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Solutions
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Class 12ChemistryChapter 2NCERTCBSE
Solutions - Class 12 Chemistry Chapter 2
📌 Quick Overview: Chapter 2 deals with solutions — homogeneous mixtures of two or more substances. You will learn about types of solutions, ways to express concentration, solubility and Henry's law, Raoult's law, ideal and non-ideal solutions, and the four colligative properties — relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure. Numericals on colligative properties and van't Hoff factor are very frequently asked in CBSE board exams.
Topics Covered:
- > Types of Solutions
- > Concentration Terms
- > Solubility of Gases (Henry's Law)
- > Raoult's Law
- > Ideal and Non-Ideal Solutions
- > Relative Lowering of Vapour Pressure
- > Elevation of Boiling Point
- > Depression of Freezing Point
- > Osmotic Pressure
- > Abnormal Molar Masses
- > van't Hoff Factor
- > Important Board Questions
1. Types of Solutions
Solution: A homogeneous mixture of two or more components is called a solution. The component present in larger quantity is the solvent and the component in smaller quantity is the solute.
| Type of Solution | Solute | Solvent | Example |
|---|---|---|---|
| Gas in Gas | Gas | Gas | Air (O2 in N2) |
| Gas in Liquid | Gas | Liquid | CO2 in water (soda water) |
| Gas in Solid | Gas | Solid | H2 in palladium |
| Liquid in Liquid | Liquid | Liquid | Ethanol in water |
| Liquid in Solid | Liquid | Solid | Mercury in amalgam |
| Solid in Liquid | Solid | Liquid | NaCl in water, sugar in water |
| Solid in Solid | Solid | Solid | Copper in gold (alloys) |
2. Concentration Terms ⭐
1. MASS PERCENTAGE (w/w):
= (Mass of solute / Mass of solution) x 100
2. VOLUME PERCENTAGE (v/v):
= (Volume of solute / Volume of solution) x 100
3. MASS BY VOLUME PERCENTAGE (w/v):
= (Mass of solute in g / Volume of solution in mL) x 100
4. PARTS PER MILLION (ppm):
= (Mass of solute / Mass of solution) x 10^6
5. MOLE FRACTION (x):
x_A = n_A / (n_A + n_B) x_B = n_B / (n_A + n_B)
x_A + x_B = 1
6. MOLARITY (M):
M = Moles of solute / Volume of solution in litres
Unit: mol/L or M
7. MOLALITY (m):
m = Moles of solute / Mass of solvent in kg
Unit: mol/kg or m
8. NORMALITY (N):
N = Equivalents of solute / Volume of solution in litres
⭐ Key Differences: Molarity vs Molality
> Molarity depends on temperature (volume changes with temperature)
> Molality is independent of temperature (uses mass of solvent, not volume)
> For dilute aqueous solutions: Molarity ≈ Molality (since density of water ≈ 1 g/mL)
Relation between Molarity and Molality:
m = (M x 1000) / (1000 x d - M x M2)
where d = density of solution (g/mL), M2 = molar mass of solute
> Molarity depends on temperature (volume changes with temperature)
> Molality is independent of temperature (uses mass of solvent, not volume)
> For dilute aqueous solutions: Molarity ≈ Molality (since density of water ≈ 1 g/mL)
Relation between Molarity and Molality:
m = (M x 1000) / (1000 x d - M x M2)
where d = density of solution (g/mL), M2 = molar mass of solute
Example: Calculate the molarity of a solution containing 5g of NaOH (M = 40 g/mol) in 250 mL of solution.
Moles of NaOH = 5/40 = 0.125 mol
Volume = 250 mL = 0.25 L
Molarity = 0.125/0.25 = 0.5 M
Moles of NaOH = 5/40 = 0.125 mol
Volume = 250 mL = 0.25 L
Molarity = 0.125/0.25 = 0.5 M
3. Solubility of Gases - Henry's Law ⭐
Henry's Law: At constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
p = KH x x
where p = partial pressure of gas, KH = Henry's law constant, x = mole fraction of gas in solution.
