Class 12 Chemistry Notes

Chapter 1 – Solutions

1. What is a Solution?

A solution is a homogeneous mixture in which the composition is the same throughout.

Components

Solvent

Present in larger quantity
Determines the physical state of the solution

Solute

Present in smaller quantity
Dissolves in the solvent

Example:

Salt + Water
Sugar + Water
Air (mixture of gases)

2. Types of Solutions

Solute	Solvent	Example
Gas	Gas	Air
Gas	Liquid	Oxygen in water
Liquid	Liquid	Alcohol in water
Solid	Liquid	Sugar in water
Gas	Solid	Hydrogen in palladium
Liquid	Solid	Dental amalgam
Solid	Solid	Brass

3. Ways to Express Concentration

(i) Mass Percentage

\text{Mass %}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times100

Used in industries.

(ii) Volume Percentage

\text{Volume %}=\frac{\text{Volume of solute}}{\text{Volume of solution}}\times100

Used for liquid-liquid mixtures.

(iii) Mass by Volume %

$\frac{\text{Mass of solute (g)}}{100\;mL\;solution}$ 100mLsolutionMass of solute (g)

Commonly used in medicines.

(iv) Parts Per Million (ppm)

Used for very dilute solutions. $ppm=\frac{\text{Amount of solute}}{\text{Amount of solution}}\times10^6$ ppm=Amount of solutionAmount of solute×106

Examples:

Drinking water impurities
Air pollution

(v) Mole Fraction

$x=\frac{\text{Moles of component}}{\text{Total moles}}$ x=Total molesMoles of component

Important Points:

No unit
Sum of mole fractions = 1

(vi) Molarity (M)

$M=\frac{\text{Moles of solute}}{\text{Volume of solution (L)}}$ M=Volume of solution (L)Moles of solute

Unit:
mol L⁻¹

Depends on temperature.

(vii) Molality (m)

$m=\frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}}$ m=Mass of solvent (kg)Moles of solute

Unit:
mol kg⁻¹

Does not depend on temperature.

4. Solubility

Solubility is the maximum amount of solute that dissolves in a given amount of solvent at a fixed temperature.

Factors affecting solubility:

Nature of Solute and Solvent

“Like dissolves like.”

Polar → Polar
Non-polar → Non-polar

Effect of Temperature

For solids:

Usually increases with temperature.

For gases:

Decreases with temperature.

Effect of Pressure

For solids:

Almost no effect.

For gases:

Solubility increases with pressure.

5. Henry’s Law

Statement:

At constant temperature, the solubility of a gas is directly proportional to its partial pressure above the solution.

Formula $P=K_Hx$ P=KHx

Where

P = Partial pressure

KH = Henry’s constant

x = Mole fraction

Applications

✔ Soft drinks are bottled under high pressure.

✔ Scuba diving

✔ High-altitude breathing problems

✔ Dissolved oxygen in lakes

6. Vapour Pressure

Vapour pressure is the pressure exerted by vapour when liquid and vapour are in equilibrium.

7. Raoult’s Law

For volatile liquids, $P=xP^\circ$ P=xP∘

Partial vapour pressure depends on mole fraction.

Total vapour pressure $P_{total}=P_1+P_2$ Ptotal=P1+P2

Applicable to ideal solutions.

8. Ideal Solution

Characteristics

✔ Obeys Raoult’s law at every concentration.

✔ Heat of mixing = 0

✔ Volume change = 0

Examples

Benzene + Toluene
n-Hexane + n-Heptane

9. Non-Ideal Solution

Does not obey Raoult’s law.

Two types

Positive Deviation

Vapour pressure increases.
A-B attraction is weaker.

Examples

Ethanol + Acetone

Negative Deviation

Vapour pressure decreases.
A-B attraction is stronger.

Examples

Chloroform + Acetone
Phenol + Aniline

10. Azeotropes

Mixtures that boil at constant temperature.

