Chemistry | Class 12 | NCERT Comprehensive Edition

SOLUTIONS

Every line. Every concept. Every problem. Decoded.

0. The Science of Mixtures

Line-by-Line Logic: NCERT begins by stating that pure substances are rare. Most biological and industrial processes occur in solutions. The utility of a solution is governed by its composition.

Substance	Composition	Critical Property / Use
Brass	Copper + Zinc	Different electrical conductivity than pure Cu
Bronze	Copper + Tin	Harder alloy for statues/tools
German Silver	Cu + Zn + Ni	Does NOT contain silver; used in flatware
Fluoride ($\ce{F^-}$)	1.0 ppm	Prevents tooth decay by strengthening enamel
Fluoride ($\ce{F^-}$)	1.5 ppm	Causes mottling (browning) of teeth
$\ce{NaF}$	High Conc.	Used as Rat Poison (Toxic to mammals)

Biological Context: Intravenous injections must be isotonic with blood plasma (matched ionic concentrations) to prevent cell damage.

1.1 Types of Solutions

Definition: A solution is a homogeneous mixture—composition and properties are uniform throughout.
Solvent: The component present in the largest quantity (determines state).
Solute: Component(s) present in smaller quantities.

Table 1.1: Comprehensive Classification

Soln Type	Solute	Solvent	NCERT Textbook Examples
Gaseous	Gas	Gas	Mixture of $\ce{O2}$ and $\ce{N2}$ (Air)
Gaseous	Liquid	Gas	Chloroform mixed with $\ce{N2}$ gas
Gaseous	Solid	Gas	Camphor in $\ce{N2}$ gas
Liquid	Gas	Liquid	Oxygen dissolved in $\ce{H2O}$
Liquid	Liquid	Liquid	Ethanol in $\ce{H2O}$
Liquid	Solid	Liquid	Glucose in $\ce{H2O}$
Solid	Gas	Solid	Solution of Hydrogen in Palladium
Solid	Liquid	Solid	Amalgam of Mercury with Sodium
Solid	Solid	Solid	Copper dissolved in Gold

1.2 Methods of Expressing Concentration

NCERT Line: Composition of a solution can be expressed qualitatively or quantitatively.

Concentration tells us how much solute is present in a given amount of solvent or solution.
It measures the strength of a solution.
Cup A: 10 g sugar + 100 mL water → Cup B: 50 g sugar + 100 mL water → Cup B is sweeter because concentration is higher.

Two Ways to Express Concentration

Type	Meaning	Examples	Board Use
Qualitative	No numbers — just description	Dilute, Concentrated	Theory only
Quantitative	Uses numbers	10%, 0.5 M, 2 molal, 5 ppm	Numericals every year

VERY IMPORTANT: NCERT gives 7 concentration units. Memorize all of them.

Unit	Symbol	Used For
Mass Percentage	w/w	Solid + liquid mixtures
Volume Percentage	v/v	Liquid + liquid mixtures
Mass by Volume %	w/v	Medicine, pharmacy, hospitals
Parts per Million	ppm	Very dilute solutions
Mole Fraction	X	Vapour pressure, gas mixtures
Molarity	M	Lab calculations (temp dependent)
Molality	m	Colligative properties (temp independent)

1. Mass Percentage (w/w)

$$\text{Mass \%} = \frac{\text{Mass of Solute}}{\text{Mass of Solution}} \times 100$$

Denominator is mass of solution — NOT solvent alone.
Solution = Solute + Solvent (always add both first).
Example: 20 g salt + 80 g water → total = 100 g → Mass % = (20/100) × 100 = 20%.
Example: 10 g sugar + 90 g water → Mass % = (10/100) × 100 = 10%.

Memory Trick: w/w = Weight by Weight = Mass by Mass.

❌ Divide by solvent mass only

✔ Divide by total solution mass (solute + solvent)

NCERT Example

Commercial bleach contains about 3.62% sodium hypochlorite by mass.

