Concentration units

Diamond is the hardest natural substance; graphite writes on paper and lubricates machines — yet both are pure carbon. The difference is entirely in how the atoms are arranged. This lesson covers the solid state from first principles: crystal classification, packing geometry, density calculation, defects, and the electrical and magnetic properties that flow from structure — a 4–8 mark guaranteed topic in JEE and NEET.

Definition — Crystalline solid: particles arranged in a long-range, repeating, ordered three-dimensional pattern (e.g., NaCl, quartz, diamond, ice).
Definition — Amorphous solid: no long-range order; structure resembles a frozen liquid (e.g., glass, rubber, plastic). Properties are isotropic (same in all directions).
Definition — Unit cell: the smallest repeating unit of a crystal lattice that, when translated in three dimensions, builds the entire crystal.

Classifying crystals by bonding

Type	Particles	Bonding forces	Example	Properties
Ionic	Cations + anions	Electrostatic	NaCl, CsCl, MgO	Hard, brittle, high MP, conducts only when molten or dissolved
Covalent (network)	Atoms	Covalent bonds	Diamond, SiO₂, AlN	Extremely hard, very high MP, poor conductor (except graphite)
Molecular	Molecules	Van der Waals, H-bonds, dipole–dipole	Ice, sugar, dry ice, iodine	Soft, low MP, poor conductors
Metallic	Cations + delocalised electrons	Metallic bond	Cu, Fe, Al, Au	Lustrous, ductile, malleable, good conductors

Why brittleness in ionic crystals? Striking NaCl shifts one layer so that like-charge ions face each other — repulsion shatters the crystal. Metals deform instead because the electron sea re-accommodates the shift.

Crystal systems and Bravais lattices

There are 7 crystal systems (classified by edge lengths a, b, c and angles α, β, γ): cubic, tetragonal, orthorhombic, monoclinic, triclinic, hexagonal, and rhombohedral (trigonal). These 7 systems give 14 Bravais lattices when body-centred and face-centred variations are included.

For exam purposes, the three cubic cells dominate.

Cubic unit cells — atoms, geometry, and packing

Simple cubic (SC):

• Atoms: only at the 8 corners. Each corner atom is shared among 8 cells → 8 × (1/8) = 1 atom per unit cell.
• Coordination number: 6 (each atom touches 4 in its plane, 1 above, 1 below).
• Edge length relation: a = 2r (atoms touch along the edge).
• Packing efficiency: 52.4% — the least efficient, lots of wasted space.
• Example: Polonium (the only element with simple cubic structure at room temperature).

Body-centred cubic (BCC):

• Atoms: 8 corners + 1 body centre → 8 × (1/8) + 1 = 2 atoms.
• Coordination number: 8 (the body-centre atom touches all 8 corner atoms).
• Edge relation: √3 · a = 4r (atoms touch along the body diagonal).
• Packing efficiency: 68%.
• Examples: Fe (at room temperature), Na, K, Cr, W.

Face-centred cubic (FCC) / cubic close packing (CCP):

• Atoms: 8 corners + 6 face centres → 8 × (1/8) + 6 × (1/2) = 4 atoms.
• Coordination number: 12 — the highest for equal spheres.
• Edge relation: √2 · a = 4r (atoms touch along the face diagonal).
• Packing efficiency: 74% — the densest possible packing for identical spheres (proved by Kepler conjecture, confirmed mathematically in 1998).
• Examples: Cu, Al, Au, Ag, Ni.

Hexagonal close packing (HCP):

• Also achieves 74% packing with coordination number 12.
• Layer sequence ABAB… (FCC has ABCABC…).
• Examples: Zn, Mg, Ti, Be.

Density of a unit cell

ρ = (Z × M) / (a³ × N_A)

where Z = atoms per unit cell, M = molar mass (g/mol), a = edge length (cm), N_A = 6.022 × 10²³ mol⁻¹.

This formula connects the macroscopic density you can measure with the microscopic structure — use it both ways: given structure → calculate density; given density → find Z (and hence identify the cell type).

