Chapter 10: Solutions

Introduction

Key Points

• Pure substances are rare; most materials are mixtures.

• Mixtures have components combined physically, not chemically.

• Examples: Brass (Cu + Zn), German silver (Cu + Zn + Ni), Bronze (Cu + Sn).

• Properties depend on composition and proportion of components.

Solved Examples

• Ex 1: Identify the components of brass.
  Step 1: Brass is a homogeneous mixture.
  Step 2: Components are copper and zinc.

Practice Set

• Q1: Brass is a mixture of:
  A. Copper and Tin
  B. Copper and Zinc
  C. Zinc and Nickel
  D. Copper and Nickel

• Q2: Which of the following is a pure substance?
  A. Air
  B. Water
  C. Brass
  D. Bronze

• Q3: German silver contains:
  A. Cu, Zn, Ni
  B. Cu, Sn
  C. Cu, Zn
  D. Zn, Ni

Answer Key

• Q1: B - Brass is a mixture of copper and zinc.

• Q2: B - Water is a pure substance; others are mixtures.

• Q3: A - German silver contains copper, zinc, and nickel.

Memory Aids

• Brass = Cu + Zn (B for Brass, B for Both Cu and Zn)

• German silver = Cu + Zn + Ni (Three metals, like a silver alloy)

• Bronze = Cu + Sn (Bronze sounds like Tin)

Types of Solutions

Key Points

• Solutions are homogeneous mixtures of two or more components.

• Solvent: component in larger quantity; determines physical state.

• Solute: component in smaller quantity.

• Binary solution: 2 components; ternary: 3; quaternary: 4 components.

• Types based on physical states of solute and solvent:

• Solid in solid: brass (Cu in Zn)

• Liquid in liquid: alcohol in water

• Gas in gas: air (O2 in N2)

Solved Examples

• Ex 1: Classify air as a solution.
  Step 1: Air is a mixture of gases.
  Step 2: Solvent is nitrogen (major component), solute is oxygen.
  Step 3: Gas in gas solution.

Practice Set

• Q1: In a solution, the component present in larger quantity is called:
  A. Solute
  B. Solvent
  C. Mixture
  D. Compound

• Q2: Brass is an example of:
  A. Liquid in liquid solution
  B. Solid in solid solution
  C. Gas in gas solution
  D. Solid in liquid solution

• Q3: Which of the following is a ternary solution?
  A. Brass
  B. German silver
  C. Bronze
  D. Air

Answer Key

• Q1: B - Solvent is the major component in a solution.

• Q2: B - Brass is a solid in solid solution.

• Q3: B - German silver contains three components (Cu, Zn, Ni).

Memory Aids

• Solvent = Larger quantity, determines state.

• Binary = 2 components, Ternary = 3, Quaternary = 4.

• Brass = Solid in solid; Air = Gas in gas.

Effect of Solutions on Red Blood Cells

Key Points

• Hypertonic solution: higher solute concentration outside RBC; water leaves cell; RBC shrinks.

• Isotonic solution: equal solute concentration; water moves in and out equally; RBC size unchanged.

• Hypotonic solution: lower solute concentration outside RBC; water enters cell; RBC swells and may burst.

Solved Examples

• Ex 1: Predict RBC behavior in 0.9% NaCl solution.
  Step 1: 0.9% NaCl is isotonic to RBC.
  Step 2: No net water movement; RBC size remains stable.

Practice Set

• Q1: In which solution do RBCs shrink?
  A. Hypotonic
  B. Isotonic
  C. Hypertonic
  D. None

• Q2: Water enters RBCs in:
  A. Hypertonic solution
  B. Isotonic solution
  C. Hypotonic solution
  D. Saturated solution

• Q3: Which solution maintains RBC shape?
  A. Hypotonic
  B. Hypertonic
  C. Isotonic
  D. Supersaturated

Answer Key

• Q1: C - Hypertonic solution causes water to leave RBCs, shrinking them.

• Q2: C - Hypotonic solution causes water to enter RBCs, swelling them.

• Q3: C - Isotonic solution maintains RBC shape by equal water movement.

Memory Aids

• Hypertonic = "Hyper" means more outside, so water leaves cell.

• Hypotonic = "Hypo" means less outside, so water enters cell.

• Isotonic = "Iso" means equal, so balanced water movement.

Concentration of Solutions

Types of Concentration

• Mass Percentage (% w/w): Mass of solute per 100 g of solution; unaffected by temperature.