Memorise — Mass %

Mass % = (Mass of Solute / Mass of Solution) × 100
Solution = Solute + Solvent

2. Volume Percentage (v/v)

$$\text{Volume \%} = \frac{\text{Volume of Component}}{\text{Total Volume}} \times 100$$

10 mL ethanol + 90 mL water → total 100 mL → Volume % = 10%.
Used mainly for liquid + liquid solutions.

NCERT Example

35% ethylene glycol is used as antifreeze in car radiators.

Memory Trick: v/v = Volume by Volume.

Memorise — Volume %

Volume % = (Volume of Solute / Volume of Solution) × 100

3. Mass by Volume % (w/v)

Definition: Mass of solute present in 100 mL solution.
5 g glucose in 100 mL solution = 5% (w/v).
Mostly used in hospitals, medicine, and pharmacy.

Memory Trick: w/v = Weight in Volume (grams per 100 mL).

Memorise — w/v %

w/v % = grams of solute per 100 mL solution

4. Parts per Million (ppm)

$$\text{ppm} = \frac{\text{Parts of Solute}}{\text{Total Parts}} \times 10^6$$

Used for very dilute solutions — pollution, water quality, blood analysis.
1 ppm = 1 part in 10,00,000 (10⁶) parts.
Sea water contains about 5.8 ppm oxygen (NCERT).
Fluoride at 1 ppm prevents tooth decay; at 1.5 ppm causes mottled teeth.

Memory Trick: ppm = Parts Per Million → Million = 10⁶.

Memorise — ppm

ppm = (Parts of Solute / Total Parts) × 10⁶

5. Mole Fraction (X) — Most Important

$$X_A = \frac{n_A}{n_A + n_B}$$

Moles of one component ÷ total moles of all components.
Example: 2 mol sugar + 8 mol water → X_sugar = 2/10 = 0.2; X_water = 8/10 = 0.8.
MOST IMPORTANT: Sum of all mole fractions always = 1. X_A + X_B = 1. For three components: X_A + X_B + X_C = 1.
Shortcut: find one mole fraction → other = 1 − X.

❌ Divide by mass

✔ Always use moles. Add all moles for denominator.

Example 1.1 — NCERT

Calculate mole fraction of ethylene glycol ($\ce{C2H6O2}$) in a solution containing 20% $\ce{C2H6O2}$ by mass.

Step 1: Basis = 100 g solution → Glycol = 20 g, Water = 80 g.
Step 2: $n_{\text{glycol}} = 20/62 = 0.322$ mol
Step 3: $n_{\text{water}} = 80/18 = 4.44$ mol
Step 4: $X_{\text{glycol}} = 0.322/(0.322+4.44) = \mathbf{0.068}$
Step 5: $X_{\text{water}} = 1 - 0.068 = \mathbf{0.932}$

Memory Trick: Fraction = Part ÷ Whole (use moles, not grams).

Memorise — Mole Fraction

X = Moles of Component / Total Moles
ΣX = 1 (all mole fractions add to 1)

6. Molarity (M) — Appears Every Year

$$M = \frac{\text{Moles of Solute}}{\text{Volume of Solution (L)}} \quad \text{Unit: mol/L or M}$$

Definition: number of moles present in 1 litre of solution.
1 mole NaCl in 1 L solution = 1 M.
Volume must be in litres. Convert: 450 mL → 0.450 L.
Use solution volume, not solvent volume.
Molarity is temperature dependent (volume changes with temperature).

Example 1.2 — NCERT

Calculate molarity of a solution containing 5 g $\ce{NaOH}$ in 450 mL solution.

Step 1: Moles $\ce{NaOH} = 5/40 = 0.125$ mol
Step 2: Volume = 450/1000 = 0.450 L
Step 3: M = 0.125/0.450 = 0.278 M

Memory Trick: Molarity = M = Measuring whole solution volume in Litres.

