | Gas | Iodine vapour in air | — |
| Solid | Liquid | Salt in water | Brine |
| Solid | Solid | Cu in Au (alloy) | Alloy / solid solution |
A solution is homogeneous at the molecular level. Binary solutions = 1 solute + 1 solvent. Nine possible combinations across states. Focus is on solid-in-liquid and gas-in-liquid.
Mass percentage (w/w%): (mass of solute / mass of solution) × 100
Volume percentage (v/v%): (volume of solute / volume of solution) × 100
Mass by volume percentage (w/v%): (mass of solute / volume of solution) × 100
Parts per million (ppm): (mass of solute / mass of solution) × 10⁶
Mole fraction (x): x_A = n_A / (n_A + n_B), and x_A + x_B = 1
Molarity (M): n_solute / V_solution(L)
Molality (m): n_solute / mass_solvent(kg)
Enter the values you have — calculate molarity, molality, mole fraction, and mass %.
| Property | Molarity (M) | Molality (m) |
|---|---|---|
| Definition | moles / L of solution | moles / kg of solvent |
| Temperature dependent? | Yes (volume changes with T) | No (mass doesn't change) |
| Common use | Lab titrations | Colligative properties |
| Units | mol/L | mol/kg |
Molarity = "sugar per full glass of tea." Molality = "sugar per spoonful of tea itself." If tea evaporates, molarity changes but molality stays same. That's why molality is used for colligative properties!
Mistake: "Molarity = moles / litres of
solvent."
Correction: It's litres of
solution. Molality is per kg of solvent. These
are the most confused terms!
7 concentration terms: mass%, volume%, w/v%, ppm, mole fraction, molarity, molality. Molarity is T-dependent; molality isn't. Use molality for colligative properties.
Solubility: The maximum amount of solute that can be dissolved in a given amount of solvent at a specific temperature to form a saturated solution.
Saturated solution: Contains maximum solute at that temperature (equilibrium between dissolved and undissolved).
Unsaturated solution: Contains less solute than saturation.
Supersaturated solution: Contains more than saturation (unstable, prepared by cooling saturated hot solution slowly).
Polar solutes dissolve in polar solvents; non-polar solutes dissolve in non-polar solvents.
Why? Dissolution requires breaking solute-solute and solvent-solvent interactions and forming solute-solvent interactions. Similar polarity → similar energies → more favorable.
Think of it as a party. Polar molecules (like water) love polar guests (like NaCl) — they mingle easily. But they don't invite oil (non-polar) to the party. Oil molecules prefer to clump together with their own kind, so they form a separate layer. Same principle for dissolution.
Unlike gases, solid solubility usually increases with temperature. But the direction depends on whether dissolution is endothermic or exothermic:
Endothermic dissolution (ΔH_solution > 0):
Exothermic dissolution (ΔH_solution < 0):
This is governed by Le Chatelier's principle: heating shifts equilibrium toward endothermic direction.
Pressure has almost no effect on solid solubility. Solids and liquids are nearly incompressible, so pressure changes volume negligibly. This is why Henry's Law applies only to gas solubility, not to solid or liquid solutes.
Lattice energy (ΔH_lattice): Energy released when gaseous ions come together to form a solid crystal (always exothermic).
Hydration energy (ΔH_hydration): Energy released when gaseous ions are surrounded by water molecules (always exothermic).
Net dissolution energy: ΔH_sol = ΔH_lattice + ΔH_hydration
Example: NaCl dissolves because lattice energy ≈ hydration energy — but entropy drives dissolution. AgCl doesn't dissolve because lattice energy ≫ hydration energy.
"Endothermic — Heat Increases" — For endothermic dissolution, heating increases solubility (KNO₃, sugar).
"Exo — Heat Exits" — For exothermic dissolution, heating decreases solubility (NaOH, CaO, Ce₂(SO₄)₃).
"Polar Party" — Polar loves polar, non-polar loves non-polar. Opposites don't attract here!
"Like dissolves like" — polar solvents dissolve polar solutes, non-polar dissolve non-polar. Solid solubility usually increases with temperature (endothermic dissolution); a few exceptions decrease (exothermic). Pressure has negligible effect. Net solubility depends on lattice energy vs hydration energy.
The solubility of a gas in a liquid depends strongly on pressure and temperature. Henry's Law quantifies the pressure effect.
Statement: At constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
Mathematically: p = K_H · x
Where: p = partial pressure of gas; K_H = Henry's Law constant; x = mole fraction of gas in solution.