Question: Copper crystallises FCC with a = 361 pm. Find its density. (M_Cu = 63.5 g/mol)

Solution:
Step 1: FCC → Z = 4.
Step 2: Convert a to cm: 361 pm = 3.61 × 10⁻⁸ cm; a³ = (3.61)³ × 10⁻²⁴ = 47.1 × 10⁻²⁴ cm³.
Step 3: ρ = (4 × 63.5) / (47.1 × 10⁻²⁴ × 6.022 × 10²³) = 254 / 28.36.
Conclusion: ρ ≈ 8.96 g/cm³, matching copper's known experimental density — confirming the FCC assignment.

Voids in close-packed structures

In FCC or HCP packing of N atoms, two types of holes appear:

• Tetrahedral void: surrounded by 4 atoms (tetrahedral arrangement); radius ratio r/R ≈ 0.225. There are 2N tetrahedral voids per N atoms.
• Octahedral void: surrounded by 6 atoms (octahedral arrangement); radius ratio r/R ≈ 0.414. There are N octahedral voids per N atoms.

Small cations occupy these voids in many ionic solids. In NaCl, Na⁺ occupies octahedral voids in a Cl⁻ FCC lattice. In ZnS (zinc blende), Zn²⁺ occupies half the tetrahedral voids in an S²⁻ FCC lattice.

Point defects

Stoichiometric (ratio preserved):

• Schottky defect: equal numbers of cations and anions are missing from their lattice sites (maintaining electrical neutrality). The crystal loses density — there are now fewer particles in the same volume. Common in NaCl, KCl, CsCl (where both ions are similar in size and can leave together).
• Frenkel defect: a cation (usually the smaller ion) leaves its lattice site and squeezes into a nearby interstitial void. No ions are missing, so density is unchanged. Common in AgCl, ZnS (large anions, small cations → cations fit into interstitials).

Non-stoichiometric (ratio shifts):

• Metal excess — F-centres: an anion vacancy is occupied by an electron to maintain neutrality. These electrons absorb visible light → crystal appears coloured. Classic example: NaCl heated in sodium vapour turns yellow (the excess Na ionises; the extra electrons occupy Cl⁻ vacancies).
• Metal deficiency: more anions than stoichiometry predicts; compensated by some cations having higher charge. Example: Fe₍₁₋ₓ₎O — some Fe²⁺ sites replaced by Fe³⁺ to balance charge.

Electrical properties

• Conductors (metals): overlapping valence and conduction bands; electrons flow freely.
• Insulators: large band gap (> ~5 eV); electrons cannot jump to the conduction band at room temperature.
• Semiconductors (Si, Ge): small band gap (~1 eV); some electrons thermally excited across.
  • n-type doping: add a pentavalent element (P, As, Sb) → one extra electron per dopant atom → majority carriers are electrons.
  • p-type doping: add a trivalent element (B, Al, Ga) → one missing electron (a "hole") per dopant → majority carriers are holes.

Magnetic properties

Type	Unpaired electrons	Behaviour	Example
Diamagnetic	None	Weakly repelled by magnets	NaCl, H₂O, Cu
Paramagnetic	Present	Weakly attracted; no permanent magnetism	O₂, Fe³⁺, Cu²⁺
Ferromagnetic	Present, aligned in domains	Strongly attracted; permanent magnets	Fe, Co, Ni, Gd
Antiferromagnetic	Present, anti-aligned equally	Net moment = 0; no attraction	MnO, FeO
Ferrimagnetic	Present, anti-aligned unequally	Net moment ≠ 0; moderate attraction	Fe₃O₄, ferrites

The Curie temperature: heating a ferromagnet above this point randomises the domains → it becomes paramagnetic. For iron, T_Curie ≈ 770 °C.

Why it matters: structure dictates every property you measure — density, hardness, melting point, conductivity, colour, and magnetism all follow from packing and bonding. This is a recurring 4–8 marks in JEE/NEET, with numerical density questions and defect identification the most common formats.

Real-world example: table salt shatters cleanly when struck because shifting ionic layers brings like-charges face to face; they repel catastrophically. Metals bend instead — the electron sea flows around the new configuration. The contrast between these two materials at a dinner table is solid-state chemistry in action.