Higher KH = lower solubility at given pressure.
p = KH x x
where p = partial pressure of gas, KH = Henry's law constant, x = mole fraction of gas in solution.
Higher KH = lower solubility at given pressure.
⭐ Applications of Henry's Law:
> Carbonated drinks (CO2 dissolved under high pressure — released when bottle opened)
> Scuba diving (at high pressure, more N2 dissolves in blood — causes "bends" on rapid ascent)
> Oxygen cylinders for patients with lung disorders
> Aquatic life needs dissolved oxygen in water
Effect of Temperature on Gas Solubility:
Solubility of gases DECREASES with increase in temperature (dissolving is exothermic).
> Carbonated drinks (CO2 dissolved under high pressure — released when bottle opened)
> Scuba diving (at high pressure, more N2 dissolves in blood — causes "bends" on rapid ascent)
> Oxygen cylinders for patients with lung disorders
> Aquatic life needs dissolved oxygen in water
Effect of Temperature on Gas Solubility:
Solubility of gases DECREASES with increase in temperature (dissolving is exothermic).
4. Raoult's Law ⭐
Raoult's Law for Volatile Solute: For a solution of two volatile liquids A and B, the partial vapour pressure of each component is proportional to its mole fraction.
pA = xA . pA* pB = xB . pB*
Total pressure: pTotal = pA + pB = xA.pA* + xB.pB*
where pA* and pB* are vapour pressures of pure A and B.
pA = xA . pA* pB = xB . pB*
Total pressure: pTotal = pA + pB = xA.pA* + xB.pB*
where pA* and pB* are vapour pressures of pure A and B.
Raoult's Law for Non-Volatile Solute: The vapour pressure of a solution is equal to the mole fraction of the solvent multiplied by the vapour pressure of the pure solvent.
p = x_solvent x p*
Since x_solvent less than 1, the vapour pressure of the solution is always less than that of the pure solvent.
p = x_solvent x p*
Since x_solvent less than 1, the vapour pressure of the solution is always less than that of the pure solvent.
5. Ideal and Non-Ideal Solutions ⭐
| Property | Ideal Solution | Non-Ideal Solution |
|---|---|---|
| Raoult's Law | Obeys at all compositions | Does NOT obey |
| Delta H mixing | Zero | Non-zero (+ or -) |
| Delta V mixing | Zero | Non-zero (+ or -) |
| A-B interactions | Same as A-A and B-B | Different from A-A and B-B |
| Examples | Benzene-toluene, n-hexane-n-heptane, ethyl bromide-ethyl iodide | See below |
Positive Deviation from Raoult's Law:
> A-B interactions weaker than A-A and B-B
> pTotal greater than expected → Delta H mixing positive (endothermic)
> Delta V mixing positive
> Examples: Acetone + CS2, Acetone + Ethanol, Ethanol + Water, Acetone + Benzene
Negative Deviation from Raoult's Law:
> A-B interactions stronger than A-A and B-B
> pTotal less than expected → Delta H mixing negative (exothermic)
> Delta V mixing negative
> Examples: Acetone + Chloroform, HNO3 + Water, HCl + Water, Acetic acid + Pyridine
> A-B interactions weaker than A-A and B-B
> pTotal greater than expected → Delta H mixing positive (endothermic)
> Delta V mixing positive
> Examples: Acetone + CS2, Acetone + Ethanol, Ethanol + Water, Acetone + Benzene
Negative Deviation from Raoult's Law:
> A-B interactions stronger than A-A and B-B
> pTotal less than expected → Delta H mixing negative (exothermic)
> Delta V mixing negative
> Examples: Acetone + Chloroform, HNO3 + Water, HCl + Water, Acetic acid + Pyridine
⭐ Azeotropes: Solutions that boil at constant temperature and have the same composition in liquid and vapour phase. They CANNOT be separated by simple distillation.
