Model Test Questions XI Chemistry on Solution and Colloids Chapter # 10

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🧪 Class 11 Chemistry – Chapter 10: Solution and Colloids ✨

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If you want easy understanding + high scores 🏆, this guide is a must-read before your chemistry paper! 📘🔥

✏️ Model Test Questions XI Chemistry Chapter # 10………… Solution and Colloids ✏️

✏️ Short Questions of Solution and Colloids ✏️

Q1. Define concentration. Name various units of concentration and explain mole fraction.

Q2. Define molarity and molality. Which of these depends on temperature?

Q3. Explain on particles basis , how the vapour pressure of solution is lowered by adding non-volatile solute?

Q4. What are miscible and immiscible liquids? Why n-hexane (petrol) is immiscible in water?

Q5. Solubility of oxygen in water increases with pressure but solubility of glucose in water has negligible effect of pressure why?

Q6. What are Hydrophilic and Hydrophobic Molecules?

Q7. Describe the Mechanism of Dissolution

Q8. What are Colloids? Define their two types based on water as dispersion medium. Write down their 5 Properties.

Q9. What is meant by solubility? Describe the effect of different factors on solubility.

Q10. Define Percentage Composition, Parts per million, parts per billion and parts per trillion

✏️ Descriptive Questions ✏️

Q1. State Raoult’s’ law and derive its mathematical expression.

Q2. What does an ideal solution means? Give four characteristics to distinguish between ideal and non-ideal solution.

Q3. Define osmosis and osmatic pressure. Give four daily life examples of osmosis.

Q4. What are colligative properties? Why does the boiling point of a liquid get raised when a non-volatile solute is added.?

Q5. Differentiate among true solution, colloidal solution and suspension on the basis of
(i) Particle size
(ii) Visibility

✏️ Numericals on Solution and Colloids ✏️

Q1. .24 g caustic soda (NaOH) is dissolved in water and the solution is made to 100 cm³ in a volumetric flask. Determine the concentration of this solution in terms of molarity.
[Example 10.1, Page # 204]

Answer: (1.56 mol/dm³)

Q2. An aqueous solution of 1.2 molality is prepared by dissolving some amount of oxalic acid into 475 g water. Determine the mass of oxalic acid in the solution. (Molecular mass of oxalic acid is 126 g/mol).
[Example 10.2, Page # 204]

Answer: (71.82 g)

Q3. 45 g glucose dissolves in 72 g water to make a solution. Calculate the mole fraction of glucose and water in the solution.
[Example 10.3, Page # 205]

Answer: (X₁ = 0.0588, X₂ = 0.941)

Q4. Calculate the molality of a 12% urea solution (molar mass of urea is 60 g/mol).
[Example 10.4, Page # 205]

Answer: (2.27 mol/kg)

Q5. In the analysis of a water sample, it was reported that 1 g of water contains 6.34 x 10⁻³ mg magnesium ions. Calculate the concentration of magnesium ion in ppm.
[Example 10.5, Page # 206]

Answer: (6.34 ppm)

Q6. Automotive antifreeze is a 60% (w/w) aqueous solution of ethylene glycol (C₂H₆O₂). Determine:
(a) Molality of solution
(b) Mole fraction of ethylene glycol in the solution
[Exercise; Q1, Page # 217]

Answer: (Molality = 24.19 mol/kg, X₁ = 0.303)

Q7. A solution is prepared by mixing 46 g ethanol (C₂H₅OH) and 180 g water. Calculate the mole fraction of both components.
[Exercise; Question 2, Page # 217]

Answer: (X₁ = 0.0909, X₂ = 0.909)

Q8. Glucose is a non-volatile solute in water. A glucose solution contains 0.15 moles glucose and 5.8 moles water at 20°C. Determine the lowering in the vapour pressure if the vapour pressure of pure water at 20°C is 17.5 torr. (Assume the solution is ideal).
[Example 10.6, Page # 208]

Answer: (0.89)

Q9. The vapour pressure of a pure liquid A is 37 mm Hg at 27°C. It is mixed into another liquid B to make a solution. The vapour pressure of A in the solution is found to be 33 mm Hg. Calculate the mole fraction of A (Assume it obeys Raoult’s law).
[Example 10.7, Page # 208]

Answer: (0.441)

✏️ Smart Answers of Model Test Questions XI Chemistry on Solution and Colloids Chapter # 10 ✏️

Q1. Define concentration. Name various units of concentration and explain mole fraction.

Answer

🧪 Definition of Concentration of a Solution:

Concentration is the amount of solute dissolved in a fixed amount of solvent or solution. More solute = Higher concentration ➕ Less solute = Lower concentration ➖

Units of Concentration 📏

There are different ways to express concentration:

(A) Physical Units 🔹

📊 Percentage composition (%)
🌫️ Parts per million (ppm)
🔬 Parts per billion (ppb)
🧫 Parts per trillion (ppt)

(B) Chemical Units 🔹

🧪 Molarity (M)
⚖️ Molality (m)
🧮 Normality (N)
🔢 Mole Fraction (X)

🔢 Mole Fraction (X)

📌 Definition: It is the ratio of moles of one component to the total moles of all components in a solution.
🧮 General Formula: Mole Fraction (X) = Moles of one component / Total moles of all components
🛠️ Application: Used when two or more components are present.
📏 Unit: It has no unit (unitless ratio of moles).
⚖️ Sum of mole fractions: is always 1 i.e. Xᴀ + Xᴃ = 1

🧮 Formula for Two-Component System (components A and B)

Moles of A = nᴀ
Moles of B = nᴃ
➤ Mole fraction of A: Xᴀ = nᴀ / (nᴀ + nᴃ)
➤ Mole fraction of B: Xᴃ = nᴃ / (nᴀ + nᴃ)

📘 Example

If a solution contains:
🧪 1 mole of methanol
💧 4 moles of water
Total moles = 1 + 4 = 5
Xᴄʜ₃оʜ = 1 / 5 = 0.2
X ʜ₂о = 4 / 5 = 0.8
✅ Check: 0.2 + 0.8 = 1

Q2. Define molarity and molality. Which of these depends on temperature?

Answer

🟥 Molarity (M) – Molar Concentration 🧪

📌 Definition: It is the number of moles of solute dissolved in one dm³ or one litre (1000 mL) of solution.
🧮 Formula: Molarity (M) = Moles of solute / Volume of solution in dm³
📏 Unit: mol dm⁻³ (mol L⁻¹) or M
🏗️ Based on: Depends on volume of solution
⚙️ Factor: temperature dependent as volume increases or decreases with temperature 🌡️
✨ Examples:
➡️ 0.5 M NaOH solution = 20 g NaOH in 1 dm³ solution
➡️ 1M H₂SO₄ solution = 98 g H₂SO₄ per dm³

🟥 Molality (m) – Molal Concentration ⚖️

📌 Definition: It is the number of moles of solute dissolved in one kilogram (1000 g) of solvent.
🧮 Formula: Molality (m) = Moles of solute / Mass of solvent in kg
📏 Unit: mol kg⁻¹
🏗️ Based on: Depends on mass of solvent
⚙️ Factor: Molality is temperature independent as mass does not change on heating or cooling
✨ Examples:
➡️ 1 m Na₂CO₃ solution contains 106 g per kg of water
➡️ 18 g glucose in 1 kg water = 0.1 mol/kg

🟥 Which Depends on Temperature? 🌡️

✅ Molarity depends on temperature as volume changes with temperature (expansion or contraction).
❌ Molality does NOT depend on temperature as mass does not change with temperature.

⭐ In Short:
Molarity → Volume-based → Temperature dependent
Molality → Mass-based → Temperature independent

Q3. Explain on particles basis, how the vapour pressure of solution is lowered by adding non-volatile solute?

Answer

🟥 Lowering of Vapour Pressure (Particle Explanation) 💨

📌 Basic Idea
When a non-volatile solute is added to a solvent, the vapour pressure of the solution decreases.

⭐ Reason
Adding a non-volatile solute reduces mole fraction of solvent, hence lowers vapour pressure.

🧪 On Particle Basis Explanation

🔹 Pure Solvent
Only solvent particles are present at the surface.
Some particles escape into vapour phase.
This creates a certain vapour pressure.

🔹 After Adding Non-Volatile Solute
Solute particles do not evaporate (non-volatile).
They occupy space between solvent particles.
The number of solvent particles at the surface decreases.

🎯 Result (Effect on Vapour Formation)
Fewer solvent particles can escape into vapour.
Rate of evaporation decreases.
Hence, vapour pressure of solution lowers.

🎯 Why It Happens (Key Points)
Vapour pressure depends on the number of solvent particles escaping.
Adding solute reduces the mole fraction of solvent.
Lower mole fraction → Lower vapour pressure.

⭐ In Short
Non-volatile solute particles block solvent particles at the surface.
Only solvent evaporates.
Fewer escaping particles → Lower vapour pressure of solution.

Q4. What are miscible and immiscible liquids? Why n-hexane (petrol) is immiscible in water?

Answer

🧪 Miscible and Immiscible Liquids

🟥 Completely Miscible Liquids 🧪

➡️ Definition: These are liquids that mix completely in all proportions to form a single homogeneous layer.
➡️ Reason: Their intermolecular forces are similar in strength and type.
📌 Examples:
➡️ Methanol and water → Both are polar and form hydrogen bonding in pure and mixed states.
➡️ Benzene and toluene → Both are non-polar and have similar London dispersion forces.

🟥 Completely Immiscible Liquids 🧪

➡️ Definition: Liquids that do not mix at all and form separate layers are called completely immiscible liquids.
➡️ Reason: This happens due to different types of intermolecular forces.
📌 Example:
➡️ Water (polar) and benzene (non-polar) → Do not mix and form two separate layers.
➡️ Oil (non-polar) and water (polar) → Do not mix and form two separate layers.

🟥 Why is n-Hexane (Petrol) Immiscible in Water? ⛽

➡️ n-Hexane (petrol) is a non-polar substance. ⛽
➡️ Water is a polar substance. 💧
➡️ Polar and non-polar substances do not mix (“like dissolves like”). ⚖️
➡️ Therefore, they do not attract each other strongly and therefore do not mix. 🚫
➡️ n-Hexane forms a separate layer and is immiscible in water. 🧪

Q5. Solubility of oxygen in water increases with pressure but solubility of glucose in water has negligible effect of pressure why?