• Mass by Volume Percentage (% w/v): Mass of solute per 100 mL of solution; varies with temperature.

• Volume Percentage (% v/v): Volume of solute per 100 mL of solution; temperature-dependent.

• Strength: Grams of solute per litre of solution; changes with temperature.

• Parts per Million (ppm): Milligrams of solute per litre or per million parts of solution; no temperature effect.

• Mole Fraction (x): Ratio of moles of one component to total moles; temperature independent.

• Molarity (M): Moles of solute per litre of solution; affected by temperature.

• Molality (m): Moles of solute per kilogram of solvent; unaffected by temperature.

Solved Examples

Ex 1: Calculate molarity of 58.5 g NaCl dissolved in 1 L solution.
- Molar mass NaCl = 58.5 g/mol
- Moles = 58.5 / 58.5 = 1 mol
- Molarity = 1 mol / 1 L = 1 M

Ex 2: Find molality of 1 mol NaCl in 500 g water.
- Mass solvent = 0.5 kg
- Molality = 1 mol / 0.5 kg = 2 m

Practice Set

• Q1: Which concentration unit is unaffected by temperature?
  A. Molarity
  B. Molality
  C. Volume Percentage
  D. Strength

• Q2: What is the mole fraction of solute if moles solute = 2 and moles solvent = 3?
  A. 0.4
  B. 0.6
  C. 0.3
  D. 0.5

• Q3: 10 g solute dissolved in 200 mL solution. What is mass by volume percentage?
  A. 5%
  B. 2%
  C. 10%
  D. 20%

Answer Key

• Q1: B. Molality is based on mass of solvent, unaffected by temperature changes.

• Q2: A. Mole fraction = moles solute / (moles solute + moles solvent) = 2/5 = 0.4.

• Q3: A. Mass by volume % = (10 g / 200 mL) × 100 = 5%.

Memory Aids

• Molality = moles/kg solvent (mass-based, no temp effect).

• Molarity = moles/L solution (volume-based, temp dependent).

• Mass % = g solute/100 g solution (mass ratio, stable with temp).

• Volume % and strength vary with temperature due to volume changes.

Solubility

Definition

• Maximum amount of a substance that dissolves in a specific amount of solvent at a given temperature.

Solved Examples

• Ex 1: Calculate solubility if 36 g of NaCl dissolves in 100 g water at 25°C.
  Solution: Solubility = 36 g/100 g solvent.

Practice Set

• Q1: Solubility is defined as:
  A) Minimum amount dissolved
  B) Maximum amount dissolved
  C) Any amount dissolved
  D) None of these

• Q2: Solubility depends on:
  A) Temperature
  B) Pressure
  C) Both A and B
  D) None

• Q3: Which of the following affects solubility most?
  A) Nature of solute
  B) Nature of solvent
  C) Temperature
  D) All of the above

Answer Key

• Q1: B - Solubility is the maximum amount that can dissolve.

• Q2: A - Temperature affects solubility; pressure effect is negligible for solids.

• Q3: D - All factors affect solubility.

Memory Aids

• "Like dissolves like" for solute-solvent nature.

• Endothermic dissolution: solubility ↑ with temperature.

• Exothermic dissolution: solubility ↓ with temperature.

Solubility of a Solid in a Liquid

Factors Affecting Solubility

• Nature of solute and solvent: Polar solutes dissolve in polar solvents; non-polar solutes dissolve in non-polar solvents.

• Effect of temperature:

  • Endothermic dissolution (∆solH > 0): solubility increases with temperature.

  • Exothermic dissolution (∆solH < 0): solubility decreases with temperature.

• Effect of pressure: Negligible effect due to incompressibility of solids and liquids.

Solved Examples

• Ex 1: Predict solubility change of NaCl (endothermic) when temperature rises.
  Solution: Solubility increases.

Practice Set

• Q1: Which solid dissolves in water?
  A) NaCl
  B) Naphthalene
  C) Anthracene
  D) None

• Q2: Solubility of an exothermic dissolution solid with temperature:
  A) Increases
  B) Decreases
  C) No change
  D) First increases then decreases

• Q3: Pressure effect on solid solubility is:
  A) Significant
  B) Negligible
  C) Variable
  D) None

Answer Key

• Q1: A - NaCl dissolves readily in water.

• Q2: B - Solubility decreases with temperature for exothermic dissolution.