Common misconception: "glass is a slowly flowing liquid — old window panes are thicker at the bottom." Glass is an amorphous solid, not a flowing liquid. The thickness variation in old panes came from historical manufacturing methods (glass was poured and one side is thicker by design), not from centuries of flow. Calling glass a "supercooled liquid" describes its disordered structure, not literal viscous flow.

"How much stuff is dissolved in how much liquid?" sounds simple, but chemistry needs several precise answers depending on which amounts and which base you choose — and picking the wrong unit is a classic exam trap. This lesson sorts out molarity, molality, mole fraction, normality and ppm, explains when each is the right tool, and shows you the worked maths.

Definition: Concentration expresses the amount of solute (the dissolved substance) in a given quantity of solution (solute + solvent) or solvent.
Definition: Solute is the substance being dissolved; solvent is the substance doing the dissolving.

The fundamental choice: per litre of solution vs per kg of solvent

This is the central fork:

• Molarity (M) = moles of solute per litre of solution → M = n ÷ V(L)
• Molality (m) = moles of solute per kilogram of solvent → m = n ÷ mass(kg)

The critical difference is temperature sensitivity:

Liquids expand when heated and contract when cooled. A litre of solution at 25 °C is not the same as a litre at 50 °C. So molarity changes with temperature — a 1 M solution at 25 °C becomes slightly less than 1 M at 50 °C because the volume grew.

Mass, on the other hand, never changes with temperature. So molality is temperature-independent — this is exactly why it is the required unit for colligative property calculations (boiling-point elevation, freezing-point depression, osmotic pressure).

Rule of thumb: Is temperature changing or does precision across temperatures matter? Use molality. Are you simply dispensing a volume at room conditions? Use molarity.

Mole fraction

Mole fraction (x_A) = moles of component A ÷ total moles in the mixture.

x_A = n_A / (n_A + n_B + …)

The mole fractions of all components always sum to 1 (Σx = 1). Mole fraction is dimensionless and is used in Raoult's law for vapour-pressure calculations in ideal solutions.

Worked example:
Question: Find the mole fraction of glucose in a solution of 18 g glucose (M = 180 g/mol) in 90 g water.
Solution:
Step 1: n(glucose) = 18 ÷ 180 = 0.1 mol. n(water) = 90 ÷ 18 = 5 mol.
Step 2: x(glucose) = 0.1 ÷ (0.1 + 5) = 0.1 ÷ 5.1.
Conclusion: x(glucose) ≈ 0.0196 (about 1.96%).

Percentage units

Three percentage expressions are used in different contexts:

• % w/w (mass/mass): (mass of solute ÷ mass of solution) × 100. Used in concentrated industrial acids (e.g. "98% H₂SO₄").
• % v/v (volume/volume): used for liquid-in-liquid mixtures. Alcoholic drinks: "40% v/v ethanol" means 40 mL ethanol per 100 mL solution.
• % w/v (mass/volume): (mass of solute in g ÷ volume of solution in mL) × 100. Dominant in medicine and pharmacy. Normal saline = 0.9% w/v = 0.9 g NaCl in every 100 mL — nurses rely on this because it lets them calculate a dose directly from the bag's volume.

Parts per million (ppm) and parts per billion (ppb)

For trace contaminants (pollutants in water, pesticides in food), percentage is too coarse a unit.

ppm = (mass of solute ÷ mass of solution) × 10⁶

For dilute aqueous solutions, 1 ppm ≈ 1 mg/L (because 1 L of dilute water ≈ 1 kg).

The WHO drinking-water limit for arsenic is 10 ppb (0.01 mg/L); the BIS standard in India is the same. Environmental monitoring and BIS/FSSAI food-safety rules cite ppm and ppb constantly — know these for exams.

Normality — the older acid-base unit

Normality (N) = gram-equivalents of solute per litre of solution.

N = gram-equivalents ÷ V(L)

Gram-equivalent weight depends on the reaction:

• Acid: equivalent weight = molar mass ÷ number of H⁺ ions released per molecule.
  • H₂SO₄: 98 ÷ 2 = 49 g/equiv; so 1 M H₂SO₄ = 2 N H₂SO₄.
  • HCl: 36.5 ÷ 1 = 36.5 g/equiv; so 1 M HCl = 1 N HCl.
• Base: equivalent weight = molar mass ÷ number of OH⁻ released.
• Redox: equivalent weight = molar mass ÷ electrons transferred.