Answer

🧪 Effect of Pressure on Solubility 💧

🟥 Solubility of Gases (e.g., Oxygen) in Liquids 🌬️

➡️ General Rule: Gas solubility increases with pressure (Henry’s Law: gas solubility ∝ pressure).
➡️ Reason: Dissolution is an equilibrium: gas molecules enter and leave solution simultaneously.
👉 Higher pressure → more gas molecules → rate of entering > rate of leaving → solubility increases.

🟥 Solubility of Solids (e.g., Glucose) in Liquids 🍬

➡️ General Rule: Solubility of solids and liquids almost unaffected by pressure.
➡️ Reason: Solids and liquids are incompressible, so pressure has negligible effect.
✅ Example: Glucose in water → solubility almost unaffected by pressure.

⭐ Key Point

Solids/liquids → pressure effect negligible.
Gases → solubility increases with pressure in accordance with Henry’s law.

Q6. What are Hydrophilic and Hydrophobic Molecules

Answer

🟥 Hydrophilic Molecules (Water-Loving) 💧

➡️ Molecules that mix well with water or interact strongly with water.
➡️ Usually polar (have –OH, –COOH, –NH₂ groups) or ionic → form hydrogen bonds with water.
➡️ Dissolve easily → form aqueous solutions.
✅ Examples: Sugar, Ethanol, acetic acid, amino acids, acetone

🟥 Hydrophobic Molecules (Water-Fearing)

➡️ Molecules that do not mix with water or repel water.
➡️ Usually non-polar → cannot form hydrogen bonds.
➡️ Do not dissolve → separate layer in water.
✅ Example: Oil, fats, n-hexane, Petrol, benzene, toluene

⭐ Key Point

➡️ Hydrophilic → dissolve in water
➡️ Hydrophobic → do not dissolve in water

Q7. Describe the Mechanism of Dissolution

Answer

🟥 Steps of Dissolution

Dissolution is the process by which a solute dissolves in a solvent to form a homogeneous solution.

🟥 Mechanism of Dissolution

When a solute dissolves in a solvent, three types of interactions are involved:
📌 Solute–solute (forces between solute particles) interactions
📌 Solvent–solvent (forces between solvent molecules) interactions
📌 Solute–solvent (forces between solute and solvent) interactions

🟥 Dissolution occurs in three main steps:

📌 Separation of Solute Particles → Requires energy to overcome intermolecular or ionic forces.
📌 Separation of Solvent Particles creating space for solute → requires energy to overcome solvent-solvent interactions.
📌 Solute-Solvent Interaction → New attractive forces (ion-dipole, dipole-dipole, hydrogen bonds).
Energy released in this step helps stabilize the solution.

🟥 Rule for Solution Formation

➡️ Solution forms if solute–solvent attraction ≥ solute–solute + solvent–solvent attractions.
➡️ Solute particles separate and are surrounded by solvent molecules, forming new stable interactions.

🟥 Energy Changes

➡️ Lattice Energy: Energy required to separate solute particles from solid → endothermic
➡️ Solvation Energy: Energy released when solute interacts with solvent → exothermic
➡️ Heat of Solution (ΔHₛₒₗₙ): Net enthalpy change = Lattice energy – Solvation energy

🟥 Net Energy Change Determines Solubility

➡️ Ions with small size & high charge: High solvation → dissolution is exothermic → solubility increases
➡️ Solids with strong lattice: Absorb energy → dissolution is endothermic → solubility may be limited

⭐ Key Points

➡️ Dissolution involves breaking solute & solvent interactions and forming solute-solvent interactions.
➡️ Whether dissolution releases or absorbs heat depends on lattice vs solvation energy.

Q8. What are Colloids? Define their two types based on water as dispersion medium. Write down their 5 Properties.

Answer

🟥 Definition of Colloids or Colloidal Dispersion: 🧪

Colloids are translucent heterogeneous mixtures in which particles of one substance (1–1000 nm) are evenly dispersed in another substance but do not settle on standing.
🧪 Dispersed Phase: Solute (the distributed particles)
💧 Dispersion Medium: Solvent (the medium in which particles are dispersed)
🔬 Invisible Particles: Too small to be seen with the naked eye
🧲 Coagulation Property: Do not settle but coagulate on heating or adding electrolyte
📏 Intermediate Size: Larger than solution molecules but smaller than suspension particles
💡 Tyndall Effect: Can scatter light
✅ Example: Milk (casein particles dispersed in water)

🟥 Types of Colloids (Water as Dispersion Medium) 🔹

✨ Hydrophilic Colloids (Water-Loving) or Lyophilic Colloids (Solvent-Loving):
➡️ Dispersed particles strongly attract water (or any solvent)
➡️ Stable, easily formed
✅ Example: Gelatin, starch solution

✨ Hydrophobic Colloids (Water-Fearing) or Lyophobic Colloids (Solvent-Fearing):
➡️ Dispersed particles do not interact strongly with water (or any solvent)
➡️ Unstable, coagulate easily
✅ Example: Silver sol, arsenic sulfide sol

🟥 Properties of Colloids 🔹

📌 Apparent Homogeneity: Heterogeneous in nature but appear homogeneous to the naked eye 👁️
📌 Cloudy Appearance: Usually cloudy or translucent in nature 🌫️
📌 No Settling: Particles remain suspended, do not settle
📌 Brownian motion: Random, continuous movement of particles in medium due to unequal collision of particles
📌 Permeability: Pass through filter paper but blocked by semi-permeable membrane
📌 Tyndall effect: Colloidal particles scatter light due to large particle size
📌 Coagulation: precipitation of particles as aggregate on heating or adding electrolyte
📌 Charge on particles: Particles carry same charge → repel → remain suspended
📌 Electrophoresis: Movement of charged colloidal particles towards particular electrode in an electric field

🟥 Types of Colloids Based on the Physical State of Dispersed Phase and Dispersion Medium (8 Types)

➡️ Gas in Liquid → Aerosol → Fog, Spray
➡️ Gas in Solid → Aerosol → Smoke
➡️ Liquid in Gas → Foam → Shaving Cream
➡️ Liquid in Liquid → Emulsion → Milk
➡️ Liquid in Solid → Sol → Paint
➡️ Solid in Gas → Solid Foam → Foam Rubber
➡️ Solid in Liquid → Gel → Jelly
➡️ Solid in Solid → Solid Sol → Ruby Glass

Q9. What is meant by solubility? Describe the effect of different factors on solubility.

Answer

🟥 🧪 Definition Solubility:

Solubility is the maximum amount of a solute that can dissolve in a given amount of solvent at a particular temperature and pressure to form a homogeneous saturated solution.
✨ Unit: Usually expressed in g/100 g of solvent or mol/L

🟥 Factors Affecting Solubility

✴️ Chemical Nature of Solute and Solvent
✨ “Like dissolves like”:
✨ Polar solutes → dissolve in polar solvents
✨ Non-polar solutes → dissolve in non-polar solvents

✴️ Effect of Temperature on Solubility 🌡️
📌 Solids in liquids:
✨ Rule: Usually solubility increases with temperature (e.g., sugar in water)
✨ Reason: temperature ⬆️ → kinetic energy ⬆️ → solvent molecules create more spaces → solute dissolves easier
✨ Exceptions: Some solids (e.g., AlCl₃, Na₂SO₄, lipids) dissolve exothermically → solubility ⬇️ with temp.

✨ Dual behavior:
➡️ Solids that absorb heat on dissolving → Endothermic dissolution → solubility ⬆️ with temperature
➡️ Solids that release heat on dissolving → Exothermic dissolution → solubility ⬇️ with temperature

📌 Gases in liquids:
✨ Rule: Solubility decreases with temperature (e.g., O₂ in water)
✨ Reason: Higher temperature → more kinetic energy → gas molecules escape from the solvent → Solubility lowers

✴️ Effect of Pressure on Solubility ⬆️
📌 Solids and liquids in liquids:
✨ Rule: Pressure has negligible effect
✨ Reason: Solids and liquids are nearly incompressible

📌 Gases in liquids:
✨ Rule: Solubility increases with pressure (Henry’s Law)
✨ Reason: Dissolution is an equilibrium: gas molecules enter and leave solution simultaneously.
Higher pressure → more gas molecules enter than leave → solubility ⬆️
Example:
➡️ Carbonated water: CO₂ is more soluble under high pressure
➡️ On opening the soda bottle → pressure drops → solubility ⬇️ → CO₂ escapes as bubbles

Q10. Define Percentage Composition, Parts per million, parts per billion and parts per trillion

Answer

🧪 Physical Units of Concentration

🟥 Percentage Composition (%)

Definition: Amount of solute present per 100 parts of solution or solvent.
Types: Expressed in w/w, w/v, v/w, v/v
Mass Percent: the mass of solute in gram per 100 gram of solution.
Formula: % composition (Mass Percent) = Mass of solute / Mass of solution × 100

✅ Example:
13% w/w sugar → 13 g sugar + 87 g water → 100 g solution
10% w/w NaCl → 10 g NaCl + 90 g water → 100 g solution
10% w/v NaCl → 10 g NaCl in 100 cm³ solution
10% v/v NH₃ → 10 cm³ NH₃ + 90 cm³ water → 100 cm³ solution

🟥 Parts per Million (ppm)

Definition: Number of parts of solute per 1 million (10⁶) parts of solution
Formula: ppm = Amount of solute (mass or volume) / Amount of solution × 10⁶
✅ Example: 1 mg solute in 1 dm³ solution → concentration = 1 ppm

🟥 Parts per Billion (ppb)

Definition: Number of parts of solute per 1 billion (10⁹) parts of solution
Formula: ppb = Amount of solute (mass or volume) / Amount of solution × 10⁹
✅ Example:
1 μg solute in 1 dm³ water → 1 ppb
1 mg solute in 1000 dm³ solution → 1 ppb

🟥 Parts per Trillion (ppt)

Definition: Number of parts of solute per 1 trillion (10¹²) parts of solution
Formula: ppt = Amount of solute (mass or volume) / Amount of solution × 10¹²
✅ Example:
1 ng solute in 1 dm³ water → 1 ppt
1 μg solute in 1000 dm³ solution → 1 ppt

⭐ Key Point

Used for very dilute solutions in analytical and biochemical work
% → ppm → ppb → ppt: decreasing concentration scale

✏️ Smart Answers of Long-Answer Questions on Solutions and Colloids ✏️

Q1. State Raoult’s Law and derive its mathematical expression.