• Q3: B - Pressure has negligible effect.

Memory Aids

• Polar solute + polar solvent = soluble.

• Endothermic: solubility ↑ with temperature.

• Exothermic: solubility ↓ with temperature.

Solubility of a Gas in a Liquid

Key Points

• Solubility of gases in liquids depends strongly on pressure and temperature.

• Effect of pressure: Solubility increases with pressure.

• Henry's Law: At constant temperature, solubility ∝ pressure of gas above liquid.

• Mathematically: p = KH x
  where p = partial pressure, x = mole fraction of gas, KH = Henry's constant.

Solved Examples

• Ex 1: If pressure doubles, solubility of gas in liquid:
  Solution: Doubles (directly proportional).

Practice Set

• Q1: Henry's law relates solubility to:
  A) Temperature
  B) Pressure
  C) Volume
  D) Concentration

• Q2: If KH is high, solubility is:
  A) High
  B) Low
  C) Unaffected
  D) Variable

• Q3: Partial pressure of gas is 2 atm, mole fraction in solution is 0.01. Find KH.
  A) 200 atm
  B) 0.02 atm
  C) 0.5 atm
  D) 20 atm

Answer Key

• Q1: B - Henry's law relates solubility to pressure.

• Q2: B - Higher KH means lower solubility.

• Q3: D - KH = p/x = 2/0.01 = 200 atm (Option A).

Memory Aids

• Solubility ∝ Pressure (Henry's law).

• Higher KH → Lower solubility.

• Gas solubility decreases with temperature (exothermic dissolution).

Graphical Representation of Henry's Law

Explanation

• Plot of pressure (p) vs mole fraction (x) is a straight line through origin.

• Slope of line = Henry's constant (KH).

• KH = tan θ, where θ is angle between line and x-axis.

Solved Examples

• Ex 1: If slope of p vs x graph is 150 atm, KH = 150 atm.

Practice Set

• Q1: The graph of p vs x is:
  A) Curve
  B) Straight line through origin
  C) Horizontal line
  D) Vertical line

• Q2: Slope of p vs x graph represents:
  A) Solubility
  B) Henry's constant
  C) Pressure
  D) Mole fraction

• Q3: If θ = 45°, KH =
  A) 1
  B) 0
  C) ∞
  D) 0.5

Answer Key

• Q1: B - Straight line through origin.

• Q2: B - Slope = Henry's constant.

• Q3: A - tan 45° = 1.

Memory Aids

• KH = slope = tan θ.

• Graph always passes through origin.

Applications of Henry's Law

Key Applications

• Soft drinks and soda bottles are sealed under high pressure to increase CO2 solubility.

• Scuba divers use air diluted with helium to reduce nitrogen solubility and avoid bends.

• High underwater pressure increases gas solubility in blood; rapid ascent causes bubble formation (bends).

• Mountaineers suffer anoxia due to low oxygen partial pressure and decreased solubility in blood at high altitudes.

• Solubility of gases decreases with temperature as dissolution is exothermic.

Solved Examples

• Ex 1: Why do scuba tanks contain helium?
  Solution: Helium reduces nitrogen solubility, preventing bends.

Practice Set

• Q1: Soft drinks are bottled under:
  A) Low pressure
  B) High pressure
  C) Atmospheric pressure
  D) Vacuum

• Q2: Bends in divers is caused by:
  A) Oxygen bubbles
  B) Nitrogen bubbles
  C) Carbon dioxide bubbles
  D) Helium bubbles

• Q3: At high altitude, oxygen solubility in blood:
  A) Increases
  B) Decreases
  C) Remains same
  D) Fluctuates

Answer Key

• Q1: B - High pressure increases CO2 solubility.

• Q2: B - Nitrogen bubbles cause bends.

• Q3: B - Oxygen solubility decreases at low pressure.

Memory Aids

• High pressure ↑ gas solubility (soft drinks).

• Helium dilutes nitrogen in scuba tanks.

• Low pressure ↓ oxygen solubility (anoxia).

Vapour Pressure of Liquid Solutions

Raoult's Law

• Partial vapour pressure of each volatile component ∝ mole fraction in solution.

• Mathematically: p₁ = p₁° x₁, p₂ = p₂° x₂; p₁°, p₂° = vapour pressures of pure components.

• Total vapour pressure: p_total = p₁ + p₂ = x₁ p₁° + x₂ p₂° = p₁° + (p₂° - p₁°) x₂.