Memorise — Molarity

M = Moles of Solute / Litre of Solution

7. Molality (m) — Extremely Important

$$m = \frac{\text{Moles of Solute}}{\text{kg of Solvent}}$$

Definition: moles of solute per kg of solvent (NOT solution).
1 mole NaCl in 1 kg water = 1 molal.
Molality is temperature independent — mass does not change with temperature.

NCERT Example

2.5 g ethanoic acid in 75 g benzene → Answer: 0.556 m

Moles acid = 2.5/60 = 0.0417 mol · Solvent = 75 g = 0.075 kg · m = 0.0417/0.075 = 0.556 m

Memory Trick: Molality → Mass → kg. Both start with M → Molality uses kg solvent.

Molarity vs Molality — Master Comparison

Feature	Molarity (M)	Molality (m)
Based on	Volume of solution	Mass of solvent
Unit	mol/L	mol/kg
Temperature	Dependent (volume changes)	Independent (mass constant)
Used in	General lab work	Colligative property numericals

Memorise — All 7 Concentration Formulas

Mass % = Solute Mass / Solution Mass × 100
Volume % = Solute Volume / Solution Volume × 100
w/v % = g of solute per 100 mL solution
ppm = Parts of Solute / Total Parts × 10⁶
Mole Fraction X = Component Moles / Total Moles · ΣX = 1
Molarity M = Moles / L Solution
Molality m = Moles / kg Solvent

30-Second Revision — Concentration

Mass % → mass/mass · Volume % → volume/volume · w/v → grams in 100 mL · ppm → trace (×10⁶) · Mole fraction → moles/total; all add to 1 · M → moles/L solution · m → moles/kg solvent · M changes with T; m does not.

Board Questions — Concentration

1 Mark: Define molarity, molality, mole fraction, ppm. State mass % formula.
2 Marks: Differentiate molarity vs molality. Why is molality temperature independent?
3 Marks: Explain all methods of expressing concentration with formulas.

1.3 Solubility Dynamics (Line‑by‑Line)

NCERT Line 1: "Like dissolves like" – polar solutes dissolve in polar solvents, non‑polar in non‑polar.

NCERT Line 2: Dissolution occurs when solute‑solvent interactions are comparable to solute‑solute and solvent‑solvent forces.

NCERT Line 3: When the rate of dissolution equals the rate of crystallisation, the solution is saturated (dynamic equilibrium).

NCERT Line 4: Solubility of gases increases with pressure (Henry’s Law) and decreases with temperature for most solids.

NCERT Line 5: Effect of temperature – solubility of solids in liquids usually rises with temperature, whereas gases show the opposite trend.

Example

Sugar in water: 100 g water at 25 °C dissolves about 211 g sugar; at 0 °C only ~180 g. Demonstrates temperature dependence.

Memory Trick: "Polar‑Polar → Perfect Pair; Non‑Polar → No Pair."

1.3.2 Solubility of Gases – Henry’s Law (Line‑by‑Line)

NCERT Line 1: Henry’s Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.

NCERT Line 2: Mathematically, $P = K_{\!H}\,x$, where $P$ is the gas pressure, $K_{\!H}$ is Henry’s constant, and $x$ is mole fraction of the gas in solution.

NCERT Line 3: $K_{\!H}$ increases with temperature, meaning gases become less soluble at higher temperatures.

Numerical

Given $K_{\!H}=0.2\,$atm for $\ce{CO2}$ at 25 °C, find the mole fraction when $P=1\,$atm.
Solution: $x = P/K_{\!H}=1/0.2 =5$ (dimensionless). In practice, $x$ is expressed per mole of solvent, so $x=5\times10^{-3}$.

Memory Trick: Higher Pressure → Higher solubility (H = Pressure).

1.4 Raoult’s Law – Full Line‑by‑Line

NCERT Line 1: For an ideal solution of volatile liquids, the partial vapour pressure of each component equals the product of its mole fraction in the solution and its vapour pressure as a pure component.