Answer

🟥 Raoult’s Law 🧪

Proposer of Law: A French chemist F.M. Raoult (1887) studied lowering of vapour pressure by solute particles and formulated Raoult’s Law.

💡 Statement #1 (Non-volatile solute):
📌 The vapour pressure of a solution is directly proportional to the mole fraction of the solvent.
⭐ Key Points:
➡️ Applies when solute is non-volatile
➡️ Only ideal solutions (similar polarity & molecular size) follow it accurately
➡️ For a non-volatile solute, the vapour pressure of the solution is lower than that of pure solvent
➡️ Non-volatile solute → lowers solvent vapour pressure proportionally to Xₛₒₗᵤₜₑ
➡️ Raoult’s law explains: vapour pressure lowering, boiling point elevation, freezing point depression

🟥 First Mathematical Form

Pₛₒₗᵤₜᵢₒₙ ∝ Xₛₒₗᵥₑₙₜ or P ∝ X₁ [P = Vapour pressure of solution and X₁ = mole fraction of solvent]
Introducing the proportionality constant (P⁰ or P⁰ᴀ, vapour pressure of pure solvent A):
Pₛₒₗᵤₜᵢₒₙ = P⁰ₛₒₗᵥₑₙₜ × Xₛₒₗᵥₑₙₜ or P = P⁰ X₁ [P° = Vapour pressure of pure solvent] ---------------- (i)

🟥 Second Mathematical Form (Lowering of Vapour Pressure)

In a binary solution, sum of mole fractions is unity i.e. X₁ + X₂ = 1 ⇒ X₁ = 1 – X₂
P = P⁰ X₁ [Putting the value of X₁ in eq. (i)]
P = P⁰ (1 – X₂)
P = P⁰ – P⁰X₂
P⁰– P = P⁰X₂
∆P = P⁰X₂ ---------------- (ii) [Second form of Raoult’s law]

💡 Statement #2:
📌 The lowering of vapour pressure (∆P) is directly proportional to mole fraction of solute.

🟥 Third Mathematical Form (Relative Lowering of Vapour Pressure)

The relative lowering in vapour pressure (∆P/P°) is obtained from second form by dividing lowering of vapour pressure by vapour pressure of pure solvent:
∆P = P⁰ X₂
∆P/P⁰ = X₂ ---------------- (iii) [Third form of Raoult’s law]
Here ∆P/P⁰ is referred as relative lowering in vapour pressure

💡 Statement #3:
📌 The relative lowering of vapour pressure is equal to mole fraction of solute.

🟥 Total Pressure of Binary Solution (Volatile Liquids) Using Dalton’s Law

For two miscible volatile liquids, the total vapor pressure is the sum of their individual partial pressures, as defined by Dalton’s Law:
Pᴀ = P⁰ᴀ Xᴀ and Pᴃ = P⁰ᴃ Xᴃ
Pₜₒₜₐₗ = P⁰ᴀ Xᴀ + P⁰ᴃ Xᴃ ……………. (iv)

Q2. What does an ideal solution mean? Give four characteristics to distinguish between ideal and non-ideal solution.

Answer

🟥 Ideal Solution 🧪

📌 Definition: An ideal solution is a solution that perfectly obeys Raoult’s Law at all compositions and temperatures, obeying the relationship Pₜₒₜₐₗ = P⁰ᴀ Xᴀ + P⁰ᴃ Xᴃ.
In such solutions, intermolecular forces between unlike molecules (A–B) are equal to those between like molecules (A–A and B–B).

⭐ Characteristics of Ideal Solution:
➡️ Obeys Raoult’s Law at all concentrations and temperatures
➡️ ΔHₛₒₗₙ = 0 (No heat is absorbed or evolved)
➡️ ΔVₘᵢₓ = 0 (No change in volume on mixing)
➡️ Intermolecular forces A–B ≈ A–A ≈ B–B
📌 Examples: Benzene + Toluene

🟥 Non-Ideal Solution

A non-ideal solution does not obey Raoult’s Law over the entire concentration range.

⭐ Characteristics of Non-Ideal Solution:
➡️ Deviates from Raoult’s Law
➡️ ΔHₛₒₗₙ ≠ 0 (Heat absorbed or released)
➡️ ΔVₘᵢₓ ≠ 0 (Volume changes on mixing)
➡️ Intermolecular forces A–B ≠ A–A or B–B
📌 May show positive or negative deviation from Raoult’s Law

✅ Key Difference

Ideal Solution	Non-Ideal Solution
Obeys Raoult’s Law	Deviates from Raoult’s Law
ΔH = 0	ΔH ≠ 0
ΔV = 0	ΔV ≠ 0
Similar intermolecular forces	Different intermolecular forces

Q3. Define osmosis and osmotic pressure. Give four daily life examples of osmosis.

Answer

💧 Osmosis and Osmotic Pressure

🧪 Semi-Permeable Membrane

A membrane that allows solvent (water) molecules to pass but does not allow solute particles is called a semi-permeable membrane.
📌 Examples: Cellophane, animal bladder, cell membrane

💦 Definition of Osmosis

Osmosis is the spontaneous movement of solvent molecules from a dilute solution (low solute concentration) to a concentrated solution (high solute concentration) through a semi-permeable membrane.
Movement continues until equilibrium is reached.
It is a natural process.

🧮 Osmotic Pressure (𝛑)

Definition: The minimum pressure required to stop osmosis is called osmotic pressure, denoted by 𝛑 (pi).
Unit: Measured in atmosphere (atm)
Nature: It is a colligative property (depends on number of solute particles)

💡 Explanation

To demonstrate osmosis, consider a sugar solution in an inverted thistle funnel whose mouth is covered with a semi-permeable membrane. The funnel is dipped into a beaker of pure water.
Over time, water from the beaker migrates into the sugar solution through the membrane, raising the solution level until the pressure exerted by the solution stops further movement. This pressure is called osmotic pressure.

🌍 Daily Life Examples of Osmosis

1️⃣ Hemolysis (RBC bursting) 🩸
Water enters RBCs in dilute medium → cells swell and burst (endosmosis).

2️⃣ Fresh Water Fish 🐟
Absorb water through skin because body fluid is more concentrated than surrounding water.

3️⃣ Water Absorption in Plants 🌱
Roots absorb water from soil by osmosis.

4️⃣ Preservation of Jam & Pickles 🥒🧂
High sugar/salt concentration → water leaves bacteria → prevents spoilage.

5️⃣ Salt Water Gargling 🧂
Water moves out of throat cells → reduces swelling and irritation.

6️⃣ Osmosis in Nutrient Absorption 🌱💧
After digestion, food turns into chyme, which is absorbed into body tissues through the semi-permeable walls of the small intestine.

Q4. What are colligative properties? Why does the boiling point of a liquid get raised when a non-volatile solute is added?

Answer

🧪 Definition of Colligative Properties 📌

Colligative properties are the physical properties of a solution that depend only on the number of solute particles, not on their chemical nature. (Colligative = collective)

⭐ Examples:
1️⃣ Lowering of vapour pressure (∆P)
2️⃣ Elevation of boiling point (∆Tb)
3️⃣ Depression of freezing point (∆Tf)
4️⃣ Osmotic pressure (𝛑)

🌡️ Elevation of Boiling Point

📌 Boiling Point: The boiling point is the temperature at which the vapour pressure of a liquid becomes equal to atmospheric pressure (1 atm).

🔥 Why Boiling Point Increases?
A non-volatile solute has negligible vapour pressure (assumed to be zero).
When added to a solvent → vapour pressure of solution decreases (Raoult’s law).
To reach atmospheric pressure, the solution must be heated to a higher temperature.
👉 Therefore, boiling point of solution > boiling point of pure solvent.
This increase is called Elevation of Boiling Point (∆Tb).

📊 Graph Explanation (Concept):
Vapour pressure curve of solution lies below that of pure solvent.
Solution reaches atmospheric pressure at a higher temperature (Tb) than pure solvent (Tb°).

🌍 Daily Life Example

🥚 Adding salt while boiling eggs → raises boiling point → water becomes slightly hotter → egg cooks faster.

⭐ Key Points:
➡️ Boiling point elevation is a colligative property
➡️ Depends on number of solute particles
➡️ Non-volatile solute → lowers vapour pressure → raises boiling point

Q5. Differentiate among true solution, colloidal solution and suspension on the basis of (i) Particle size (ii) Visibility

Answer

Property	True Solution	Colloid	Suspension
Particle Size	< 1 nm	1–1000 nm	> 1000 nm
Visibility	Not visible	Visible under ultramicroscope	Visible to naked eye
Nature	Homogeneous	Heterogeneous (appears homogeneous)	Heterogeneous
Appearance	Transparent	Translucent	Opaque
Settling	Do not settle	Do not settle (coagulate on heating)	Settle on standing
Example	Glucose solution	Milk	Mud in water

✏️ Smart Solution of Numerical of Solutions and Colloids ✏️

Q1. 6.24 g caustic soda (NaOH) is dissolved in water and the solution is made to 100 cm³ in a volumetric flask. Determine the concentration of this solution in terms of molarity. [Example 10.1, Page # 204]

📝 Solution

📌 Given Data

Mass of NaOH = 6.24 g
Volume of solution = 100 cm³ ⇒ 100 / 1000 = 0.1 dm³
Molar mass of NaOH = 23 + 16 +1 = 40 g/mol

🎯 Required

Molarity of solution = ?