Alternative form: x = p / K_H → doubling pressure doubles solubility.
Gas solubility decreases as temperature increases. Dissolution of a gas is generally exothermic, so heating shifts equilibrium toward the gas escaping.
Mistake: "Higher K_H means more soluble."
Correction:
x = p / K_H → larger K_H means smaller x → lower
solubility. High K_H = low solubility.
Henry's Law: p = K_H · x. Gas solubility decreases with increasing temperature. Applications: soda, diving, aquarium, industrial gas storage.
Vapour pressure (p°): Pressure exerted by vapour of a liquid in equilibrium with its liquid state in a closed container at a given temperature.
Volatile liquids: High vapour pressure (acetone, ether, alcohol).
Non-volatile liquids: Low vapour pressure (water, oil).
For two volatile liquids A and B:
p_A = x_A · p°_A; p_B = x_B · p°_B
P_total = p_A + p_B = x_A p°_A + x_B p°_B
For a non-volatile solute: p_total = p_solvent = x_solvent · p°_solvent
For a solution with a non-volatile solute: (p° − p) / p° = x_solute
This is one of the four colligative properties.
Imagine a crowded bus (pure solvent). Some people get off (evaporate) at a certain rate. Now some seats are occupied by people who never get off (non-volatile solute). Fewer can exit per minute — the "exit pressure" drops. That's what non-volatile solute does to vapour pressure.
Raoult's Law: p_A = x_A p°_A. For non-volatile solute, (p°−p)/p° = x_solute. Vapour pressure decreases when solute is added — first colligative property.
Definition: Solutions that obey Raoult's Law at all concentrations and temperatures.
Conditions:
Examples: Benzene + Toluene, n-Hexane + n-Heptane, Bromoethane + Chloroethane.
Positive deviation (p > p_Raoult):
Negative deviation (p < p_Raoult):
At a specific composition, non-ideal solutions form a mixture that boils at constant temperature without changing composition — like a pure substance.
Minimum boiling azeotropes (positive deviation): Ethanol-water (95.6% ethanol, BP 78.1°C), Acetone-CS₂.
Maximum boiling azeotropes (negative deviation): HNO₃-water (68% HNO₃, BP 120.5°C), Chloroform-acetone.
Azeotropes cannot be separated by simple distillation.
Ideal solutions obey Raoult's Law (ΔH=0, ΔV=0). Positive deviation (ΔH>0) → higher VP → min-boiling azeotropes. Negative deviation (ΔH<0) → lower VP → max-boiling azeotropes. Azeotropes can't be separated by simple distillation.
Solid solution: A homogeneous solid mixture where one solid (solute) is uniformly distributed in another solid (solvent). Commonly called alloys when both are metals.
Two main types based on how atoms arrange:
Solute atoms replace solvent atoms in the lattice. Requires atoms of comparable size.
Conditions (Hume-Rothery rules):
Examples:
Small solute atoms occupy interstitial spaces (gaps) between larger solvent atoms. Requires solute atoms much smaller than solvent atoms.
Conditions:
Examples:
Alloying dramatically changes properties:
Solid solutions are of two types: substitutional (solute replaces solvent — similar-sized atoms; e.g., brass) and interstitial (small solute in gaps; e.g., steel). Alloying changes hardness, strength, corrosion resistance, and other properties — fundamental to metallurgy.
Properties of dilute solutions that depend only on the number of solute particles, not their identity.
(p° − p) / p° = x_solute
For dilute solutions: (p° − p)/p° ≈ n_solute / n_solvent
ΔT_b = T_b − T°_b = K_b · m
K_b values (K·kg/mol): Water 0.52, Benzene 2.53, Chloroform 3.63, CCl₄ 5.03.
ΔT_f = T°_f − T_f = K_f · m
K_f values (K·kg/mol): Water 1.86, Benzene 5.12, Chloroform 4.68, Camphor 39.7.
π = C · R · T
Or: π = (n/V)RT = (w/M)RT/V
Best method for molar mass of macromolecules (proteins, polymers) — even small concentrations give measurable pressure.
Choose solvent, enter molality — see all four colligative effects.