> Maximum boiling azeotrope: negative deviation (HNO3 + Water boiling at 120.5°C)
> Minimum boiling azeotrope: positive deviation (Ethanol + Water boiling at 78.1°C)
> Maximum boiling azeotrope: negative deviation (HNO3 + Water boiling at 120.5°C)
> Minimum boiling azeotrope: positive deviation (Ethanol + Water boiling at 78.1°C)
6. Colligative Properties ⭐ (Most Important)
Colligative Properties are properties that depend only on the NUMBER of solute particles in a solution, and NOT on the nature of the solute.
The four colligative properties are:
1. Relative Lowering of Vapour Pressure (RLVP)
2. Elevation of Boiling Point (Tb)
3. Depression of Freezing Point (Tf)
4. Osmotic Pressure (pi)
The four colligative properties are:
1. Relative Lowering of Vapour Pressure (RLVP)
2. Elevation of Boiling Point (Tb)
3. Depression of Freezing Point (Tf)
4. Osmotic Pressure (pi)
6.1 Relative Lowering of Vapour Pressure (RLVP) ⭐
p* - p = Delta p = vapour pressure lowering
RLVP = (p* - p) / p* = x_solute = n_B / (n_A + n_B)
For dilute solutions (n_B very small):
RLVP = (p* - p) / p* = n_B / n_A = (w_B x M_A) / (M_B x w_A)
where:
p* = vapour pressure of pure solvent
p = vapour pressure of solution
n_A = moles of solvent, n_B = moles of solute
w = mass, M = molar mass
Example: The vapour pressure of water at 25°C is 23.8 mm Hg. Find vapour pressure of solution when 36g of glucose (M = 180) is dissolved in 90g of water (M = 18).
n_solute (glucose) = 36/180 = 0.2 mol
n_solvent (water) = 90/18 = 5 mol
x_solute = 0.2/(0.2+5) = 0.2/5.2 = 0.0385
(p* - p)/p* = 0.0385
p* - p = 0.0385 x 23.8 = 0.916 mm Hg
p = 23.8 - 0.916 = 22.88 mm Hg
n_solute (glucose) = 36/180 = 0.2 mol
n_solvent (water) = 90/18 = 5 mol
x_solute = 0.2/(0.2+5) = 0.2/5.2 = 0.0385
(p* - p)/p* = 0.0385
p* - p = 0.0385 x 23.8 = 0.916 mm Hg
p = 23.8 - 0.916 = 22.88 mm Hg
6.2 Elevation of Boiling Point ⭐
Delta Tb = Tb - Tb* = Kb x m
where:
Delta Tb = elevation in boiling point
Kb = ebullioscopic constant (molal elevation constant)
m = molality of solution
Kb = (R x Tb*^2 x M_A) / (1000 x Delta H_vap)
Molar mass of solute:
M_B = (Kb x w_B x 1000) / (Delta Tb x w_A)
Example: 1.8g of glucose (M=180) is dissolved in 100g of water. Kb for water = 0.52 K kg/mol. Find elevation in boiling point.
m = (1.8/180) / (100/1000) = 0.01/0.1 = 0.1 mol/kg
Delta Tb = Kb x m = 0.52 x 0.1 = 0.052 K
m = (1.8/180) / (100/1000) = 0.01/0.1 = 0.1 mol/kg
Delta Tb = Kb x m = 0.52 x 0.1 = 0.052 K
6.3 Depression of Freezing Point ⭐
Delta Tf = Tf* - Tf = Kf x m
where:
Delta Tf = depression in freezing point
Kf = cryoscopic constant (molal depression constant)
m = molality of solution
Kf = (R x Tf*^2 x M_A) / (1000 x Delta H_fus)
Molar mass of solute:
M_B = (Kf x w_B x 1000) / (Delta Tf x w_A)
⭐ Kf and Kb values to remember:
For Water: Kf = 1.86 K kg/mol, Kb = 0.52 K kg/mol, Tb* = 100°C, Tf* = 0°C
For Benzene: Kf = 5.12 K kg/mol, Kb = 2.53 K kg/mol
Applications of Depression of Freezing Point:
> Antifreeze in car radiators (ethylene glycol added to water)
> Salting icy roads in cold countries
> Sea water freezes at lower temperature than pure water
For Water: Kf = 1.86 K kg/mol, Kb = 0.52 K kg/mol, Tb* = 100°C, Tf* = 0°C
For Benzene: Kf = 5.12 K kg/mol, Kb = 2.53 K kg/mol
Applications of Depression of Freezing Point:
> Antifreeze in car radiators (ethylene glycol added to water)
> Salting icy roads in cold countries
> Sea water freezes at lower temperature than pure water
Example: 2g of benzoic acid (M=122) in 25g of benzene shows a depression of 1.62 K. Find Kf of benzene.