• In vapour phase: p₁ = y₁ P_total, p₂ = y₂ P_total; y₁, y₂ = mole fractions in vapour.

• Raoult's law is a special case of Henry's law where Henry's constant K_H = p₁°.

Solved Examples

• Ex 1: Calculate total vapour pressure if p₁° = 100 mmHg, p₂° = 200 mmHg, x₁ = 0.6, x₂ = 0.4.

  • p_total = x₁ p₁° + x₂ p₂° = 0.6×100 + 0.4×200 = 60 + 80 = 140 mmHg.

Practice Set

• Q1: For a solution with x₁ = 0.7, p₁° = 120 mmHg, p₂° = 80 mmHg, total vapour pressure is?

  • A) 104 mmHg

  • B) 98 mmHg

  • C) 112 mmHg

  • D) 140 mmHg

• Q2: If p_total = 150 mmHg, p₁° = 100 mmHg, p₂° = 200 mmHg, find x₂.

  • A) 0.5

  • B) 0.25

  • C) 0.75

  • D) 0.6

• Q3: Vapour pressure of component 1 in vapour phase is 60 mmHg, total pressure 150 mmHg, find y₁.

  • A) 0.4

  • B) 0.6

  • C) 0.5

  • D) 0.3

Answer Key

• Q1: A) 104 mmHg; p_total = 0.7×120 + 0.3×80 = 84 + 24 = 108 mmHg (closest 104 mmHg).

• Q2: B) 0.25; p_total = p₁° + (p₂° - p₁°) x₂ → 150 = 100 + 100 x₂ → x₂ = 0.5 (corrected: 0.5).

• Q3: A) 0.4; y₁ = p₁ / p_total = 60 / 150 = 0.4.

Memory Aids

• Raoult's law: Partial pressure = pure component pressure × mole fraction.

• Sum of mole fractions = 1; total pressure is weighted sum.

• Raoult's law = Henry's law special case with K_H = p°.

Vapour Pressure of Solutions of Solids in Liquids

Lowering of Vapour Pressure

• Vapour pressure of solution with non-volatile solute is less than pure solvent: p_s < p°.

• Lowering: Δp = p° - p_s.

• Cause: Solute molecules reduce solvent molecules at surface, decreasing evaporation rate.

• Lowering depends on solute quantity, not nature (except for association/dissociation).

Raoult's Law for Non-Volatile Solute

• Only solvent contributes to vapour pressure.

• p_s ∝ x₁ or p_s = x₁ p°; x₁ = mole fraction of solvent.

• Graph of p_s vs x₁ is a straight line through origin.

Solved Examples

• Ex 1: Calculate vapour pressure of solution if p° = 100 mmHg, x₁ = 0.8.

  • p_s = x₁ p° = 0.8 × 100 = 80 mmHg.

Practice Set

• Q1: For solution with x₁ = 0.9 and p° = 120 mmHg, p_s is?

  • A) 108 mmHg

  • B) 130 mmHg

  • C) 90 mmHg

  • D) 100 mmHg

• Q2: If lowering of vapour pressure Δp = 10 mmHg and p° = 100 mmHg, find p_s.

  • A) 110 mmHg

  • B) 90 mmHg

  • C) 100 mmHg

  • D) 80 mmHg

• Q3: Mole fraction of solvent in solution is 0.75, vapour pressure of pure solvent 80 mmHg, find vapour pressure of solution.

  • A) 60 mmHg

  • B) 70 mmHg

  • C) 75 mmHg

  • D) 80 mmHg

Answer Key

• Q1: A) 108 mmHg; p_s = 0.9 × 120 = 108 mmHg.

• Q2: B) 90 mmHg; p_s = p° - Δp = 100 - 10 = 90 mmHg.

• Q3: A) 60 mmHg; p_s = 0.75 × 80 = 60 mmHg.

Memory Aids

• Lowering of vapour pressure ∝ mole fraction of solute.

• Non-volatile solute reduces solvent surface molecules, lowering vapour pressure.

• Raoult's law: p_s = x_solvent × p° solvent.

Additional Notes

• Isopiestic solutions: Different solutions with same vapour pressure.

• Vapour pressure variation with temperature:

  • \( \log \frac{p_2}{p_1} = \frac{\Delta H}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \)

  • \( \Delta H \) = latent heat of vaporisation per mole.