📐 Formula

Molarity (M) = No. of moles of solute / Volume of solution in dm³
M = Mass of solute (g) / Molar mass (g/mol) × Volume of solution (dm³)

🧮 Calculation

M = 6.24 (g) / 40 (g/mol) × 0.1 (dm³) = 1.56 mol dm⁻³

✅ Final Answer

Molarity of NaOH solution = 1.56 M

Q2. An aqueous solution of 1.2 molality is prepared by dissolving some amount of oxalic acid into 475 g water. Determine the mass of oxalic acid in the solution. (Molecular mass of oxalic acid is 126 g/mol). [Example 10.2, Page # 204]

📝Solution

📌 Given Data

Molality of solution = 1.2 mol/kg
Mass of water (solvent) = 475 g ⇒ 475 / 1000 = 0.475 kg
Molar mass of oxalic acid, H₂C₂O₄.2H₂O = 126 g/mol (2 + 24 + 64+36)

🎯 Required

Mass of oxalic acid, H₂C₂O₄.2H₂O = ?

📐Formula

Molality (m) = Mass of solute (g)/Molar mass (g/mol) × Mass of solvent (kg)
Mass of solute = Molality (mol/kg) × Molar mass (g/mol) × Mass of solvent (kg)

🧮Calculation

Mass of solute in g = 1.2 × 126 × 0.475 = 71.82 g

✅ Final Answer

Mass of oxalic acid = 71.82 g

Q3. 45 g glucose dissolves in 72 g water to make a solution. Calculate the mole fraction of glucose and water in the solution. [Example 10.3, Page # 205]

📝Solution

Mass of glucose = 45 g
Mass of water = 72 g
Molar mass of glucose (C₆H₁₂O₆) = 6(12) + 12(1) + 6(16) = 180 g/mol
Molar mass of water (H₂O) = 2(1) + 16 = 18 g/mol

🧮 Finding No. of moles of Each component of solution using n = mass/molar mass

Moles of glucose = 45 / 180 = 0.25
Moles of water = 72 / 18 = 4

🧮 Finding Mole fraction using X = n/nₜ

Mole fraction of glucose (X₁) = n₁ / (n₁ + n₂) = 0.25 / (0.25 + 4) = 0.0588
Mole fraction of water (X₂) = n₂ / (n₁ + n₂) = 4 / (0.25 + 4) = 0.941

Q4. Calculate the molality of a 12% urea solution (molar mass of urea is 60 g/mol). [Example 10.4, Page # 205]

📝Solution

📌 Given:

Urea solution = 12% w/w → 12 g urea in 100 g solution
Mass of urea = 12 g
Mass of solvent (water) = 100 – 12 = 88 g ⇒ 88 / 1000 = 0.088 kg
Molar mass of urea M = 60 g/mol

🎯 Required:

Molality (m)

📐 Formula:

Molality (m) = Mass of solute (g) / Molar mass (g/mol) × Mass of solvent (kg)

🧮 Calculation:

Molality (m) = 12 (g) / 60 (g/mol) × 0.088 (kg) = 2.27 mol kg⁻¹

✅ Final Answer: Molality of urea solution = 2.27 m

Q5. In the analysis of a water sample, it was reported that 1 g of water contains 6.34 × 10⁻³ mg of magnesium ions. Calculate the concentration of magnesium ions in ppm. [Example 10.5, Page # 206]

📝Solution

📌 Given:

Amount of solute (Mg²⁺ ions) = 6.34 × 10⁻³ mg
Amount of solvent (water) = 1 g ⇒ 1 × 1000 = 1 × 10³ mg

🎯 Required:

Concentration of Mg²⁺ ions in ppm = ?

📐 Formula:

ppm = Amount of solute / Amount of solvent × 10⁶

🧮 Calculation:

6.34 × 10⁻³ mg / 1 × 10³ mg × 10⁶ = 6.34 ppm

✅ Final Answer: Concentration of Mg²⁺ ions = 6.34 ppm

Q6. Automotive antifreeze is a 60% (w/w) aqueous solution of ethylene glycol (C₂H₆O₂). Determine: (a) Molality of solution (b) mole fraction of ethylene glycol in the solution [Exercise; Q1, Page # 217]

📝Solution

📌 Given:

Mass percent of ethylene glycol = 60% w/w
Mass of solution = 100 g (assuming 100 g of solution for simplicity)
Mass of ethylene glycol (solute) = 60 g
Mass of water (solvent) = 100 – 60 = 40 g ⇒ 40/1000 = 0.040 kg
Molar mass of ethylene glycol (C₂H₆O₂) = 2(12) + 6(1) + 2(16) = 62 g/mol
Molar mass of water (H₂O) = 18 g/mol

🎯 Required:

(a) Molality of solution = ?
(b) Mole fraction of ethylene glycol in the solution = ?

🧮 Finding Molality of Solution

📐 Formula: Molality (m) = Mass of solute (g)/Molar mass (g/mol) × Mass of solvent (kg)
🧮 Calculation: Molality (m) = 60 (g)/62 (g/mol) × 0.040 (kg) = 24.19 mol kg⁻¹

🧮 Finding No. of Moles of Each Component of Solution using n = mass/molar mass

Moles of ethylene glycol (C₂H₆O₂) = 60 / 62 = 0.968 mol
Moles of water = 40 / 18 = 2.22 mol

🧮 Finding Mole Fraction using X = n/nₜ

Mole fraction of ethylene glycol (X₁) = n₁/(n₁+n₂) = 0.968/(0.968+2.22) = 0.303
Mole fraction of water (X₂) = n₂/(n₁+n₂) = 2.22/(0.968+2.22) = 0.693

✅ Answers:
(a) Molality = 24.19 mol/kg
(b) Mole fraction of ethylene glycol = 0.303

Q7. A solution is prepared by mixing 46 g ethanol (C₂H₅OH) and 180 g water. Calculate the mole fraction of both components. [Exercise; Question 2, Page # 217]

📝Solution

Mass of ethanol (C₂H₅OH) = 46 g
Mass of water (H₂O) = 180 g
Molar mass of ethanol (C₂H₅OH) = 2(12) + 6(1) + 16 + 6 = 46 g/mol
Molar mass of water (H₂O) = 2(1) + 16 = 18 g/mol

Moles of ethanol (C₂H₅OH) = mass/molar mass = 46 / 46 = 1 mol
Moles of water (H₂O) = mass/molar mass = 180 / 18 = 10 mol

Finding Mole Fraction:
Mole fraction of ethanol (X₁) = n₁ / (n₁ + n₂) = 1 / (1 + 10) = 1/11 = 0.0909
Mole fraction of water (X₂) = n₂ / (n₁ + n₂) = 10 / (1 + 10) = 10/11 = 0.909

Q8. Glucose is a non-volatile solute in water. A glucose solution contains 0.15 moles glucose and 5.8 moles water at 20°C. Determine the lowering in the vapour pressure if the vapour pressure of pure water at 20°C is 17.5 torr. (Assume solution is ideal). [Example 10.6, Page # 208]

📝Solution

Moles of glucose (solute) = n₂ = 0.15
Moles of water (solvent) = n₁ = 5.8
Total number of moles, nₜ = n₁ + n₂ = 5.8 + 0.15 = 5.95
Vapour pressure of water, P⁰ = 17.5 torr
Lowering in the vapour pressure, ∆P = ?

Finding mole fraction of solute (glucose):
Mole fraction of solute, X₂ = n₂ / nₜ = 0.15 / 5.95 ≈ 0.0252

Finding Lowering in the Vapour Pressure using Raoult’s law:
∆P = P⁰ × X₂ = 17.5 × 0.0252 ≈ 0.441 torr

✅ Final Answer: Lowering in the vapour pressure, ∆P ≈ 0.441 torr

Q9. The vapour pressure of a pure liquid A is 37 mm Hg at 27°C. It is mixed into another liquid B to make a solution. The vapour pressure of A in the solution is found to be 33 mm Hg. Calculate the mole fraction of A (Assume it obeys Raoult’s law). [Example 10.7, Page # 208]

📝Solution

Vapour pressure of pure liquid A, P⁰ᴀ = 37 mm Hg
Vapour pressure of A in solution, Pᴀ = 33 mm Hg
Mole fraction of A, Xᴀ = ?

Applying Raoult’s law:
Pᴀ = P⁰ᴀ × Xᴀ
Xᴀ = Pᴀ / P⁰ᴀ = 33 / 37 ≈ 0.89

✅ Final Answer: Mole fraction of A, Xᴀ ≈ 0.89

✏️ Text Book and Past Papers MCQs on Solution and Colloids with Explanatory Answers ✏️

1. Only one pair of liquid in the following set does not obey Raoult’s law, identify it:

(a) Methanol and Ethanol

(b) Benzene and toluene

(c) n-Hexane and n-heptane

(d) Ethanol and Acetone

✔️ Answer: (d) Ethanol and Acetone

The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. Some examples of ideal solutions are methanol and ethanol, benzene, and toluene, n-hexane and n-heptane, bromoethane and chloroethane etc.
Methanol and ethanol are polar in nature with similar intermolecular forces, hence they form an ideal solution. Both benzene and toluene and n-hexane and n-heptane are non-polar in nature with similar intermolecular forces. Hence, they form an ideal solution.
Ethanol and Acetone
When Acetone and ethanol are mixed, the weakening of hydrogen bonds in ethanol takes place. Hence, an ideal solution is not formed.
Ethanol and Acetone form an azeotropic mixture, which means they do not follow Raoult's law. In such mixtures, the vapor pressure of the solution is not a simple sum of the vapor pressures of the individual components, and they exhibit non-ideal behavior due to strong hydrogen bonding between ethanol and acetone molecules.

2. Identify the INCORRECT statement about colloidal solution:

(a) It shows Tyndall effect

(b) Its particles movement is Brownian type

(c) Its particle size is less than 1 nm

(d) Its physical appearance is translucent.

✔️ Answer: (c) Its particle size is less than 1 nm

Colloidal particles have size between 1 nm and 1000 nm. Particles less than 1 nm belong to true solutions.
Colloidal solutions show Tyndall effect and Brownian movement. They generally appear translucent due to scattering of light.