Four colligative properties: (p°−p)/p° = x_solute; ΔT_b = K_b·m; ΔT_f = K_f·m; π = CRT. All depend on number of particles. Used for molar mass determination and membrane biology.
i = observed colligative property / calculated (expected) property
Or: i = actual number of particles / number of formula units dissolved
For non-electrolytes (glucose, urea): i = 1
For electrolytes that dissociate into n ions: i ≈ n (if 100% dissociated)
For associating molecules (dimers): i < 1
| Substance | Type | Van't Hoff factor (i) |
|---|---|---|
| Glucose, Urea, Sucrose | Non-electrolyte | 1 |
| NaCl | Na⁺ + Cl⁻ | 2 |
| KCl | K⁺ + Cl⁻ | 2 |
| CaCl₂ | Ca²⁺ + 2Cl⁻ | 3 |
| Al₂(SO₄)₃ | 2Al³⁺ + 3SO₄²⁻ | 5 |
| Acetic acid in benzene | Dimer (CH₃COOH)₂ | ~0.5 |
For dissociation into n ions: i = 1 + (n − 1)α
For association into n molecules: i = 1 − α(1 − 1/n)
For dimerization (n=2): i = 1 − α/2
Example: acetic acid dimer, i=0.6: α = 2(1−0.6) = 0.8 → 80% associated
Calculated molar mass differs from true molar mass:
Example: NaCl in water shows M_obs ≈ 29 g/mol (close to 58.5/2 = 29.25) due to dissociation.
Mistake: "Van't Hoff factor is always an
integer."
Correction: In real solutions,
dissociation is often incomplete:
"DISA — Dissociate Increases Small mass, Associate Adds mass"
DISA = Dissociation → increase in particle count → smaller calculated molar mass.
Association → fewer particles → larger calculated molar mass.
Van't Hoff factor i corrects colligative properties for dissociation (i > 1) and association (i < 1). For non-electrolytes i = 1. For n-ion electrolyte i ≈ n. Calculated molar mass is abnormal when dissociation/association occurs.
"MP3-VM" — Mass%, PPM, Volume%, Molality, Mole fraction
Wait, better: "MVPP-MF-Mola"
"Endo — Heat Increases" (KNO₃, sugar)
"Exo — Heat Decreases" (NaOH, Ce₂(SO₄)₃)
"Gases ALWAYS go down with heat" — universal rule.
"VP down, BP up, FP down, π up"
Or remember: "1 down, 2 up, 3 down, 4 up"
"DISA — Dissociate IncreaseS Smaller mass, Associate Adds mass"
Dissociation → more particles → smaller calculated M
Association → fewer particles → larger calculated M
"Water is easiest: 0.52 and 1.86" — remember K_b ≈ 0.5, K_f ≈ 1.86 (nearly 2)
"Camphor is crazy: 39.7" — huge K_f (used for molar mass of unknowns)
"Raoult is for Liquids (VP), Henry is for Gases (Solubility)"
Raoult: p_A = x_A · p°_A (vapour pressure of liquid)
Henry: p = K_H · x (solubility of gas)
"Zero-Zero-Equal" — ΔH = 0, ΔV = 0, forces equal
Ideal solution = no heat + no volume change + equal intermolecular forces
"Positive → Minimum" (Positive deviation → Minimum boiling azeotrope)
"Negative → Maximum" (Negative deviation → Maximum boiling azeotrope)
Remember: Positive (p > p_ideal) → low boiling. Negative (p < p_ideal) → high boiling.
Mnemonics make recall faster in exams. Use "VP down, BP up, FP down, π up" for colligative properties, "Endo-Heat Increases" for temperature-solubility, and "DISA" for Van't Hoff factor effects.
Q1. Calculate mole fraction of glucose in a solution of 18 g glucose (M=180) and 180 g water (M=18).
n_glucose = 18/180 = 0.1 mol
n_water = 180/18 = 10 mol
x_glucose = 0.1/(0.1+10) = 0.0099
Q2. 18 g glucose is dissolved in 1 kg water. At what temperature will it boil? (K_b = 0.52)
n = 18/180 = 0.1; m = 0.1/1 = 0.1 mol/kg
ΔT_b = 0.52 × 0.1 = 0.052 K
T_b = 100 + 0.052 = 100.052°C
Q3. Osmotic pressure of 0.1 M glucose solution at 300 K? (R = 0.0821)
π = CRT
π = 0.1 × 0.0821 × 300
π = 2.463 atm
Glucose (i=1); for NaCl (i≈1.8): π ≈ 4.43 atm
Q4. A solution has VP 23.2 mm Hg at 25°C. Pure water VP = 23.8 mm Hg. Find mole fraction of solute.