m = (2/122) / (25/1000) = 0.01639/0.025 = 0.656 mol/kg
Kf = Delta Tf / m = 1.62 / 0.656 = 2.47 K kg/mol
m = (2/122) / (25/1000) = 0.01639/0.025 = 0.656 mol/kg
Kf = Delta Tf / m = 1.62 / 0.656 = 2.47 K kg/mol
6.4 Osmotic Pressure ⭐
Osmosis: The flow of solvent molecules from a region of lower solute concentration to a region of higher solute concentration through a semipermeable membrane is called osmosis.
Osmotic Pressure (pi): The excess pressure applied on the solution side to stop osmosis is called osmotic pressure.
Osmotic Pressure (pi): The excess pressure applied on the solution side to stop osmosis is called osmotic pressure.
pi = CRT = (n/V)RT
where:
pi = osmotic pressure (atm or Pa)
C = molar concentration (mol/L)
R = gas constant (0.0821 L atm/mol/K)
T = temperature in Kelvin
Molar mass:
M_B = (w_B x R x T) / (pi x V)
⭐ Types of Solutions based on Osmotic Pressure:
> Isotonic: Same osmotic pressure (0.9% NaCl = blood plasma)
> Hypertonic: Higher osmotic pressure than reference
> Hypotonic: Lower osmotic pressure than reference
Reverse Osmosis: When pressure greater than osmotic pressure is applied on solution side, solvent flows from solution to pure solvent. Used in water purification (RO purifiers).
> Isotonic: Same osmotic pressure (0.9% NaCl = blood plasma)
> Hypertonic: Higher osmotic pressure than reference
> Hypotonic: Lower osmotic pressure than reference
Reverse Osmosis: When pressure greater than osmotic pressure is applied on solution side, solvent flows from solution to pure solvent. Used in water purification (RO purifiers).
Example: 200 mL of a solution contains 1.26g of a polymer. Osmotic pressure at 300K = 2.57 x 10^(-3) atm. Find molar mass of polymer.
M = wRT/(pi x V) = (1.26 x 0.0821 x 300) / (2.57 x 10^(-3) x 0.2)
= 31.03 / (5.14 x 10^(-4))
= 60,370 g/mol (about 60,000 g/mol)
M = wRT/(pi x V) = (1.26 x 0.0821 x 300) / (2.57 x 10^(-3) x 0.2)
= 31.03 / (5.14 x 10^(-4))
= 60,370 g/mol (about 60,000 g/mol)
7. Abnormal Molar Mass and van't Hoff Factor ⭐
When solutes dissociate (like NaCl → Na+ + Cl-) or associate (like acetic acid dimerisation), the actual number of particles differs from expected. This causes abnormal colligative properties.