NCERT Line 2: $p_i = x_i\,p_i^{\circ}$, where $p_i$ is the partial pressure, $x_i$ is mole fraction, and $p_i^{\circ}$ is vapour pressure of the pure component.

NCERT Line 3: The total vapour pressure is the sum of the partial pressures: $p_{\text{total}} = \sum_i x_i p_i^{\circ}$.

Example 1.5

Mix 25.5 g $\ce{CHCl3}$ ($p^{\circ}=200\,$mm Hg) with 40 g $\ce{CH2Cl2}$ ($p^{\circ}=415\,$mm Hg). Calculate total vapour pressure.

Solution steps shown earlier remain valid; final $p_{\text{total}}=347.9\,$mm Hg.

Memory Trick: Raoult → “Root” of mole fraction times pure pressure.

1.5 Ideal & Non‑Ideal Solutions – Detailed Line‑by‑Line

NCERT Line 1: An ideal solution obeys Raoult’s law for all components across the whole composition range.

NCERT Line 2: Positive deviation occurs when intermolecular forces between unlike molecules are weaker than those between like molecules, giving higher vapour pressure than predicted.

NCERT Line 3: Negative deviation occurs when unlike interactions are stronger, leading to lower vapour pressure.

Azeotrope Example

95 % ethanol‑water mixture shows a minimum boiling azeotrope (negative deviation).

Memory Trick: Positive → Pressure ↑, Negative → Pressure ↓.

❌ Assume all solutions are ideal.

✔ Check for deviation before applying Raoult’s law.

1.3 Solubility Dynamics

Like Dissolves Like: Polar solutes dissolve in polar solvents. Naphthalene (non-polar) dissolves in benzene but not in water.

Dissolution vs Crystallisation: When rate of dissolution = rate of crystallisation, the solution is Saturated. This is a state of dynamic equilibrium.

Effect of Pressure: Pressure has no significant effect on solubility of solids in liquids because solids/liquids are highly incompressible.

1.3.2 Solubility of Gases (Henry's Law)

Henry's Law: The partial pressure of the gas in vapour phase ($p$) is proportional to the mole fraction of the gas ($x$) in the solution.

$$p = K_H \cdot x$$

CRITICAL CONCEPT: Higher $K_H$ at a given pressure means lower solubility. $K_H$ values for $\ce{N2}$ and $\ce{O2}$ increase with temperature, meaning gases are less soluble in hot water. (Aquatic species prefer cold water).

Intext 1.6

$\ce{H2S}$, a toxic gas with rotten egg smell, is used for qualitative analysis. If solubility of $\ce{H2S}$ in water at STP is $0.195\text{ m}$, calculate Henry’s law constant.

Logic: $0.195\text{ mol}$ in $1\text{kg}$ solvent ($55.5\text{ mol}$). $x_{\ce{H2S}} = 0.195 / (0.195+55.5) = 0.0035$.
At STP, $P = 0.987 \text{ bar}$. $K_H = 0.987 / 0.0035 = 282 \text{ bar}$.

1.4 Vapour Pressure of Liquid Solutions

Raoult's Law (Volatile components): For a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.

$$p_1 = p_1^\circ x_1 \quad ; \quad p_{\text{total}} = p_1^\circ x_1 + p_2^\circ x_2$$

Example 1.5

Vapour pressure of chloroform ($\ce{CHCl3}$) and dichloromethane ($\ce{CH2Cl2}$) at 298 K are $200\text{ mm Hg}$ and $415\text{ mm Hg}$. Calculate total V.P. of solution prepared by mixing $25.5\text{g}$ $\ce{CHCl3}$ and $40\text{g}$ $\ce{CH2Cl2}$.

1. Moles $\ce{CHCl3} = 0.213$; Moles $\ce{CH2Cl2} = 0.47$.
2. $x_{\ce{CH2Cl2}} = 0.47 / 0.683 = 0.688$.
3. $P_{\text{total}} = 200 + (415 - 200) \times 0.688 = 347.9 \text{ mm Hg}$.
4. Vapour phase composition: $y_i = p_i / P_{\text{total}}$. Vapour is richer in the more volatile component ($\ce{CH2Cl2}$).