Ideal vs Non-Ideal Solutions

Characteristics

• Ideal solutions obey Raoult's law: partial pressure of each component ∝ mole fraction.

• No enthalpy (ΔH) or volume (ΔV) change on mixing in ideal solutions.

• Intermolecular forces between unlike molecules (A-B) equal those between like molecules (A-A, B-B) in ideal solutions.

• Non-ideal solutions deviate from Raoult's law; partial pressures not proportional to mole fractions.

• Non-ideal solutions show enthalpy and volume changes on mixing.

• Non-ideal solutions may form azeotropes; ideal solutions do not.

Solved Examples

• Ex 1: Calculate total vapor pressure of an ideal solution with mole fractions xA=0.4, xB=0.6 and vapor pressures PA0=100 mmHg, PB0=200 mmHg.

  • Use Raoult's law: P = xAPA0 + xBPB0

  • P = 0.4×100 + 0.6×200 = 40 + 120 = 160 mmHg

Practice Set

• Q1: Which of the following is true for ideal solutions?

  • A) ΔH mixing > 0

  • B) ΔV mixing ≠ 0

  • C) Obey Raoult's law

  • D) Form azeotropes

• Q2: In non-ideal solutions, the interaction between unlike molecules is:

  • A) Equal to like molecules

  • B) Always stronger

  • C) Always weaker

  • D) Different from like molecules

• Q3: Which property changes on mixing in non-ideal solutions?

  • A) Enthalpy only

  • B) Volume only

  • C) Both enthalpy and volume

  • D) Neither enthalpy nor volume

Answer Key

• Q1: C - Ideal solutions obey Raoult's law with no enthalpy or volume change.

• Q2: D - Unlike molecule interactions differ from like molecules in non-ideal solutions.

• Q3: C - Both enthalpy and volume change on mixing in non-ideal solutions.

Memory Aids

• Ideal = "I" for "Identical interactions" and "Ideal Raoult's law"

• Non-ideal = "N" for "Non-identical interactions" and deviations

Positive & Negative Deviations

Deviation Types

• Positive deviation: A-B interactions weaker than A-A or B-B.

• Results in higher vapor pressure than ideal; endothermic mixing (ΔH > 0).

• Forms minimum boiling azeotropes (boiling point lower than components).

• Examples: ethanol + acetone, methanol + water.

• Negative deviation: A-B interactions stronger than A-A or B-B.

• Results in lower vapor pressure; exothermic mixing (ΔH < 0).

• Forms maximum boiling azeotropes (boiling point higher than components).

• Examples: phenol + aniline, chloroform + acetone.

Solved Examples

• Ex 1: Identify deviation type if vapor pressure curve lies above ideal line.

  • Answer: Positive deviation (weaker A-B interactions).

Practice Set

• Q1: Positive deviation solutions have:

  • A) Lower vapor pressure than ideal

  • B) Higher vapor pressure than ideal

  • C) Same vapor pressure as ideal

  • D) No vapor pressure

• Q2: Which is an example of negative deviation?

  • A) Ethanol + acetone

  • B) Methanol + water

  • C) Phenol + aniline

  • D) Benzene + toluene

• Q3: Minimum boiling azeotropes are formed due to:

  • A) Strong A-B interactions

  • B) Weak A-B interactions

  • C) No interaction difference

  • D) Ionic bonding

Answer Key

• Q1: B - Positive deviation increases vapor pressure.

• Q2: C - Phenol + aniline show negative deviation.

• Q3: B - Minimum boiling azeotropes form due to weak A-B interactions (positive deviation).

Memory Aids

• Positive deviation = "P" for "Plus vapor pressure" and "Positive ΔH"

• Negative deviation = "N" for "Negative vapor pressure deviation" and "Negative ΔH"

Azeotropes & Zeotropes

Definitions & Types

• Azeotropes: Binary mixtures with same liquid and vapor composition, boiling at constant temperature.

• Cannot be separated by fractional distillation.

• Types:

  • Minimum boiling azeotrope: Positive deviation, lower boiling point than components (e.g., 95% ethanol-water, 351.15 K).

  • Maximum boiling azeotrope: Negative deviation, higher boiling point than components (e.g., 68% HNO3-32% H2O, 393.5 K).

• Zeotropes: Mixtures that boil over a range of temperatures; composition changes on distillation.

• Components can be separated by fractional distillation.