(p°−p)/p° = x_solute
(23.8−23.2)/23.8 = 0.6/23.8
x_solute = 0.0252
2.5 mol% solute ≈ 1.4 molal
Q5. 2.5 g non-electrolyte in 100 g benzene. BP rises by 0.42°C. Find molar mass. (K_b = 2.53)
ΔT_b = K_b × (w_B × 1000)/(M_B × w_A)
0.42 = 2.53 × (2.5 × 1000)/(M_B × 100)
0.42 M_B = 63.25; M_B = 150.6 g/mol
m = 2.5/150.6 × 1000/100 = 0.166; ΔT_b = 2.53 × 0.166 = 0.42 ✓
Non-electrolyte → i = 1. M ≈ 150 g/mol
Q6. 1.5 g solute in 100 g water freezes at −0.31°C. Find molar mass. (K_f = 1.86)
ΔT_f = 0 − (−0.31) = 0.31 K
ΔT_f = K_f × (w_B × 1000)/(M_B × w_A)
0.31 = 1.86 × 15/M_B
0.31 M_B = 27.9; M_B = 90 g/mol
M ≈ 90 → could be oxalic acid, lactic acid, etc.
Q7. NaCl solution freezes at −0.36°C. Degree of dissociation? (K_f = 1.86, m = 0.1)
ΔT_f = i × K_f × m
0.36 = i × 1.86 × 0.1
i = 0.36/0.186 = 1.935
For NaCl (n=2): i = 1 + α
α = i − 1 = 0.935 = 93.5%
Check: with i = 1.935, ΔT_f = 1.935 × 1.86 × 0.1 = 0.36 ✓
93.5% of NaCl molecules dissociate into Na⁺ and Cl⁻. Remaining 6.5% stay as ion pairs.
For 0.1m NaCl at 0°C, observed i would be ~1.87 in practice. Close to our calculated value.
Selected exercise questions from NCERT Class 12 Chemistry Chapter 1. These are commonly asked in board exams. Click each to reveal the solution.
Solution: A solution is a homogeneous mixture of two or more substances, whose composition can be varied within limits. Particles are of the size of molecules or ions (< 1 nm).
Types (9 combinations):
Solution: Hydrogen dissolves in palladium (H₂ in Pd) to form a solid solution. This is exploited in fuel cells and hydrogen storage. Another example: hydrogen in platinum (used in catalytic hydrogenation).
Solution: Let moles of each = x. Mass of Na₂CO₃ = 106x, mass of NaHCO₃ = 84x. Total mass = 106x + 84x = 190x = 1 g → x = 1/190 = 5.26×10⁻³ mol.
HCl required: For Na₂CO₃ (2 mol HCl per mol) = 2x. For NaHCO₃ (1 mol HCl per mol) = x. Total HCl needed = 3x = 3 × 5.26×10⁻³ = 1.58×10⁻² mol.
Volume of 0.1 M HCl = 1.58×10⁻² / 0.1 = 0.158 L = 158 mL
Solution: Let x_A be mole fraction of A in liquid, x_B = 1 − x_A.
P_total = x_A · p°_A + x_B · p°_B
600 = 450 x_A + 700(1 − x_A)
600 = 450 x_A + 700 − 700 x_A
600 = 700 − 250 x_A
250 x_A = 100 → x_A = 0.4, x_B = 0.6
Vapour phase:
y_A = (x_A · p°_A) / P_total = (0.4 × 450)/600 = 180/600 = 0.3
y_B = (x_B · p°_B) / P_total = (0.6 × 700)/600 = 420/600 = 0.7
Answer: Liquid: x_A = 0.4, x_B = 0.6. Vapour: y_A = 0.3, y_B = 0.7
Solution: After mixing, total volume = 200 mL = 0.2 L.
Moles of urea = 0.1 × 0.25 = 0.025 mol
Moles of glucose = 0.1 × 0.1 = 0.01 mol
Total moles of solute = 0.035 mol (both are non-electrolytes, i = 1)
Total molarity = 0.035 / 0.2 = 0.175 M
π = CRT = 0.175 × 0.0821 × 300 = 4.31 atm
Positive deviation (p > p_Raoult): Observed vapour pressure is higher than predicted. ΔH_mix > 0 (endothermic — heat absorbed). A-B interactions are weaker than A-A and B-B. Molecules escape more easily → higher VP.
Negative deviation (p < p_Raoult): Observed VP is lower than predicted. ΔH_mix < 0 (exothermic — heat released). A-B interactions are stronger than A-A and B-B. Molecules held tightly → lower VP.
Relation: Positive deviation ↔ ΔH > 0. Negative deviation ↔ ΔH < 0. Ideal solution (no deviation) ↔ ΔH = 0.