1.5 Ideal and Non-Ideal Solutions

Feature	Ideal Solution	Positive Deviation	Negative Deviation
Raoult's Law	Obeys strictly	$p > \text{Predicted}$	$p < \text{Predicted}$
Interactions	$A-B \approx A-A, B-B$	$A-B < A-A, B-B$	$A-B > A-A, B-B$
$\Delta H_{\text{mix}}$	Zero	Positive (Endo)	Negative (Exo)
$\Delta V_{\text{mix}}$	Zero	Positive (Expansion)	Negative (Contraction)
Examples	Benzene + Toluene	Ethanol + Acetone	Phenol + Aniline

Azeotropes: Binary mixtures having same composition in liquid and vapour phase.
1. Minimum Boiling: Show large positive deviation (e.g., $95\%$ ethanol).
2. Maximum Boiling: Show large negative deviation (e.g., $68\%$ nitric acid).

1.6 Colligative Properties

Colligative properties depend solely on the number of solute particles relative to total particles — not on their chemical identity.

Four colligative properties: Relative lowering of V.P., Elevation of B.P., Depression of F.P., Osmotic pressure.
Only works for non-volatile solutes (for V.P. lowering).
For electrolytes, always multiply by van't Hoff factor i.

1.6.1 Relative Lowering of Vapour Pressure

$$\frac{p_1^\circ - p_1}{p_1^\circ} = x_2 = \frac{n_2}{n_1 + n_2}$$

Adding non-volatile solute replaces solvent molecules at the surface → fewer escape → vapour pressure decreases.
Find molar mass: $\frac{\Delta p}{p^\circ} = \frac{w_2 M_1}{M_2 w_1}$

Memorise

Relative Lowering: $(p^\circ - p)/p^\circ = X_{\text{solute}}$
Raoult's Law: $P = X P^\circ$

1.6.2 Elevation of Boiling Point (ΔTb)

NCERT Concept: A liquid boils when its vapour pressure becomes equal to the external (atmospheric) pressure.

What is Boiling?

At low temperature → low vapour pressure → no boiling.
Heat increases → more molecules escape → vapour pressure rises.
When vapour pressure = atmospheric pressure → bubbles form throughout → boiling.
Boiling point = temperature at which vapour pressure of liquid equals external atmospheric pressure. (1-mark definition)
Water boils at 100°C because at 1 atm its vapour pressure reaches 1 atm at 100°C.

Why Does Boiling Point Increase with Solute?

Beaker A: Pure water → B.P. = 100°C.
Beaker B: Water + sugar (non-volatile) → sugar lowers vapour pressure of water.
At 100°C, solution V.P. < 1 atm → cannot boil yet → must heat more.
Eventually V.P. reaches 1 atm at higher temperature → B.P. increases (e.g. 102°C).

Definition: The increase in boiling point of a solvent after adding a non-volatile solute is called Elevation of Boiling Point.

$$\Delta T_b = T_b - T_b^\circ \qquad \text{Example: } 102 - 100 = 2°\text{C}$$

Molecular Reason

Pure water: many molecules escape → high vapour pressure → boils at 100°C.
Sugar solution: sugar blocks surface → fewer water molecules escape → lower V.P. → need higher T.

$$\Delta T_b = K_b \cdot m \quad \text{(with electrolytes: } \Delta T_b = i K_b m\text{)}$$

K_b = Molal Elevation Constant = Ebullioscopic Constant.
Definition: elevation in B.P. when 1 mole solute dissolved in 1 kg solvent.
Units: K kg mol⁻¹ — memorize this.
More particles (higher molality) → greater V.P. lowering → higher boiling point.
Derivation: $\Delta T_b \propto m$ → introduce constant → $\Delta T_b = K_b m$.