van't Hoff Factor (i):
i = (Observed colligative property) / (Theoretical colligative property)
i = (Normal molar mass) / (Observed molar mass)
i = (Actual number of particles after dissociation/association) / (Number of formula units dissolved)
van't Hoff Factor (i):
i = (Observed colligative property) / (Theoretical colligative property)
i = (Normal molar mass) / (Observed molar mass)
i = (Actual number of particles after dissociation/association) / (Number of formula units dissolved)
Modified colligative property formulas:
RLVP: (p* - p)/p* = i x x_B
Delta Tb = i x Kb x m
Delta Tf = i x Kf x m
pi = i x CRT
| Solute | Dissociation/Association | Value of i |
|---|---|---|
| NaCl | Na+ + Cl- (2 ions) | i = 2 |
| KCl | K+ + Cl- (2 ions) | i = 2 |
| MgCl2 | Mg2+ + 2Cl- (3 ions) | i = 3 |
| AlCl3 | Al3+ + 3Cl- (4 ions) | i = 4 |
| K2SO4 | 2K+ + SO4^2- (3 ions) | i = 3 |
| Glucose, Urea | No dissociation | i = 1 |
| Acetic acid in benzene | Dimerisation (2 molecules → 1) | i = 0.5 |
⭐ Degree of Dissociation (alpha) and van't Hoff Factor:
For electrolyte AxBy → x A^y+ + y B^x- (n ions total)
i = 1 + (n-1) x alpha
alpha = (i-1)/(n-1)
For association: n molecules → 1 associated unit
i = 1 - alpha(1 - 1/n) = 1 - alpha + alpha/n
For electrolyte AxBy → x A^y+ + y B^x- (n ions total)
i = 1 + (n-1) x alpha
alpha = (i-1)/(n-1)
For association: n molecules → 1 associated unit
i = 1 - alpha(1 - 1/n) = 1 - alpha + alpha/n
Example: Find the freezing point of a solution of 0.1 mol of NaCl in 1 kg of water. Kf = 1.86 K kg/mol. (Assume complete dissociation)
NaCl → Na+ + Cl- → i = 2
m = 0.1 mol/kg
Delta Tf = i x Kf x m = 2 x 1.86 x 0.1 = 0.372 K
Freezing point = 0 - 0.372 = -0.372°C
NaCl → Na+ + Cl- → i = 2
m = 0.1 mol/kg
Delta Tf = i x Kf x m = 2 x 1.86 x 0.1 = 0.372 K
Freezing point = 0 - 0.372 = -0.372°C
8. Important Board Exam Questions
Q1. State and explain Henry's Law. Mention two applications.
Henry's Law states that at constant temperature, the partial pressure of a gas (p) above a solution is directly proportional to the mole fraction (x) of the gas in solution: p = KH x x.
Applications:
(i) Carbonated beverages: CO2 is dissolved under high pressure. When bottle is opened, pressure decreases and CO2 escapes as bubbles.
(ii) Scuba diving: At deep sea, high pressure causes more N2 to dissolve in blood. On rapid ascent (decompression), N2 escapes as bubbles in blood causing "bends" — a painful and dangerous condition. To avoid this, air is replaced with helium-oxygen mixture.
Applications:
(i) Carbonated beverages: CO2 is dissolved under high pressure. When bottle is opened, pressure decreases and CO2 escapes as bubbles.
(ii) Scuba diving: At deep sea, high pressure causes more N2 to dissolve in blood. On rapid ascent (decompression), N2 escapes as bubbles in blood causing "bends" — a painful and dangerous condition. To avoid this, air is replaced with helium-oxygen mixture.
Q2. What is meant by positive and negative deviation from Raoult's law? Give one example of each.
Positive Deviation: The observed vapour pressure is MORE than predicted by Raoult's law. Occurs when A-B molecular interactions are weaker than A-A and B-B interactions. Delta H mixing is positive (endothermic), Delta V mixing is positive.
Example: Acetone + Ethanol
Negative Deviation: The observed vapour pressure is LESS than predicted by Raoult's law. Occurs when A-B interactions are stronger than A-A and B-B interactions. Delta H mixing is negative (exothermic), Delta V mixing is negative.