The key titration shortcut: N₁V₁ = N₂V₂ (normalities × volumes of acid and base at equivalence point).

Why normality matters: older lab manuals, pharmacopoeia entries and some exam questions still use it. A "10 N H₂SO₄" solution is 5 M — confusing if you haven't converted.

Worked examples

Worked example 1:
Question: Find the molarity of 5.85 g of NaCl dissolved in 500 mL of solution. (Molar mass NaCl = 58.5 g/mol)
Solution:
Step 1: n(NaCl) = 5.85 ÷ 58.5 = 0.1 mol.
Step 2: V = 500 mL = 0.5 L. M = 0.1 ÷ 0.5.
Conclusion: Molarity = 0.2 M.

Worked example 2 (molality of same solution):
Question: Using the same solution (density ≈ 1 g/mL, so total mass ≈ 500 g), find the molality.
Solution:
Step 1: Mass of solvent = 500 − 5.85 = 494.15 g = 0.494 kg.
Step 2: m = 0.1 ÷ 0.494.
Conclusion: Molality ≈ 0.202 m — almost equal to molarity because this is a dilute aqueous solution. They diverge in concentrated solutions.

Shortcut for dilute aqueous solutions

For any dilute aqueous solution (density ≈ 1 g/mL):

Molarity ≈ Molality ≈ mole fraction × 55.5

(55.5 mol is the approximate molar concentration of pure water.) This breaks down for concentrated solutions or non-aqueous solvents — never apply it there.

Common misconception: Students treat molarity and molality as interchangeable. They are nearly equal only for dilute aqueous solutions (density ≈ 1 g/mL, solvent mass ≈ solution mass). For concentrated sulphuric acid or any organic solvent, the two diverge sharply — and only molality stays valid when temperature changes, since it is built on mass, not volume.

Real-world example: A saline drip labelled "0.9% w/v Normal Saline" contains 0.9 g NaCl per 100 mL. The ICU nurse uses % w/v, not molarity, to calculate how much salt a patient receives per hour — straightforward from the drip rate without any mole arithmetic. Meanwhile the laboratory chemist titrating that same salt solution uses normality and N₁V₁ = N₂V₂ to find its exact strength.

Cover the answer, recall, then check. 12 cards on concentration units for JEE Main.

Q1. Define molarity (M) and its unit.
A1. Moles of solute per litre of solution: M = n_solute / V(L). Unit: mol L⁻¹. Temperature-dependent (volume changes with T).

Q2. Define molality (m) and its unit.
A2. Moles of solute per kilogram of solvent: m = n_solute / mass_solvent(kg). Unit: mol kg⁻¹. Temperature-independent.

Q3. Why is molality preferred for colligative properties?
A3. It uses mass of solvent, which does not change with temperature, unlike molarity's volume.

Q4. Define mole fraction of a component.
A4. x_A = n_A / (n_A + n_B + …). Sum of all mole fractions = 1. Dimensionless, temperature-independent.

Q5. Relationship between molarity, density (d, g/mL) and molar mass for a solution.
A5. M = (10 × d × %w/w) / M_solute, where %w/w is mass percent of solute.

Q6. Define ppm (parts per million).
A6. ppm = (mass of solute / mass of solution) × 10⁶. Used for very dilute solutions (trace impurities).

Q7. Which concentration units are temperature-dependent, which are not?
A7. Temperature-dependent: molarity, normality, %w/v (all use volume). Independent: molality, mole fraction, %w/w.

Q8. Define normality (N).
A8. Gram-equivalents of solute per litre of solution. N = M × n-factor (valency factor). Unit: eq L⁻¹.

Q9. 1 M H₂SO₄ equals how many normal?
A9. 2 N (n-factor = 2 for H₂SO₄ as an acid, since it furnishes 2 H⁺).

Q10. Convert molality m to mole fraction of solute (solvent molar mass M_A).
A10. x_solute = (m × M_A/1000) / (1 + m × M_A/1000). For water, M_A/1000 = 0.018.