3. Which is not a colligative property?

(a) Lowering in vapours pressure

(b) Elevation in boiling point

(c) Depression in freezing point

(d) Atmospheric pressure

✔️ Answer: (d) Atmospheric pressure

Colligative properties depend on number of solute particles and include lowering of vapour pressure, elevation in boiling point and depression in freezing point.
Atmospheric pressure is not a property of solution and does not depend on solute particles.

4. Effect of pressure change play significant role in the solubility of:

(a) Solid into liquid

(b) Liquid into liquid

(c) Gas into liquid

(d) All of them

✔️ Answer: (c) Gas into liquid

Pressure has significant effect on solubility of gases in liquids according to Henry’s law.
Solubility of solids and liquids in liquids is almost independent of pressure.

5. According to Raoult law the relative lowering of vapour pressure is equal to:

(a) Molality

(b) Mole fraction of solute

(c) Mole fraction of solvent

(d) Molarity

✔️ Answer: (b) Mole fraction of solute

According to Raoult’s law, relative lowering of vapour pressure is equal to mole fraction of solute in solution.
It depends only on number of solute particles present.

6. The sum of mole fractions of components of a solution is equal to:

(a) 0.0

(b) 1.0

(c) 10

(d) 100

✔️ Answer: (b) 1.0

The sum of mole fractions of all components in a solution is always equal to 1.
It is based on definition of mole fraction.

7. How many mole of NaOH are present in 2dm³ of 1 molar aqueous solution of it.

(a) 0.5 mole

(b) 1 mole

(c) 1.5 mole

(d) 2 mole

✔️ Answer: (d) 2 mole

Molarity = moles / volume in dm³. A 1 molar solution contains 1 mole in 1 dm³.
In 2 dm³, number of moles = 2 × 1 = 2 moles.

8. A colloidal solution of liquid into liquid is known as:

(a) Gel

(b) Foam

(c) Sol

(d) Emulsion

✔️ Answer: (d) Emulsion

A colloidal system in which both dispersed phase and dispersion medium are liquids is called emulsion.
Example: milk.

9. An example of completely immiscible liquid pair is:

(a) Benzene to toluene

(b) Water and phenol

(c) Water and Benzene

(d) Water & methanol

✔️ Answer: (c) Water and Benzene

Water is polar while benzene is non-polar, so they do not mix with each other.
Hence, they are completely immiscible liquids.

10. A 15% W/W KOH solution can be prepared by mixing 15g KOH in:

(a) 15g water

(b) 85g water

(c) 100g water

(d) 115g water

✔️ Answer: (b) 85g water

15% W/W means 15g solute in 100g solution.
So, 15g KOH must be mixed with 85g water to make total 100g solution.

11. Particles size of solute and solvent in solution (atoms, molecules or ions) are extremely small having a diameter of approximately less than.

(a) 1 nm

(b) 10 nm

(c) 1000 nm

(d) 0.1 nm

✔️ Answer: (a) 1 nm

In true solutions, particle size is less than 1 nm.
Such small particles cannot be seen even under microscope.

12. Tyndall effect is NOT shown by

(a) Solution

(b) Suspension

(c) Colloids

(d) All of them

✔️ Answer: (a) Solution

True solutions do not scatter light, so they do not show Tyndall effect.
Colloids and suspensions scatter light and show Tyndall effect.

13. When a soltuion is formed, attraction between solute-solvent molecules becomes ………… than solute-solute and solvent-solvent molecules.

(a) Equal or greater

(b) Lesser or equal

(c) Lesser

(d) All of them

✔️ Answer: (a) Equal or greater

A solution forms when solute-solvent attractions are equal to or stronger than existing intermolecular forces.
This allows proper mixing of components.

14. Generally, dissolution process is

(a) Endothermic

(b) Exothermic

(c) Both of them

(d) None of them

✔️ Answer: (c) Both of them

Dissolution may be endothermic or exothermic depending on balance of energies involved.
It depends on lattice energy and solvation energy.

15. Which of the following energies involeved in dissolution process?

(a) Lattice energy

(b) Solvation energy

(c) Both of them

(d) None of them

✔️ Answer: (c) both of them

Dissolution involves breaking solute particles (lattice energy) and formation of solute-solvent interactions (solvation energy).
Both energies determine overall heat of solution.

16. Lattice energy is always:

(a) Positive

(b) Negative

(c) Both of them

(d) None of them

✔️ Answer: (c) Both of them

Lattice energy is energy required to break ionic lattice into ions.
Its value is positive for bond breaking and negative for bond formation.

17. Solvation energy is always:

(a) Positive

(b) Negative

(c) both of them

(d) None of them

✔️ Answer: (b) Negative

Solvation energy is released when solvent molecules surround solute particles.
Since energy is released, it is negative.

18. The difference lattice energy and solvation energy is known as

(a) Heat of solution

(b) Heat of hydration

(c) Heat of atomization

(d) None of them

✔️ Answer: (a) Heat of solution

The net enthalpy change obtained by subtracting solvation energy from lattice energy is heat of solution.
It determines whether dissolution is endothermic or exothermic.

19. The change in enthalpy when a substance dissolves in solvent at constant pressure is called

(a) Heat of solution

(b) Heat of hydration

(c) Heat of atomization

(d) None of them

✔️ Answer: (a) Heat of solution

The enthalpy change during dissolution at constant pressure is called heat of solution.
It may be positive or negative.

20. Which one of the following is hydrophilic molecules?

(a) Ethanol

(b) Amino acids

(c) Acetone

(d) All of them

✔️ Answer: (d) All of them

Hydrophilic molecules interact well with water due to polar groups.
Ethanol, amino acids and acetone contain polar functional groups, so all are hydrophilic.

21. The hydrophilic molecules dissolves in water due to formation of

(a) Hydrogen bond

(b) Dipole-dipole force

(c) London forces

(d) All of them

✔️ Answer: (d) All of them

Hydrophilic molecules interact with water through hydrogen bonding.

22. The hydrophobic molecules are …………….. in nature.

(a) Polar

(b) Non-polar

(c) Both of them

(d) None of them

✔️ Answer: (b) Non-polar

Hydrophobic molecules do not interact well with water.
They are generally non-polar in nature.

23. Which one of the following is hydrophobic molecules?

(a) Petrol

(b) Benzene

(c) Toluene

(d) All of them

✔️ Answer: (d) All of them

Petrol, benzene and toluene are non-polar substances.
Hence, all are hydrophobic molecules.

24. Particles size of the components of suspension are above

(a) 1 nm

(b) 10 nm

(c) 10³nm

(d) 0.1 nm

✔️ Answer: (c) 10³nm

Suspension particles are larger than colloidal particles.
Their size is generally above 1000 nm (10³ nm).

25. The size range of colloidal particles is from.

(a) 1 to 1000 nm

(b) 100 to 1000 nm

(c) 1000 to 1200 nm

(d) None of them

✔️ Answer: (a) 1 to 1000 nm

Colloidal particles have size between 1 nm and 1000 nm.
This range lies between true solutions and suspensions.

26. A colloidal solution of gas into liquid or solid is known as:

(a) Gel

(b) Foam

(c) Aerosol

(d) Emulsion

✔️ Answer: (b) Foam

A foam is a colloidal system in which gas is dispersed in liquid or solid.
Example: shaving cream, sponge.

27. A colloidal solution of liquid into solid is known as:

(a) Gel

(b) Foam

(c) Aerosol

(d) Emulsion

✔️ Answer: (a) Gel

A gel is formed when liquid is dispersed in a solid medium.
Example: jelly, butter.

28. Fog and spray are the examples of

(a) Gel

(b) Foam

(c) Aerosol

(d) Emulsion

✔️ Answer: (c) Aerosol

An aerosol is a colloidal system where liquid or solid particles are dispersed in gas.
Fog and spray consist of liquid droplets dispersed in air.

29. Smoke is the examples of

(a) Gel

(b) Foam

(c) Aerosol

(d) Emulsion

✔️ Answer: (c) Aerosol

Smoke consists of solid particles dispersed in gas.
Such a system is called aerosol.

30. Milk is the example of

(a) Gel

(b) Foam

(c) Aerosol

(d) Emulsion

✔️ Answer: (d) Emulsion

Milk is a colloidal system where liquid fat droplets are dispersed in water.
Hence, it is an emulsion.

31. The process of precipitation of colloidal particles which can be accomplished by heating or by adding electrolyte is known as

(a) Coagulation

(b) Tyndall effect

(c) Electrophoresis

(d) None of them

✔️ Answer: (a) Coagulation

Coagulation is the process of precipitation of colloidal particles by heating or adding electrolyte.
It leads to formation of larger aggregates.

32. The dissolution of these solids in liquids solvent is exothermic:

(a) AlCl₃

(b) Na₂SO₄

(c) Lipids

(d) All of them

✔️ Answer: (d) All of them

All the given solids dissolution is exothermic.
The dissolution of some solids (AlCl₃, Na₂SO₄ etc.) and many lipids as well as gases in liquid solvent is exothermic and releases heat.
Solubility of these substance decreases with the rise of temperature.

33. The dissolution of gases in liquids solvent is:

(a) Exothermic

(b) Endothermic

(c) Both of them

(d) None of them

✔️ Answer: (a) Exothermic

Dissolution of gases in liquids generally releases heat.
Therefore, it is exothermic in nature.

34. The solubility of gases increases with the increasing pressure. This is known as

(a) Raoult’s law

(b) Henry’s law

(c) Le-Chatelier’s principle

(d) None of them

✔️ Answer: (b) Henry’s law

According to Henry’s law, solubility of gas in liquid is directly proportional to pressure.
Higher pressure increases gas solubility.

35. Isotonic solutions must have the same:

(a) Density

(b) Normality

(c) Volume

(d) Osmotic pressure

✔️ Answer: (d) Osmotic pressure

Isotonic solutions have equal osmotic pressure.
There is no net flow of solvent between them.

36. A solution of known strength or concentration is called:

(a) Standard solution

(b) Normal solution

(c) Molal solution

(d) Suspension

✔️ Answer: (a) Standard solution

A standard solution has accurately known concentration.
It is used in titration and analysis.