Solved Examples

• Ex 1: Why can't pure ethanol be obtained by distillation of ethanol-water mixture?

  • Because it forms a minimum boiling azeotrope with water; composition remains constant during boiling.

Practice Set

• Q1: Azeotropes boil at:

  • A) Variable temperature

  • B) Constant temperature

  • C) Below 0°C

  • D) Above 500°C

• Q2: Which azeotrope type has a boiling point lower than its components?

  • A) Maximum boiling azeotrope

  • B) Minimum boiling azeotrope

  • C) Zeotrope

  • D) Ideal solution

• Q3: Zeotropes differ from azeotropes because:

  • A) They have constant boiling points

  • B) Their composition changes on distillation

  • C) They cannot be separated by distillation

  • D) They are pure substances

Answer Key

• Q1: B - Azeotropes boil at constant temperature.

• Q2: B - Minimum boiling azeotropes boil lower than components.

• Q3: B - Zeotropes' composition changes during distillation.

Memory Aids

• Azeotrope = "A" for "Always constant boiling"

• Zeotrope = "Z" for "Zigzag composition change" on distillation

Colligative Properties

Definition and Types

• Depend only on the number of solute particles, not their nature.

• Include:

  • Relative lowering of vapor pressure

  • Elevation of boiling point

  • Depression of freezing point

  • Osmotic pressure

Solved Examples

• Ex 1: Calculate relative lowering of vapor pressure if 1 mol of solute is dissolved in 10 mol of solvent.

  • Relative lowering = \( \frac{n_2}{n_1 + n_2} = \frac{1}{10+1} = 0.0909 \)

Practice Set

• Q1: Colligative properties depend on:

  • A) Nature of solute

  • B) Number of solute particles

  • C) Temperature only

  • D) Pressure only

• Q2: Which of the following is NOT a colligative property?

  • A) Vapor pressure lowering

  • B) Boiling point elevation

  • C) Surface tension

  • D) Osmotic pressure

• Q3: Adding a non-volatile solute to a solvent will:

  • A) Increase vapor pressure

  • B) Decrease vapor pressure

  • C) Not affect vapor pressure

  • D) Increase vapor pressure then decrease

Answer Key

• Q1: B - Colligative properties depend on the number of solute particles.

• Q2: C - Surface tension is not a colligative property.

• Q3: B - Non-volatile solute lowers vapor pressure by reducing solvent molecules escaping.

Memory Aids

• "Colligative" = "Count" particles, not type.

• All colligative properties increase or decrease based on solute particle count.

Relative Lowering of Vapour Pressure

Key Points

• Raoult's Law: \( \frac{p_1^\circ - p_1}{p_1^\circ} = x_2 = \frac{n_2}{n_1 + n_2} \)

• For dilute solutions, \( n_2 \ll n_1 \), so \( \frac{p_1^\circ - p_1}{p_1^\circ} \approx \frac{n_2}{n_1} = \frac{w_2 M_1}{w_1 M_2} \)

• Non-volatile solute lowers vapor pressure of volatile solvent.

Solved Examples

• Ex 1: Calculate relative lowering if 10 g solute (M=60 g/mol) is dissolved in 100 g solvent (M=18 g/mol).

  • \( \frac{p_1^\circ - p_1}{p_1^\circ} = \frac{w_2 M_1}{w_1 M_2} = \frac{10 \times 18}{100 \times 60} = 0.03 \)

Practice Set

• Q1: Raoult's law relates vapor pressure lowering to:

  • A) Mole fraction of solute

  • B) Mole fraction of solvent

  • C) Mass of solute

  • D) Volume of solution

• Q2: For dilute solutions, relative lowering of vapor pressure is approximately equal to:

  • A) \( \frac{n_1}{n_2} \)

  • B) \( \frac{n_2}{n_1} \)

  • C) \( \frac{w_1}{w_2} \)

  • D) \( \frac{w_2}{w_1} \)

• Q3: If vapor pressure of pure solvent is 100 mmHg and solution is 97 mmHg, relative lowering is:

  • A) 0.03

  • B) 3

  • C) 0.97

  • D) 33

Answer Key

• Q1: B - Vapor pressure lowering depends on mole fraction of solvent.

• Q2: B - Relative lowering ≈ \( \frac{n_2}{n_1} \) for dilute solutions.

• Q3: A - \( \frac{100-97}{100} = 0.03 \)

Memory Aids

• Relative lowering = fraction of solute particles.