Q11. State the dilution equation.
A11. M₁V₁ = M₂V₂ (moles of solute conserved on dilution). Same form for normality: N₁V₁ = N₂V₂.

Q12. Molarity of pure water?
A12. ≈ 55.5 M (1000 g ÷ 18 g mol⁻¹ per litre, density ≈ 1 g/mL).

Concentration units quantify "how much solute in how much solution/solvent." JEE Main tests unit interconversion (molarity ↔ molality ↔ mole fraction) and dilution every year, and the topic underpins solutions, colligative properties and titrations. The recurring trap is temperature dependence.

The five workhorse units

Unit	Definition	T-dependent?
Molarity (M)	mol solute / L solution	Yes
Molality (m)	mol solute / kg solvent	No
Mole fraction (x)	n_component / n_total	No
Normality (N)	g-equiv solute / L solution	Yes
% w/w, ppm	mass ratios	No

Master conversion: M = (10 × d × %w/w)/M_solute, with density d in g/mL. Dilution: M₁V₁ = M₂V₂.

Why molality wins for colligative properties

Volume expands with temperature, so molarity drifts; mass of solvent does not, so molality and mole fraction are temperature-independent — that is why ΔTb, ΔTf and osmotic-pressure formulae are built on them.

Exam Tricks & Tips

• 🎯 Temperature-independent = uses mass/moles (molality, mole fraction, %w/w); T-dependent = uses volume (molarity, normality, %w/v). This single rule answers many one-liners.
• 🎯 Memorise M = 10 d (%w/w)/M_solute — the fastest route from a labelled bottle (density + mass %) to molarity.
• 🎯 Normality = Molarity × n-factor. For H₂SO₄ acid n = 2, for Na₂CO₃ base n = 2, for KMnO₄ (acidic) n = 5.
• 🎯 Pure water is 55.5 M — handy for very dilute-solution approximations.
• 🎯 For dilution, only solvent is added, so moles of solute stay fixed: M₁V₁ = M₂V₂.
• ❌ Common mistake: using mass of solution where molality needs mass of solvent — subtract the solute mass first.

Expected exam pattern

"A 20% (w/w) solution of density d has what molarity?", "convert molarity to molality/mole fraction," or a dilution/titration numerical using M₁V₁ = M₂V₂. Frequently single-numerical answers.

Quick recap

Molarity uses litres of solution (T-dependent); molality and mole fraction use solvent mass/moles (T-independent). Use M = 10 d (%w/w)/M for bottle data, N = M × n-factor, and M₁V₁ = M₂V₂ for dilution.

Key formulas

• Molarity M = moles solute / litres solution (mol/L); temperature-dependent.
• Molality m = moles solute / kg solvent; temperature-independent.
• Mole fraction xᵢ = nᵢ/n_total; Σx = 1.
• Mass % = (mass solute/mass solution)×100; ppm = (mass solute/mass solution)×10⁶.
• Normality N = gram-equivalents/L = M × n-factor.
• Dilution: M₁V₁ = M₂V₂.
• Relation: molarity = (10 × d × %w/w)/M_molar (d = density g/mL).

Usage: use molality (not molarity) for colligative properties since it is temperature-independent.

Worked Example

Problem: A glucose solution is 10% by mass (w/w) and has a density of 1.20 g/mL. Find its (a) molarity and (b) mole fraction of glucose. (Molar mass of glucose = 180 g/mol, water = 18 g/mol.)

Solution:
Take 100 g of solution as a basis. It contains 10 g glucose and 90 g water.

Moles: glucose = 10/180 = 0.0556 mol; water = 90/18 = 5.0 mol.

(a) Molarity needs the solution volume:
Volume = mass/density = 100 g / 1.20 g·mL⁻¹ = 83.3 mL = 0.0833 L
Molarity = moles of solute / litres of solution = 0.0556 / 0.0833 ≈ 0.667 M

(b) Mole fraction of glucose:
x_glucose = 0.0556 / (0.0556 + 5.0) = 0.0556 / 5.0556 ≈ 0.011

Answer: Molarity ≈ 0.67 M; mole fraction of glucose ≈ 0.011.