Numerical

Molality = 0.5 m, K_b = 0.52 K kg mol⁻¹. Find ΔT_b.

$\Delta T_b = K_b m = 0.52 \times 0.5 = \mathbf{0.26\ K}$

Everyday Applications

Cooking rice: salt raises boiling point slightly → water becomes slightly hotter.
Car radiators: coolants (ethylene glycol) raise B.P. → prevents overheating.

Memory Trick: B before F → Boiling → K_b, Freezing → K_f. Ebullition = Boiling → Ebullioscopic constant.

❌ Sugar itself boils · ❌ Use molarity instead of molality · ❌ Use K_f for boiling

✔ Only solvent boils · ✔ Always use molality (m) · ✔ Use K_b for ΔT_b

Memorise — Boiling Point Elevation

Boiling: Vapour Pressure = Atmospheric Pressure
ΔT_b = T_b − T_b°
ΔT_b = K_b × m (or iK_bm for electrolytes)
K_b = Ebullioscopic Constant · Units: K kg mol⁻¹
m = Moles of Solute / kg of Solvent

Board Questions — ΔTb

1 Mark: Define boiling point. Define elevation in boiling point. State ΔT_b = K_bm.
2 Marks: Why does B.P. increase after adding non-volatile solute? Define ebullioscopic constant.
3 Marks: Explain elevation with graph. Write and explain ΔT_b = K_bm.
5 Marks: Explain phenomenon + graph + derive ΔT_b = K_bm + solve numerical.

1.6.3 Depression of Freezing Point (ΔTf)

What is Freezing?

Freezing = liquid changes to solid (water → ice).
Freezing point = temperature at which liquid and solid coexist in equilibrium (dynamic equilibrium).
At 0°C and 1 atm: water ⇌ ice at equal rates → freezing point = 0°C.

Why Does Freezing Point Decrease?

Pure water: molecules arrange into crystal easily → freezes at 0°C.
Sugar solution: sugar particles block water molecules from forming ice crystals.
Ice formation becomes difficult → must cool more → freezing point decreases (e.g. −2°C).

Definition: The decrease in freezing point after adding a non-volatile solute is called Depression of Freezing Point.

$$\Delta T_f = T_f^\circ - T_f \qquad \text{Example: } 0 - (-2) = 2°\text{C}$$

$$\Delta T_f = K_f \cdot m \quad \text{(with electrolytes: } \Delta T_f = i K_f m\text{)}$$

K_f = Cryoscopic Constant = Molal Depression Constant.
Definition: depression in F.P. when 1 mole solute dissolved in 1 kg solvent.
Units: K kg mol⁻¹ (same as K_b).
More particles → harder crystal formation → lower freezing point.

Numerical

K_f = 1.86 K kg mol⁻¹, molality = 0.5 m. Find ΔT_f.

$\Delta T_f = K_f m = 1.86 \times 0.5 = \mathbf{0.93\ K}$

Everyday Applications

Antifreeze in cars: ethylene glycol lowers freezing point → water doesn't freeze in winter.
Salt on snowy roads: salt lowers F.P. → ice melts → safer roads.
Ice cream making: salt added to ice → F.P. drops below 0°C → mixture gets colder → ice cream freezes faster.

Elevation vs Depression — Comparison

Feature	Elevation (ΔT_b)	Depression (ΔT_f)
Effect	B.P. increases	F.P. decreases
Formula	ΔT_b = K_bm	ΔT_f = K_fm
Constant	Ebullioscopic (K_b)	Cryoscopic (K_f)
Reason	Lower V.P. needs more heating	Solute interferes with crystal formation

Memory Trick: F → Falls. Freezing point Falls. F → K_f.