Example: Acetone + Chloroform (due to H-bonding between them)
Example: Acetone + Ethanol
Negative Deviation: The observed vapour pressure is LESS than predicted by Raoult's law. Occurs when A-B interactions are stronger than A-A and B-B interactions. Delta H mixing is negative (exothermic), Delta V mixing is negative.
Example: Acetone + Chloroform (due to H-bonding between them)
Q3. 1.5g of an unknown compound is dissolved in 100g of water. The freezing point of the solution is -0.186°C. Find the molar mass of the compound. (Kf of water = 1.86 K kg/mol)
Delta Tf = 0 - (-0.186) = 0.186 K
m = Delta Tf / Kf = 0.186/1.86 = 0.1 mol/kg
Moles of solute = m x mass of solvent in kg = 0.1 x 0.1 = 0.01 mol
Molar mass = mass/moles = 1.5/0.01 = 150 g/mol
m = Delta Tf / Kf = 0.186/1.86 = 0.1 mol/kg
Moles of solute = m x mass of solvent in kg = 0.1 x 0.1 = 0.01 mol
Molar mass = mass/moles = 1.5/0.01 = 150 g/mol
Q4. Define osmotic pressure. How is it used to determine the molar mass of macromolecules?
Osmotic pressure is the excess pressure that must be applied on the solution side to stop the flow of solvent through a semipermeable membrane (to stop osmosis).
For macromolecules (polymers, proteins): Even small amounts produce measurable osmotic pressure. Using pi = CRT = (w/MV)RT, we can find M = wRT/(pi x V). Osmotic pressure is preferred over other colligative properties for large molecules because even at very low concentrations, osmotic pressure is large enough to measure accurately, while other colligative property changes are too small to measure.
For macromolecules (polymers, proteins): Even small amounts produce measurable osmotic pressure. Using pi = CRT = (w/MV)RT, we can find M = wRT/(pi x V). Osmotic pressure is preferred over other colligative properties for large molecules because even at very low concentrations, osmotic pressure is large enough to measure accurately, while other colligative property changes are too small to measure.
Q5. A 0.6% (w/v) solution of urea (M = 60) is isotonic with a 1.8% (w/v) solution of glucose (M = 180). Explain why.
For isotonic solutions, osmotic pressure must be equal: pi1 = pi2
C1 = (0.6g/100mL) x (1000/60) = 0.1 mol/L for urea
C2 = (1.8g/100mL) x (1000/180) = 0.1 mol/L for glucose
Since C1 = C2 and both are non-electrolytes (i=1), their osmotic pressures are equal: pi = iCRT. Both solutions are isotonic. ✓
C1 = (0.6g/100mL) x (1000/60) = 0.1 mol/L for urea
C2 = (1.8g/100mL) x (1000/180) = 0.1 mol/L for glucose
Since C1 = C2 and both are non-electrolytes (i=1), their osmotic pressures are equal: pi = iCRT. Both solutions are isotonic. ✓
Q6. The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order: Acetic acid less than Trichloroacetic acid less than Trifluoroacetic acid. Explain.
Depression in freezing point is a colligative property that depends on the number of solute particles. Greater dissociation → more particles → greater depression.
Fluorine is the most electronegative element, so CF3COOH (trifluoroacetic acid) is the strongest acid and dissociates the most → maximum i → maximum Delta Tf.
CCl3COOH (trichloroacetic acid) dissociates more than CH3COOH (acetic acid) due to Cl being more electronegative than H.
Hence: Acetic acid less than Trichloroacetic acid less than Trifluoroacetic acid.
Fluorine is the most electronegative element, so CF3COOH (trifluoroacetic acid) is the strongest acid and dissociates the most → maximum i → maximum Delta Tf.
CCl3COOH (trichloroacetic acid) dissociates more than CH3COOH (acetic acid) due to Cl being more electronegative than H.
Hence: Acetic acid less than Trichloroacetic acid less than Trifluoroacetic acid.