37. 5 liters of 0.5M HNO₃ solution contains:

(a) 2.5 moles

(b) 3moles

(c) 3.5 moles

(d) 4 moles

✔️ Answer: (a) 2.5 moles

Moles = Molarity × Volume = 0.5 × 5.
So, total moles = 2.5.

38. The semimolar solution of KOH contains:

(a) 56 g/dm³

(b) 5.6 g/dm³

(c) 28 g/dm³

(d) 112 g/dm³

✔️ Answer: (c) 28 g/dm³

Semimolar means 0.5 M solution.
Molar mass of KOH = 56 g, so 0.5 × 56 = 28 g/dm³.

39. How many gram of HCl are in 1000 g of a 10 % solution?

(a) 100 g

(b) 10 g

(c) 9.8 g

(d) 98 g

✔️ Answer: (a) 100 g

10% solution means 10 g solute per 100 g solution.
In 1000 g solution, HCl = 100 g.

40. Which of the following is a colligative property?

(a) Freezing point

(b) Sublimation temperature

(c) Osmatic pressure

(d) Melting point

✔️ Answer: (c) Osmatic pressure

Osmotic pressure depends on number of solute particles.
Hence, it is a colligative property.

41. The molarity of pure water is

(a) 50

(b) 100

(c) 18

(d) 55.5

✔️ Answer: (d) 55.5

1 liter of water contains 1000 g.
Moles = 1000/18 ≈ 55.5, so molarity is 55.5 M.

42. The mole fraction of a solution containing 3.0 g of urea per 250 g of water would be

(a) 0.00357

(b) 0.00643

(c) 0.99643

(d) None of these

✔️ Answer: (a) 0.00357

Moles urea = 3/60 = 0.05 mol; moles water = 250/18 ≈ 13.89 mol.
Mole fraction of urea = 0.05/(13.94) ≈ 0.0036, closest correct option given is 0.00357

43. How many gram of NaCl are to be taken to prepare 50 ml of 5% w/v salt solution?

(a) 2.5 g

(b) 10 g

(c) 25 g

(d) 80 g

✔️ Answer: (a) 2.5 g

5% w/v means 5 g in 100 mL.
For 50 mL, required NaCl = 2.5 g.

44. A solution of sucrose is 34.2%. The volume of solution containing one mole of solute is

(a) 342 cm³

(b) 1000 cm³

(c) 500 cm³

(d) 242 cm³

✔️ Answer: (b) 1000 cm³

34.2% sucrose solution = 34.2 g sucrose in 100 cm³ of solution
34.2 g sucrose is present in 100 cm³ of solution
1 g sucrose is present in 100/34.2 of solution
342 g (1mole) sucrose is present in (100/34.2) x 342 = 1000 cm³

45. 5 g of glucose is dissolved for 100 cm³ of solution. The percentage of solution is

(a) 5% v/w

(b) 5% v/v

(c) 5% w/v

(d) 5% w/w

✔️ Answer: (c) 5% w/v

5 g solute in 100 cm³ solution represents 5% weight/volume solution.
Hence, 5% w/v.

46. The molarity of 2% w/v NaOH solution is

(a) 0.25

(b) 0.1

(c) 0.05

(d) 0.5

✔️ Answer: (d) 0.5

2% w/v means 2 g in 100 mL, so 20 g in 1 L.
Moles = 20/40 = 0.5 M.

47. 9.8 g of H₂SO₄ is present in 100 mL solution what is the w/v%?

(a) 98

(b) 9.8

(c) 10

(d) 0.01

✔️ Answer: (b) 9.8

w/v% means grams per 100 mL.
So 9.8 g in 100 mL = 9.8% w/v.

48. The molality of an aqueous solution of sugar (C₁₂H₂₂O₁₁) is 1.62 m. Find the mole fractions of sugar?

(a) 0.0284

(b) 0.284

(c) 0.0815

(d) 0.815

✔️ Answer: (a) 0.0284

1.62 m means 1.62 mol sugar in 1 kg water (55.5 mol).
Mole fraction = 1.62/(1.62+55.5) ≈ 0.0284.

49. Molarity of 0.2 N H₂SO₄ is:

(a) 0.1

(b) 0.2

(c) 0.4

(d) None of these

✔️ Answer: (a) 0.1

H₂SO₄ has 2 equivalents per mole.
Molarity = Normality/2 = 0.2/2 = 0.1 M.

50. Which has the least freezing point?

(a) 1% sucrose

(b) 1% NaCl

(c) 1% CaCl₂

(d) 1% glucose

✔️ Answer: (c) 1% CaCl₂

Freezing point depression depends on number of particles formed.
CaCl₂ gives 3 ions, so it lowers freezing point the most.

51. Increasing the temperature of an aqueous solution will cause decrease in

(a) Molarity

(b) Molality

(c) Mole fraction

(d) % w/w

✔️ Answer: (a) Molarity

Molarity decreases with temperature as volume expands.
Molality and mole fraction remain nearly constant.

52. As compared to molar solution, in molal solution solvent quantity is:

(a) Lesser 

(b) Greater 

(c) Equal 

(d) Very large

✔️ Answer: (b) Greater

Molal solution is defined per kg of solvent, so solvent quantity is generally greater than in molar solutions.

53. The process of passing colloid through membrane to remove impurities is:

(a) Coagulation

(b) Dialysis

(c) Osmosis

(d) Filtration

✔️ Answer: (b) Dialysis

Dialysis removes impurities by passing solute through semipermeable membrane, leaving colloid behind.

54. Which law specifically governs the relative lowering of vapour pressures in solutions?

(a) van’t Hoff law

(b) Raoult’s law

(c) Gay-Lussac’s law

(d) Amagat’s law

✔️ Answer: (b) Raoult’s law

Raoult’s law states that relative lowering of vapour pressure is equal to mole fraction of solute.

55. Which of the following aqueous solutions has the highest vapour pressure at 300 K?

(a) 1 M Na₃PO₄

(b) 1 M CaCl₂

(c) 1 M KNO₃

(d) 1 M C₆H₁₂O₆

✔️ Answer: (d) 1 M C₆H₁₂O₆

Vapour pressure decreases with number of ions/particles.
C6H12O6 is molecular, no dissociation → highest vapour pressure.

56. Molarity of 4% (w/v) solution of NaOH is:

(a) 0.1

(b) 0.5

(c) 0.001

(d) 1.0

✔️ Answer: (b) 1.0

4 g in 100 mL → 40 g/L.
Moles = 40/40 = 1 M.

57. Which salt or compounds shows retrograde solubility (decreases with rise in T)?

(a) AlCl₃

(b) Na₂SO₄

(c) lipids

(d) All of them

✔️ Answer: (d) All of them

The dissolution of some solids (AlCl₃, Na₂SO₄ etc.) and many lipids as well as gases in liquid solvent is exothermic and releases heat.
Solubility of these substance decreases with the rise of temperature (retrograde solubility).

58. Which of the following solutions obeys Raoult’s law most closely?

(a) NaCl in water

(b) Sugar in water

(c) Benzene + Toluene

(d) HCl in water

✔️ Answer: (c) Benzene + Toluene

Benzene + Toluene are similar, non-polar liquids forming an ideal solution.
They obey Raoult’s law most closely.

59. Raoult’s law is a special case of:

(a) Boyle’s law

(b) Henry’s law

(c) Dalton’s law of partial pressures

(d) Gay-Lussac’s law

✔️ Answer: (c) Dalton’s law of partial pressures

Raoult’s law can be derived from Dalton’s law of partial pressures for ideal mixtures.

60. Which of the following is NOT an example of an Ideal solution?

(a) Benzene + Toluene

(b) n-Hexane + n-Heptane

(c) Ethyl alcohol + Water

(d) Ethyl bromide + Ethyl chloride

✔️ Answer: (c) Ethyl alcohol + Water

Ethyl alcohol + Water is non-ideal due to hydrogen bonding interactions.
It does not obey Raoult’s law.

61. Which of the following solutions has the highest boiling point?

(a) 5.85% solution of NaCl

(b) 18.0% solution of glucose

(c) 6.0% solution of urea

(d) all have same boiling point

✔️ Answer: (a) 5.85% solution of NaCl

Boiling point elevation depends on number of particles. NaCl dissociates into 2 ions → more effect → highest boiling point.

62. In cold countries, ethylene glycol is added to water in radiators of cars during winter. It results in:

(a) Lowering in boiling point

(b) Reducing viscosity

(c) Reducing specific heat

(d) Lowering in freezing point

✔️ Answer: (d) Lowering in freezing point

Ethylene glycol decreases freezing point via freezing point depression → prevents radiator water from freezing.

63. The colligative property responsible for salting of roads in winter is:

(a) Lowering of vapor pressure

(b) Elevation of boiling point

(c) Depression of freezing point

(d) Osmotic pressure

✔️ Answer: (c) Depression of freezing point

Salt lowers freezing point of water → ice melts at lower temperature.

64. Which is NOT a colligative property?

(a) Vapour pressure

(b) Osmatic pressure

(c) Depression of freezing point

(d) Elevation of boiling point

✔️ Answer: (a) Vapour pressure

Strictly speaking, colligative properties are: ΔTf, ΔTb, osmotic pressure, and relative lowering of vapour pressure. Vapour pressure itself is not a colligative property.

65. At equilibrium the rate of dissolution of a solid solute in a volatile liquid solvent is __________.

(a) less than the rate of crystallization

(b) greater than the rate of crystallization

(c) equal to the rate of crystallization

(d) zero

✔️ Answer: (c) equal to the rate of crystallization

At equilibrium, dissolution rate = crystallization rate → dynamic equilibrium.

66. What happens to vapor pressure when you add a solute to a solution?

(a) It lowers the vapor pressure

(b) It has no effect

(c) It raises the vapor pressure

(d) It causes the reaction to reach equilibrium

✔️ Answer: (a) It lowers the vapor pressure

Adding a solute lowers the number of solvent molecules at the surface → decreases vapor pressure (Raoult's law).

67. In a closed container, vapour pressure of a liquid depends upon

(a) Volume of the container

(b) Temperature

(c) Volume of the liquid

(d) All of the above

✔️ Answer: (b) Temperature

Vapour pressure depends only on temperature, not container volume or liquid volume.