❌ ΔT_f = T_f − T_f° · ❌ Use K_b for freezing · ❌ Sugar freezes

✔ ΔT_f = T_f° − T_f · ✔ Use K_f · ✔ Sugar only blocks crystal formation

Memorise — Freezing Point Depression

Freezing point: Liquid ⇌ Solid at equilibrium
ΔT_f = T_f° − T_f
ΔT_f = K_f × m (or iK_fm)
K_f = Cryoscopic Constant · Units: K kg mol⁻¹

Board Questions — ΔTf

1 Mark: Define freezing point. Define depression in freezing point. Define cryoscopic constant.
2 Marks: Why does sugar lower F.P.? State relation between ΔT_f and molality.
5 Marks: Derive ΔT_f = K_fm + molecular reason + numerical.

1.6.4 Osmosis & Osmotic Pressure

Osmosis: The flow of solvent molecules from a region of lower concentration to higher concentration through a semipermeable membrane (SPM).

Semipermeable membrane: Allows only solvent molecules to pass, not solute particles.

Only the solvent moves during osmosis — never the solute.
Osmosis stops when concentrations equalize OR when osmotic pressure is applied.
Osmotic pressure (π): the pressure required to stop osmosis.

$$\Pi = CRT = \frac{n_2}{V}RT \qquad \text{(electrolytes: } \Pi = iCRT\text{)}$$

C = molarity (mol/L), R = 0.0821 L atm mol⁻¹ K⁻¹, T must be in Kelvin.
Rearranged for molar mass: $M = \frac{wRT}{\Pi V}$
Derivation: πV = nRT → π = (n/V)RT = CRT.
Advantage: osmotic pressure method works even for very dilute solutions (large π value, easy to measure).

Reverse Osmosis

Apply pressure greater than π on the solution side.
Solvent flows from solution → pure solvent side (opposite of osmosis).
Used for desalination of sea water and RO water purification.

Isotonic, Hypertonic, Hypotonic

Type	Concentration vs Cell	Water Movement	Effect on Cell
Isotonic	Same concentration	No net movement	Normal (IV saline = isotonic with blood)
Hypertonic	Higher than cell	Water OUT of cell	Cell shrinks (crenation)
Hypotonic	Lower than cell	Water INTO cell	Cell swells / bursts

Memory Tricks:
ISO → Same · Hyper → High concentration → Water Out · Hypo → Low concentration → Water In

Memorise — Osmosis

π = CRT (or iCRT for electrolytes)
M = wRT / πV
Reverse osmosis: pressure > π → desalination
Always use Kelvin for T in π = CRT

Board Questions — Osmosis

1 Mark: Define osmosis, osmotic pressure, semipermeable membrane, reverse osmosis.
2 Marks: Osmosis vs reverse osmosis. Isotonic vs hypertonic vs hypotonic.
3 Marks: Explain osmosis with diagram. Derive π = CRT.
5 Marks: Derive π = CRT + explain reverse osmosis + numerical.

1.7 Abnormal Molar Mass & van't Hoff Factor

NCERT Concept: When solute associates or dissociates in solution, the observed colligative property differs from the calculated value — giving abnormal molar mass.

Dissociation (electrolytes): more particles → observed colligative property increases → apparent molar mass decreases → i > 1.
Association (e.g. acetic acid in benzene): fewer particles → observed property decreases → apparent molar mass increases → i < 1.
Example: $\ce{NaCl} \to \ce{Na+ + Cl-}$ → i = 2 (complete dissociation).
Example: $\ce{CH3COOH}$ in benzene → dimerises → i = 0.5.

$$i = \frac{\text{Normal (Expected) Molar Mass}}{\text{Observed (Abnormal) Molar Mass}} = \frac{\text{Normal Colligative Property}}{\text{Observed Colligative Property}}$$

Dissociation — Degree of Dissociation (α)

$$i = 1 + (n - 1)\alpha$$

n = number of ions/particles formed on dissociation.
α = degree of dissociation (0 to 1).
Example: $\ce{NaCl}$ → n = 2 → i = 1 + α. If complete dissociation (α = 1) → i = 2.
Example: $\ce{CaCl2} \to \ce{Ca^{2+} + 2Cl-}$ → n = 3 → i = 1 + 2α.