9. Key Formulas at a Glance
CONCENTRATION TERMS:
Molarity M = moles of solute / volume of solution (L)
Molality m = moles of solute / mass of solvent (kg)
Mole fraction = n_A / (n_A + n_B)
HENRY'S LAW: p = KH x x
RAOULT'S LAW: p = xA.pA* + xB.pB*
RLVP = (p*-p)/p* = x_solute = n_B/(n_A+n_B)
COLLIGATIVE PROPERTIES:
RLVP: (p*-p)/p* = i.x_B
Boiling: Delta Tb = i.Kb.m
Freezing: Delta Tf = i.Kf.m
Osmotic: pi = i.C.R.T
MOLAR MASS FROM:
Boiling: M_B = (1000 x Kb x w_B) / (Delta Tb x w_A)
Freezing: M_B = (1000 x Kf x w_B) / (Delta Tf x w_A)
Osmotic: M_B = (w_B x R x T) / (pi x V)
VAN'T HOFF FACTOR:
i = 1 + (n-1).alpha [for dissociation]
alpha = (i-1)/(n-1)
WATER CONSTANTS:
Kf = 1.86 K kg/mol, Kb = 0.52 K kg/mol
10. MCQ Practice (1 Mark)
1. Which of the following concentration terms is independent of temperature?
(a) Molarity (b) Normality (c) Molality (d) Volume percentage
Answer: (c) Molality — uses mass of solvent, not volume
2. Henry's law constant KH for CO2 is more than that for He at 298K. This means:
(a) CO2 is more soluble than He (b) He is more soluble than CO2 (c) Both are equally soluble (d) None
Answer: (b) He is more soluble — higher KH means lower solubility
3. Which colligative property is most suitable for determination of molar mass of macromolecules?
(a) Elevation of boiling point (b) Depression of freezing point (c) Osmotic pressure (d) RLVP
Answer: (c) Osmotic pressure — gives measurable values even at very low concentrations
4. The van't Hoff factor for MgCl2 assuming complete dissociation is:
(a) 1 (b) 2 (c) 3 (d) 4
Answer: (c) 3 — MgCl2 → Mg2+ + 2Cl- = 3 ions
5. Azeotropic mixture of HCl and water boils at 108.6°C. This is an example of:
(a) Positive deviation (b) Negative deviation (c) Ideal solution (d) No deviation
Answer: (b) Negative deviation — maximum boiling azeotrope
6. The osmotic pressure of 0.1 M NaCl solution at 27°C is approximately:
(a) 2.46 atm (b) 4.92 atm (c) 1.23 atm (d) 0.82 atm
Answer: (b) 4.92 atm — pi = iCRT = 2 x 0.1 x 0.0821 x 300 = 4.92 atm
11. Exam Tips
- Always convert units before calculating: mass in grams, volume in litres or mL as required, temperature in Kelvin for osmotic pressure.
- For colligative property numericals: first check if the solute is electrolyte or non-electrolyte — if electrolyte, multiply by van't Hoff factor i.
- Molality is preferred for colligative property calculations (not molarity) because it doesn't change with temperature.
- Remember: higher KH = LOWER solubility of gas (not higher) — this is a common confusion in MCQs.
- Positive deviation: vapour pressure increases (A-B forces weaker). Negative deviation: vapour pressure decreases (A-B forces stronger).
- Molar mass from freezing point depression formula: M_B = (1000 x Kf x w_B) / (Delta Tf x w_A) — memorise this, it appears in almost every board exam.
- Osmotic pressure is used for polymers and proteins because even tiny concentrations give measurable osmotic pressure values.
- Isotonic solutions: same osmotic pressure = same molar concentration (for non-electrolytes).
✅ Summary: Solutions is a numerical-heavy chapter. Master the four colligative properties and their formulas, especially the molar mass determination formulas. Always remember to apply van't Hoff factor for electrolytes. Henry's law, Raoult's law, ideal vs non-ideal solutions, and azeotropes are the theory portion frequently tested. Practice at least 5 numericals on each colligative property before the exam.
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