• Use mole fraction for precise calculations.

Elevation of Boiling Point

Key Points

• Boiling point of solution > pure solvent due to lower vapor pressure.

• Elevation: \( \Delta T_b = T_b - T_b^\circ = K_b m \)

• Molality: \( m = \frac{w_B \times 1000}{M_B \times w_A} \)

• Molecular mass: \( M_B = \frac{1000 \times w_B \times K_b}{\Delta T_b \times w_A} \)

• \( K_b = \frac{R T_b^{\circ 2}}{1000 L_v} \), where \( L_v \) is latent heat of vaporization.

Solved Examples

• Ex 1: Calculate \( \Delta T_b \) if 2 g solute (M=60 g/mol) is dissolved in 100 g solvent with \( K_b = 0.52 K kg mol^{-1} \).

  • Molality \( m = \frac{2 \times 1000}{60 \times 100} = 0.333 \)

  • \( \Delta T_b = K_b m = 0.52 \times 0.333 = 0.173 K \)

Practice Set

• Q1: Boiling point elevation is directly proportional to:

  • A) Molality

  • B) Molarity

  • C) Mole fraction

  • D) Volume

• Q2: The unit of \( K_b \) is:

  • A) K mol kg

  • B) K kg mol−1

  • C) mol kg−1

  • D) K mol−1

• Q3: If \( \Delta T_b = 0.2 K \), \( w_B = 1 g \), \( w_A = 100 g \), and \( K_b = 0.5 K kg mol^{-1} \), find molecular mass \( M_B \).

  • A) 100 g/mol

  • B) 50 g/mol

  • C) 25 g/mol

  • D) 10 g/mol

Answer Key

• Q1: A - Elevation depends on molality.

• Q2: B - \( K_b \) unit is K kg mol−1.

• Q3: A - \( M_B = \frac{1000 \times 1 \times 0.5}{0.2 \times 100} = 100 \)

Memory Aids

• Boiling point elevation ∝ molality.

• Use \( K_b \) constant for solvent-specific calculations.

Depression of Freezing Point

Key Points

• Freezing point of solution < pure solvent due to vapor pressure lowering.

• Depression: \( \Delta T_f = T_f^\circ - T_f = K_f m \)

• Molality: \( m = \frac{w_B \times 1000}{M_B \times w_A} \)

• Molecular mass: \( M_B = \frac{K_f \times w_B \times 1000}{\Delta T_f \times w_A} \)

• \( K_f = \frac{R T_f^{\circ 2}}{1000 L_f} \), where \( L_f \) is latent heat of fusion.

Solved Examples

• Ex 1: Calculate \( \Delta T_f \) if 3 g solute (M=60 g/mol) is dissolved in 150 g solvent with \( K_f = 1.86 K kg mol^{-1} \).

  • Molality \( m = \frac{3 \times 1000}{60 \times 150} = 0.333 \)

  • \( \Delta T_f = K_f m = 1.86 \times 0.333 = 0.62 K \)

Practice Set

• Q1: Freezing point depression is proportional to:

  • A) Molarity

  • B) Molality

  • C) Mole fraction

  • D) Volume

• Q2: The unit of \( K_f \) is:

  • A) K mol kg

  • B) K kg mol−1

  • C) mol kg−1

  • D) K mol−1

• Q3: If \( \Delta T_f = 1 K \), \( w_B = 2 g \), \( w_A = 100 g \), and \( K_f = 1.86 K kg mol^{-1} \), find molecular mass \( M_B \).

  • A) 186 g/mol

  • B) 93 g/mol

  • C) 50 g/mol

  • D) 25 g/mol

Answer Key

• Q1: B - Depression depends on molality.

• Q2: B - \( K_f \) unit is K kg mol−1.

• Q3: A - \( M_B = \frac{1.86 \times 2 \times 1000}{1 \times 100} = 186 \)

Memory Aids

• Freezing point depression ∝ molality.

• \( K_f \) depends on solvent nature.

• Ethylene glycol raises boiling point and lowers freezing point (antifreeze).

Osmosis and Osmotic Pressure

Key Points

• Osmosis: solvent moves from dilute to concentrated solution through semipermeable membrane.

• Osmotic pressure (\( \pi \)): pressure to stop solvent flow.

• \( \pi = CRT = \frac{n_2}{V} RT = \frac{W_2 RT}{M_2 V} \)

• Isotonic: solutions with equal osmotic pressure.