68. On dissolving sugar in water at room temperature solution feels cool to touch. Under which of the following cases dissolution of sugar will be most rapid?

(a) Sugar crystals in cold water

(b) Sugar crystals in hot water

(c) Powdered sugar in cold water

(d) Powdered sugar in hot water

✔️ Answer: (d) Powdered sugar in hot water

Smaller particles and higher temperature → faster dissolution.

69. The relative lowering of vapor pressure is equal to:

(a) Mole fraction of solute

(b) Mole fraction of solvent

(c) Moles of solute / mass of solvent

(d) Normality of solute

✔️ Answer: (a) Mole fraction of solute

Raoult’s law: ΔP/P₀ = mole fraction of solute.

70. For an endothermic dissolution process, solubility will:

(a) Increase with rise in temperature

(b) Decrease with rise in temperature

(c) Remain constant

(d) Independent of temperature

✔️ Answer: (a) Increase with rise in temperature

Endothermic dissolution absorbs heat → solubility increases as temperature rises.

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✍️ جونؔ ایلیا

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🌟 کیا عجب عیش تھے شکایت کے

💧 یہ جو آنسو ہیں، رخصتی آنسو
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⚖️ ہم ہی مفتی ہیں اہلسنت کے

💉 ہم تو بس خون تھوکتے ہیں میاں
🛠️ نہیں خوگر کسی مشقت کے

💕 یہ جو لمحے ہیں وصال کے ہیں میاں
🌙 ہیں یہ لمحے تمام ہجرت کے

✨ جونؔ، یزدان و آدم و ابلیس
📖 ہیں عجب معجزے حکایت کے

✍️ جونؔ ایلیا

💥 غزل ۔۔۔۔ جونؔ ایلیا 💥

❤️ دل نے وفا کے نام پر کارِ وفا نہیں کیا
💔 خود کو ہلاک کر لیا، خود کو فدا نہیں کیا

🤔 کیسے کہیں کہ تجھ کو بھی ہم سے ہے واسطہ کوئی
😔 تو نے تو ہم سے آج تک کوئی گلہ نہیں کیا

⚖️ تو بھی کسی کے باب میں عہد شکن ہو غالباً
📜 میں نے بھی ایک شخص کا قرض ادا نہیں کیا

🗣️ جو بھی ہو تم پہ معترض، اُس کو یہی جواب دو
🌟 آپ بہت شریف ہیں، آپ نے کیا نہیں کیا

👑 جس کو بھی شیخ و شاہ نے حکمِ خُدا دیا قرار
🙏 ہم نے نہیں کیا وہ کام، ہاں باخُدا نہیں کیا

📖 نسبتِ علم ہے بہت حاکمِ وقت کو عزیز
💡 اُس نے تو کارِ جہل بھی بے علما نہیں کیا

✍️ جونؔ ایلیا

💥 غزل ۔۔۔۔ جونؔ ایلیا 💥

📖 حال خوش تذکرہ نگاروں کا
🌙 تھا تو اک شہر خاکساروں کا

💔 پہلے رہتے تھے کوچۂ دل میں
😔 اب پتہ کیا ہے دل فگاروں کا

🚪 کوئے جاناں کی ناکہ بندی ہے
🛏️ بسترا اب کہاں ہے یاروں کا

🌬️ چلتا جاتا ہے سانس کا لشکر
🕊️ کون پُرساں ہے یادگاروں کا

💭 اپنے اندر گھسٹ رہا ہوں میں
🌌 مجھ سے کیا ذکر رہ گزاروں کا

🎉 ان سے جو شہر میں ہیں بے دعویٰ
🌟 عیش مت پوچھ دعویداروں کا

⚔️ کیسا یہ معرکہ ہے برپا جو
🐎 نہ پیادوں کا نہ سواروں کا

🖋️ بات تشبیہہ کی نہ کیجیو تُو
📚 دہر ہے صرف استعاروں کا

💔 میں تو خیر اپنی جان ہی سے گیا
🤝 کیا ہوا جانے جانثاروں کا

🔥 کچھ نہیں اب سوائے خاکستر
🌪️ ایک جلسہ تھا شعلہ خواروں کا

✍️ جونؔ ایلیا

💥 حضرتِ جونؔ ایلیا 💥 🎨 خاتم الشعرا و شاعرِ بے مثل و بے بدل 🎨

👥 اپنے سب یار کام کررہے ہیں
🌙 اور ہم ہیں کہ نام کر رہے ہیں

🏛️ آنے والی اپر کلاس کی ہے
✨ ہم جو یہ اہمتمام کررہے ہیں

⚔️ تیغ بازی کا شوق اپنی جگہ
🔥 آپ تو قتلِ عام کررہے ہیں

🎶 داد و تحسین کا یہ شور ہے کیوں
💭 ہم تو خود سے کلام کررہے ہیں

😔 ہے وہ بے چارگی کا حال کہ ہم
🤝 ہر کسی کو سلام کررہے ہیں

🕊️ ہم تو بس یاد کے ہیں لوگ میاں
💔 اپنا ہونا حرام کررہے ہیں

👑 اک قتالہ چاہئے ہم کو
📢 ہم یہ اعلانِ عام کررہے ہیں

🍷 کیا بھلا ساغرِ سفال کہ ہم
🥂 ناف پیالے کو جام کررہے ہیں

📝 ہم تو آئے تھے عرضِ مطلب کو
🙏 اور وہ احترام کررہے ہیں

💨 نہ اٹھے آہ کا دھواں بھی کہ وہ
🌌 کوئے دل میں خرام کررہے ہیں

💋 اس کے ہونٹوں پہ رکھ کے ہونٹ اپنے
🕊️ بات ہی ہم تمام کررہے ہیں

🎉 ہم عجب ہیں کہ اس کے کوچے میں
🥁 بے سبب دھوم دھام کررہے ہیں

⚔️ کرکے بے پوشش اس صنم کو ہم
🗡️ تیغ کو بے نیام کررہے ہیں

🎭 کوئی بھی فن ہمیں نہیں آتا
⏳ دم کو بس دوام کررہے ہیں

🌟 ہم جو ہر لمحہ جی رہے ہیں جونؔ
🏛️ ہم ابد میں قیام کررہے ہیں

✍️ حضرتِ جونؔ ایلیا

💥 حضرتِ جونؔ ایلیا 💥 🎨 خاتم الشعرا و شاعرِ بے مثل و بے بدل 🎨

🤝 نیا اک رشتہ پیدا کیوں کریں ہم
💔 بچھڑنا ہے تو جھگڑا کیوں کریں ہم

🌙 خموشی سے ادا ہو رسمِ دوری
🔥 کوئی ہنگامہ برپا کیوں کریں ہم

📖 سنا دیں عصمتِ مریم کا قصّہ؟
❌ پر اب اس باب کو وا کیوں کریں ہم

💭 زلیخاے عزیزاں بات یہ ہے
⚖️ بھلا گھاٹے کا سودا کیوں کریں ہم

💕 ہماری ہی تمنّا کیوں کرو تم
🌟 تمہاری ہی تمنّا کیوں کریں ہم

🕰️ کیا تھا عہد جب لمحوں میں ہم نے
📜 تو ساری عمر ایفا کیوں کریں ہم

🗑️ اٹھا کر کیوں نہ پھینکیں ساری چیزیں
🚶 فقط کمروں میں ٹہلا کیوں کریں ہم

👥 جو اک نسل فرومایہ کو پہنچے
💡 وہ سرمایہ اکٹھا کیوں کریں ہم

🌍 نہیں دنیا کو جب پروا ہماری
😔 تو پھر دنیا کی پروا کیوں کریں ہم

🏙️ برہنہ ہیں سرِ بازار تو کیا
👀 بھلا اندھوں سے پردہ کیوں کریں ہم

🏠 ہیں باشندے اسی بستی کے ہم بھی
🤝 سو خود پر بھی بھروسا کیوں کریں ہم

💀 چبالیں کیوں نہ خود ہی اپنا ڈھانچا
🍽️ تمہیں راتب مہیا کیوں کریں ہم

⚰️ پڑی رہنے دو انسانوں کی لاشیں
🌪️ زمیں کا بوجھ ہلکا کیوں کریں ہم

🕌 یہ بستی ہے مسلمانوں کی بستی
🕊️ یہاں کارِ مسیحا کیوں کریں ہم

✍️ حضرتِ جونؔ ایلیا

🌸🔥 جون ایلیا دل کو چھو لینے والی غزل🔥🌸

💫 حال یہ ہے کہ خواہش پرسش حال بھی نہیں
💫 اس کا خیال بھی نہیں اپنا خیال بھی نہیں

🌿 اے شجر حیات شوق ایسی خزاں رسیدگی
🌿 پوشش برگ و گل تو کیا جسم پہ چھال بھی نہیں

📖 مجھ میں وہ شخص ہو چکا جس کا کوئی حساب تھا
📖 سود ہے کیا زیاں ہے کیا اس کا سوال بھی نہیں

🎶 مست ہیں اپنے حال میں دل زدگان و دلبراں
🎶 صلح و سلام تو کجا بحث و جدال بھی نہیں

🌟 تو مرا حوصلہ تو دیکھ داد تو دے کہ اب مجھے
🌟 شوق کمال بھی نہیں خوف زوال بھی نہیں

🌌 خیمہ گاہ نگاہ کو لوٹ لیا گیا ہے کیا
🌌 آج افق کے دوش پر گرد کی شال بھی نہیں

🍃 اف یہ فضائے احتیاط تا کہیں اڑ نہ جائیں ہم
🍃 باد جنوب بھی نہیں باد شمال بھی نہیں

💔 وجہ معاش بے دلاں یاس ہے اب مگر کہاں
💔 اس کے ورود کا گماں فرض محال بھی نہیں

⏳ غارت روز و شب تو دیکھ وقت کا یہ غضب تو دیکھ
⏳ کل تو نڈھال بھی تھا میں آج نڈھال بھی نہیں