Solution: Let original VP = p°, new VP = 0.8 p°.
Relative lowering: (p° − p)/p° = 0.2 = x_solute
x_solute = n_solute / (n_solute + n_octane) = 0.2
n_octane = 114/114 = 1 mol (M_octane = 114 g/mol)
Let n_solute = w/40. Then:
(w/40)/((w/40) + 1) = 0.2
w/40 = 0.2 × (w/40 + 1)
w/40 = 0.005w + 0.2
w/40 − 0.005w = 0.2
w(1/40 − 1/200) = 0.2
w × (5−1)/200 = 0.2
w × 4/200 = 0.2
w = 0.2 × 200/4 = 10 g
Solution: π = CRT = (w/M) × (1/V) × RT
Method 1: Direct proportion. Since T, R, V fixed and same solute:
π ∝ w → π₂/π₁ = w₂/w₁ = 18/36 = 0.5
π₂ = 4.98 × 0.5 = 2.49 bar
Method 2: Directly: M_glucose = 180 g/mol, V = 1 L, T = 300 K, R = 0.083 bar·L/(mol·K)
Moles = 18/180 = 0.1 mol, C = 0.1 M
π = 0.1 × 0.083 × 300 = 2.49 bar ✓
Solution: Mass of solute in first solution = 300 × 0.25 = 75 g
Mass of solute in second solution = 400 × 0.40 = 160 g
Total solute = 75 + 160 = 235 g
Total solution = 300 + 400 = 700 g
Mass% = (235/700) × 100 = 33.57%
Solution: M(CH₃CH₂CHClCOOH) = 122.5 g/mol
Moles = 10/122.5 = 0.0816 mol
Molality = 0.0816 × 1000/250 = 0.3264 m
Weak acid dissociates partially:
Let α = degree of dissociation. K_a = C·α²/(1−α) ≈ C·α² (if α small)
1.4×10⁻³ = 0.3264 × α² → α² = 4.29×10⁻³ → α = 0.0655
Van't Hoff factor: i = 1 + α = 1 + 0.0655 = 1.0655
ΔT_f = i · K_f · m = 1.0655 × 1.86 × 0.3264 = 0.647 K
Freezing point = 0 − 0.647 = −0.647°C
Van't Hoff factor (i): The ratio of observed colligative property to the calculated colligative property, assuming no association or dissociation.
(a) Association: Two or more molecules combine into one. Number of particles decreases. i < 1. E.g., acetic acid in benzene (dimer) → i ≈ 0.5.
(b) Dissociation: One molecule splits into two or more ions. Number of particles increases. i > 1. E.g., NaCl → i ≈ 2, CaCl₂ → i ≈ 3.
For non-electrolytes (no change): i = 1
| K_b (water) | 0.52 K·kg/mol | Ebullioscopic constant |
| K_f (water) | 1.86 K·kg/mol | Cryoscopic constant |
| K_b (benzene) | 2.53 K·kg/mol | — |
| K_f (benzene) | 5.12 K·kg/mol | — |
| K_f (camphor) | 39.7 K·kg/mol | Largest K_f — ideal for molar mass |
| K_H (CO₂ in water, 298 K) | 1.67 × 10³ bar | Higher solubility |
| K_H (O₂ in water, 298 K) | 4.6 × 10⁴ bar | Low solubility |
| R | 0.0821 L·atm/(mol·K) | or 8.314 J/(mol·K) |
| 1 atm | 760 mm Hg | = 1.013 bar |
| M(water) | 18 g/mol | Most common solvent |
| M(glucose) | 180 g/mol | Non-electrolyte reference |
| M(NaCl) | 58.5 g/mol | i ≈ 2 |
| M(CaCl₂) | 111 g/mol | i ≈ 3 |
| w_A, w_B | Mass of solvent (A) and solute (B) |
| M_A, M_B | Molar masses of solvent and solute |
| n_A, n_B | Moles of solvent and solute |
| x_A, x_B | Mole fractions |
| p°_A, p_A | VP of pure A and partial pressure |
| K_H | Henry's Law constant |
| K_b, K_f | Molal elevation/depression constants |
| ΔT_b, ΔT_f | Change in boiling/freezing point |
| π | Osmotic pressure |
| C | Molarity (mol/L) |
| i | Van't Hoff factor |
| α | Degree of dissociation or association |
| R | Gas constant (0.0821 or 8.314) |
(A) Both true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
The formula or definition is shown on every card for quick revision before exams.