Association

$$i = 1 - \frac{\alpha(n-1)}{n}$$

n = number of molecules that associate to form one aggregate.
Acetic acid dimerises: 2 molecules → 1 dimer → n = 2.

Modified Colligative Formulas (with i)

Memorise — All Formulas with i

Relative Lowering: $(p^\circ - p)/p^\circ = i \cdot X_{\text{solute}}$
ΔT_b = i K_b m
ΔT_f = i K_f m
π = i C R T
i = Normal / Observed (molar mass or colligative property)
Dissociation: i = 1 + (n − 1)α
Association: i = 1 − α(n−1)/n

Memory Tricks:
Dissociation → Divide → i > 1 · Association → Attach → i < 1 · Never reverse the i formula

Board Questions — van't Hoff Factor

1 Mark: Define van't Hoff factor. Define abnormal molar mass.
2 Marks: Association vs dissociation. Effect on i value.
3 Marks: Explain abnormal molar mass with examples.
5 Marks: Explain van't Hoff factor + derive i = 1 + (n−1)α + numerical.

NCERT Intext Question Bank

Read this before tomorrow's exam. These are the most important in-text concepts with board-ready answers.

Q1. Why does solubility of gases increase with pressure?

Henry's Law: $P = K_H X$. As pressure ↑ → more gas molecules forced into liquid → solubility increases.

Q2. Why are cold drinks bottled under high pressure?

High pressure → more CO₂ dissolves → bottle stays fizzy. When opened → pressure drops → CO₂ escapes as bubbles.

Q3. Why do mountaineers suffer at high altitudes?

Pressure decreases → less O₂ dissolves in blood → less O₂ reaches tissues → breathing difficulty (hypoxia).

Q4. Why are scuba divers affected by decompression sickness?

At depth, pressure is high → more N₂ dissolves in blood. Rising quickly → pressure drops suddenly → N₂ bubbles form → pain and blockage ("the bends").

Q5. Why does adding non-volatile solute lower vapour pressure?

Solute particles replace solvent at surface → fewer solvent molecules escape → vapour pressure decreases.

Final Comprehensive Exercise Bank

The following is a curated logic guide for the most critical NCERT Back Exercises (1.1 - 1.41).

Q 1.4: Conc. $\ce{HNO3}$ is $68\%$ by mass. Density is $1.504\text{ g/mL}$. Find Molarity.

Logic: Assume $100\text{g}$ soln. Mass acid = $68\text{g}$. Moles = $68/63 = 1.079$. Vol = $100/1.504 = 66.49\text{ mL}$. $M = (1.079 / 66.49) \times 1000 = 16.23 \text{ M}$.

Q 1.20: $5\%$ sugar solution (mass) has F.P. $271\text{K}$. Find F.P. of $5\%$ glucose in water. (Pure water F.P. = $273.15\text{K}$).

Logic: Use Sugar $\Delta T_f = 2.15$ to find $K_f$. $\Delta T_f = K_f \cdot m$. Then apply that $K_f$ to Glucose molality. $\Delta T_f$ (Glucose) = $4.09$. F.P. = $273.15 - 4.09 = 269.06\text{ K}$.

Q 1.40: Amount of $\ce{CaCl2}$ ($i = 2.47$) in $2.5\text{L}$ water so $\Pi = 0.75\text{ atm}$ at $27^\circ\text{C}$.

Logic: $\Pi = i CRT \implies 0.75 = 2.47 \times (n/2.5) \times 0.0821 \times 300$. Solve for $n$, then multiply by molar mass of $\ce{CaCl2}$ ($111\text{ g/mol}$). Mass $\approx 3.42\text{ g}$.

Quick Atomic Mass Sheet

Elem	Mass	Elem	Mass	Elem	Mass
H	1	Na	23	S	32
C	12	Mg	24	Cl	35.5
N	14	Al	27	K	39
O	16	P	31	Ca	40