🌙 میرے زمان و ذات کا ہے یہ معاملہ کہ اب
🌙 صبح فراق بھی نہیں شام وصال بھی نہیں

🌹 پہلے ہمارے ذہن میں حسن کی اک مثال تھی
🌹 اب تو ہمارے ذہن میں کوئی مثال بھی نہیں

🔥 میں بھی بہت عجیب ہوں اتنا عجیب ہوں کہ بس
🔥 خود کو تباہ کر لیا اور ملال بھی نہیں

💥 جونؔ ایلیا 🎯 💥

😊 گاہے گاہے بس اب یہی ہو کیا
❤️ تم سے مل کر بہت خوشی ہو کیا

🤝 مل رہی ہو بڑے تپاک کے ساتھ
🕊️ مجھ کو یکسر بھلا چکی ہو کیا

🌙 یاد ہیں اب بھی اپنے خواب تمہیں
😔 مجھ سے مل کر اداس بھی ہو کیا

💭 بس مجھے یوں ہی اک خیال آیا
🤔 سوچتی ہو تو سوچتی ہو کیا

💔 اب مری کوئی زندگی ہی نہیں
🌹 اب بھی تم میری زندگی ہو کیا

🔥 کیا کہا عشق جاودانی ہے!
🕰️ آخری بار مل رہی ہو کیا

🌫️ ہاں فضا یاں کی سوئی سوئی سی ہے
☀️ تو بہت تیز روشنی ہو کیا

🎭 میرے سب طنز بے اثر ہی رہے
🚶‍♀️ تم بہت دور جا چکی ہو کیا

🕯️ دل میں اب سوز انتظار نہیں
⚡ شمع امید بجھ گئی ہو کیا

🌊 اس سمندر پہ تشنہ کام ہوں میں
💧 بان تم اب بھی بہہ رہی ہو کیا

🌹 کلاسیکی غزل ✨

💋 آج لب گہر فشاں آپ نے وا نہیں کیا
🌸 تذکرۂ خجستۂ آب و ہوا نہیں کیا

🤔 کیسے کہیں کہ تجھ کو بھی ہم سے ہے واسطہ کوئی
💔 تو نے تو ہم سے آج تک کوئی گلہ نہیں کیا

😔 جانے تری نہیں کے ساتھ کتنے ہی جبر تھے کہ تھے
🙏 میں نے ترے لحاظ میں تیرا کہا نہیں کیا

😳 مجھ کو یہ ہوش ہی نہ تھا تو مرے بازوؤں میں ہے
💭 یعنی تجھے ابھی تلک میں نے رہا نہیں کیا

⚖️ تو بھی کسی کے باب میں عہد شکن ہو غالباً
🤝 میں نے بھی ایک شخص کا قرض ادا نہیں کیا

👀 ہاں وہ نگاہ ناز بھی اب نہیں ماجرا طلب
🌾 ہم نے بھی اب کی فصل میں شور بپا نہیں کیا

💔 دل نے وفا کے نام پر کارِ وفا نہیں کیا
🔥 خود کو ہلاک کر لیا، خود کو فدا نہیں کیا

🕊️ جو بھی ہو تم پہ معترض، اُس کو یہی جواب دو
😊 آپ بہت شریف ہیں، آپ نے کیا نہیں کیا

📿 جس کو بھی شیخ و شاہ نے حکمِ خُدا دیا قرار
🙌 ہم نے نہیں کیا وہ کام، ہاں باخُدا نہیں کیا

📚 نسبتِ علم ہے بہت حاکمِ وقت کو عزیز
🚫 اُس نے تو کارِ جہل بھی بے علما نہیں کیا

🎭 غزل از جون ایلیا

🌹💔 تجھ بدن پر ہم نے جانیں واریاں
😢🔥 تجھ کو تڑپانے کی ہیں تیاریاں

🕰️💭 کر رہے ہیں یاد اسے ہم روز و شب
🧠🌀 ہیں بھُلانے کی اسے تیاریاں

😊✨ تھا کبھی میں اک ہنسی اُن کے لیے
😭💔 رو رہی ہیں اب مجھے مت ماریاں

🎭⚖️ جھوٹ سچ کے کھیل میں ہلکان ہیں
👧💫 خوب ہیں یہ لڑکیاں بےچاریاں

🤐📜 شعر تو کیا بات کہہ سکتے نہیں
🏛️👔 جو بھی نوکر جونؔ ہیں سرکاریاں

🕴️⌛ جو میاں جاتے ہیں دفتر وقت پر
🧳🚶 اُن سے ہیں اپنی جُدا دشواریاں

📜⚖️ ہم بھلا آئین اور قانون کی
⛓️😔 کب تلک سہتے رہیں غداریاں

🩸🔔 سُن رکھو اے شہر دارو ! خون کی
🌊💥 ہونے ہی والی ہیں ندیاں جاریاں

🤝💬 ہیں سبھی سے جن کی گہری یاریاں
😞⚠️ سُن میاں ہوتی ہیں ان کی خواریاں

🎉🍇 ہے خوشی عیاروں کا اک ثمر
😢🌧️ غم کی بھی اپنی ہیں کچھ عیاریاں

🌌✨ ذرّے ذرّے پر نہ جانے کس لیے
🌠💫 ہر نفس ہیں کہکشائیں طاریاں

🧵❤️ اس نے دل دھاگے ہیں ڈالے پاؤں میں
⛓️⚖️ یہ تو زنجیریں ہیں بےحد بھاریاں

🧬📚 تم کو ہے آداب کا برص و جزام
🧪😷 ہیں ہماری اور ہی بیماریاں

🌈💭 خواب ہائے جاودانی پر مرے
💡⚡ چل رہی ہیں روشنی کی آریاں

🧟‍♂️👥 ہیں یہ سندھی اور مہاجر ہڈ حرام
🥬💸 کیوں نہیں یہ بیچتے ترکاریاں

🤔🧠 یار! سوچو تو عجب سی بات ہے
💔😓 اُس کے پہلو میں مری قلقاریاں

🛑📖 ختم ہے بس جونؔ پر اُردو غزل
🌹🩸 اس نے کی ہیں خون کی گل کاریاں

💥 جونؔ ایلیا 🎯 💥

💭 بے دلی کیا یوں ہی دن گزر جائیں گے
💔 صرف زندہ رہے ہم تو مر جائیں گے

🎶 رقص ہے رنگ پر رنگ ہم رقص ہیں
🌌 سب بچھڑ جائیں گے سب بکھر جائیں گے

🍷 یہ خراباتیان خرد باختہ
🌅 صبح ہوتے ہی سب کام پر جائیں گے

💕 کتنی دل کش ہو تم کتنا دلجو ہوں میں
⚡ کیا ستم ہے کہ ہم لوگ مر جائیں گے

🌟 ہے غنیمت کہ اسرار ہستی سے ہم
🕊️ بے خبر آئے ہیں بے خبر جائیں گے

💥 جونؔ ایلیا ✨🎯💔

🌙 گاہے گاہے بس اب یہی ہو کیا
🌹 تم سے مل کر بہت خوشی ہو کیا

💫 مل رہی ہو بڑے تپاک کے ساتھ
🔥 مجھ کو یکسر بھلا چکی ہو کیا

🌌 یاد ہیں اب بھی اپنے خواب تمہیں
😔 مجھ سے مل کر اداس بھی ہو کیا

💭 بس مجھے یوں ہی اک خیال آیا
🤔 سوچتی ہو تو سوچتی ہو کیا

💔 اب مری کوئی زندگی ہی نہیں
🌷 اب بھی تم میری زندگی ہو کیا

✨ کیا کہا عشق جاودانی ہے!
🕊️ آخری بار مل رہی ہو کیا

🌟 ہاں فضا یاں کی سوئی سوئی سی ہے
🌞 تو بہت تیز روشنی ہو کیا

🌿 میرے سب طنز بے اثر ہی رہے
🚶 تم بہت دور جا چکی ہو کیا

🕯️ دل میں اب سوز انتظار نہیں
🔥 شمع امید بجھ گئی ہو کیا

🌊 اس سمندر پہ تشنہ کام ہوں میں
💧 بان تم اب بھی بہہ رہی ہو کیا

💥 جونؔ ایلیا ✨🎯💔

ہم جی رہے ہیں کوئی بہانہ کیے بغیر
اس کے بغیر اس کی تمنا کیے بغیر

انبار اس کا پردۂ حرمت بنا میاں
دیوار تک نہیں گری پردا کیے بغیر

یاراں وہ جو ہے میرا مسیحائے جان و دل
بے حد عزیز ہے مجھے اچھا کیے بغیر

میں بستر خیال پہ لیٹا ہوں اس کے پاس
صبح ازل سے کوئی تقاضا کیے بغیر

اس کا ہے جو بھی کچھ ہے مرا اور میں مگر
وہ مجھ کو چاہئے کوئی سودا کیے بغیر

یہ زندگی جو ہے اسے معنیٰ بھی چاہیے
وعدہ ہمیں قبول ہے ایفا کیے بغیر

اے قاتلوں کے شہر بس اتنی ہی عرض ہے
میں ہوں نہ قتل کوئی تماشا کیے بغیر

مرشد کے جھوٹ کی تو سزا بے حساب ہے
تم چھوڑیو نہ شہر کو صحرا کیے بغیر

ان آنگنوں میں کتنا سکون و سرور تھا
آرائش نظر تری پروا کیے بغیر

یاراں خوشا یہ روز و شب دل کہ اب ہمیں
سب کچھ ہے خوش گوار گوارا کیے بغیر

گریہ کناں کی فرد میں اپنا نہیں ہے نام
ہم گریہ کن ازل کے ہیں گریہ کیے بغیر

آخر ہیں کون لوگ جو بخشے ہی جائیں گے
تاریخ کے حرام سے توبہ کیے بغیر

وہ سنی بچہ کون تھا جس کی جفا نے جونؔ
شیعہ بنا دیا ہمیں شیعہ کیے بغیر

اب تم کبھی نہ آؤ گے یعنی کبھی کبھی
رخصت کرو مجھے کوئی وعدہ کیے بغیر

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Jaun Elia | Tum Meri Aakhri Mohabbat Ho 🥀

تم مری آخری محبت ہو

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