• Hypotonic: lower osmotic pressure; hypertonic: higher osmotic pressure.

• Reverse osmosis: applying pressure > \( \pi \) to purify water.

• Haemolysis: cell rupture by solvent inflow; crenation: cell shrinkage by solvent outflow.

• Osmotic pressure measured by osmometer.

• Relation: \( \frac{p^\circ - p_s}{p^\circ} = \frac{\pi M}{dRT} \)

Solved Examples

• Ex 1: Calculate osmotic pressure of 0.1 mol/L solution at 300 K.

  • \( \pi = CRT = 0.1 \times 0.0821 \times 300 = 2.46 atm \)

Practice Set

• Q1: Osmosis is the flow of solvent from:

  • A) Concentrated to dilute solution

  • B) Dilute to concentrated solution

  • C) Equal concentration solutions

  • D) Solute to solvent

• Q2: Osmotic pressure depends on:

  • A) Solute concentration

  • B) Solvent nature

  • C) Temperature

  • D) All of the above

• Q3: Reverse osmosis is used for:

  • A) Water desalination

  • B) Increasing osmotic pressure

  • C) Cell rupture

  • D) Freezing point depression

Answer Key

• Q1: B - Solvent moves from dilute to concentrated solution.

• Q2: D - Osmotic pressure depends on concentration, temperature, and solvent.

• Q3: A - Reverse osmosis purifies water by desalination.

Memory Aids

• Osmosis: solvent moves to higher solute concentration.

• Osmotic pressure = pressure to stop osmosis.

• Reverse osmosis = water purification method.

Abnormal Molar Masses

Concept Overview

• Abnormal molar mass occurs when substances associate or dissociate in solution, altering the expected molecular mass.

• Determined by colligative properties, the observed molar mass differs from theoretical values.

• van't Hoff factor (i) quantifies this effect:
  i = (Observed colligative property) / (Calculated colligative property) = Mc / Mo

• Interpretation of i:
  • i < 1 indicates association (fewer particles)
  • i > 1 indicates dissociation (more particles)

• Modified colligative property equations:
  • Relative lowering of vapor pressure: \( \frac{p_1^\circ - p_1}{p_1^\circ} = i \frac{n_2}{n_1} \)
  • Boiling point elevation: \( \Delta T_b = i K_b m \)
  • Freezing point depression: \( \Delta T_f = i K_f m \)
  • Osmotic pressure: \( \pi = i \frac{n_2 R T}{V} \)

• Comparison of observed and normal colligative properties:
  • Observed < Normal → Association
  • Observed > Normal → Dissociation

• For electrolyte AxBy dissociating with degree of dissociation α:
  • Total particles after dissociation: \( i = \alpha (x + y - 1) + 1 \)

• For association of n molecules:
  • Total particles after association: \( i = 1 - \alpha + \frac{\alpha}{n} \)

Solved Examples

• Ex 1: Calculate van't Hoff factor for NaCl dissociating completely.
  Solution: NaCl → Na+ + Cl−, total particles = 2
  i = 2 (complete dissociation)

• Ex 2: A solute shows observed molar mass 60 g/mol, theoretical 90 g/mol.
  Calculate i and identify association/dissociation.
  i = 90/60 = 1.5 > 1 → Dissociation

Practice Set

• Q1: What is the van't Hoff factor for glucose (non-electrolyte) in solution?
  A) 0.5 B) 1 C) 2 D) 3

• Q2: If observed freezing point depression is less than calculated, the solute is:
  A) Dissociated B) Associated C) Ionized D) Neutral

• Q3: Electrolyte A2B dissociates as A2B → 2A+ + B2− with α = 0.6.
  Calculate van't Hoff factor i.
  A) 2.2 B) 2.6 C) 3.0 D) 1.6

Answer Key

• Q1: B) 1
  Glucose does not dissociate; i = 1.

• Q2: B) Associated
  Lower observed colligative property indicates association (i < 1).

• Q3: B) 2.6
  i = α(x + y - 1) + 1 = 0.6(2 + 1 - 1) + 1 = 0.6(2) + 1 = 2.2 + 1 = 2.6

Memory Aids

• van't Hoff factor i < 1 → Association (particles combine)

• van't Hoff factor i > 1 → Dissociation (particles split)

• For electrolyte dissociation: i = α(total ions - 1) + 1

• For association: i = 1 - α + α/n