Cannot be separated completely by fractional distillation.

Types

Minimum boiling

Example:
Ethanol + Water

Maximum boiling

Example:
Nitric acid + Water

11. Colligative Properties

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

They are:

Relative lowering of vapour pressure
Elevation of boiling point
Depression in freezing point
Osmotic pressure

12. Relative Lowering of Vapour Pressure

$\frac{P^\circ-P}{P^\circ}=x_{solute}$ P∘P∘−P=xsolute

Used for finding molar mass.

13. Elevation of Boiling Point

Adding a non-volatile solute raises the boiling point.

Formula $\Delta T_b=K_bm$ ΔTb=Kbm

Where

Kb = Ebullioscopic constant

m = Molality

14. Depression in Freezing Point

Adding a non-volatile solute lowers the freezing point.

Formula $\Delta T_f=K_fm$ ΔTf=Kfm

Where

Kf = Cryoscopic constant

15. Osmosis

Movement of solvent molecules through a semipermeable membrane from dilute solution to concentrated solution.

16. Osmotic Pressure

Formula $\pi=CRT$ π=CRT

Also, $\pi=\frac{nRT}{V}$ π=VnRT

Applications

Reverse osmosis (RO)
Preservation of food
Medical saline solutions

17. Van’t Hoff Factor (i)

Accounts for association or dissociation of solute particles.

Formula $i=\frac{\text{Observed colligative property}}{\text{Calculated colligative property}}$ i=Calculated colligative propertyObserved colligative property

i > 1 → Dissociation

i < 1 → Association

i = 1 → No association/dissociation

Examples

NaCl → i ≈ 2

CaCl₂ → i ≈ 3

Glucose → i = 1

Important Formula Sheet

Mass % = (Mass of solute / Mass of solution) × 100
Volume % = (Volume of solute / Volume of solution) × 100
Mole Fraction = Moles of component / Total moles
Molarity = Moles / Volume (L)
Molality = Moles / Mass of solvent (kg)
Henry’s Law: $P=K_Hx$ P=KHx
Raoult’s Law: $P=xP^\circ$ P=xP∘
Relative lowering of vapour pressure = $(P^\circ-P)/P^\circ$ (P∘−P)/P∘
Elevation of boiling point = $K_bm$ Kbm
Depression in freezing point = $K_fm$ Kfm
Osmotic pressure = $CRT$ CRT

Section A – MCQs (Part 1: Core Concepts)

Section A: Top 30 MCQs (Part 1)

1. A solution is a:

A) Heterogeneous mixture
B) Homogeneous mixture
C) Suspension
D) Colloid

Ans: B

2. The component present in larger amount in a solution is called:

A) Solute
B) Solvent
C) Saturated solution
D) Diluent

Ans: B

3. Molarity is defined as:

A) Moles of solute/kg solvent
B) Moles of solute/L solution
C) Mass of solute/L solvent
D) Mole fraction of solute

Ans: B

4. Molality is independent of:

A) Pressure
B) Temperature
C) Mass
D) Moles

Ans: B

5. Which concentration term has no unit?

A) Molarity
B) Molality
C) Mole fraction
D) ppm

Ans: C

6. The sum of mole fractions of all components is:

A) 0
B) 1
C) 10
D) Depends on solvent

Ans: B

7. Solubility of gases generally decreases with:

A) Increase in pressure
B) Decrease in temperature
C) Increase in temperature
D) Increase in surface area

Ans: C

8. Henry’s law relates:

A) Pressure and gas solubility
B) Temperature and pressure
C) Volume and temperature
D) Mass and volume

Ans: A

9. Henry’s law equation is:

A) P = xP°
B) P = KHx
C) π = CRT
D) PV=nRT

Ans: B

10. Carbon dioxide is dissolved in soft drinks under:

A) Low pressure
B) High pressure
C) Vacuum
D) Zero pressure

Ans: B

11. According to Raoult’s law, partial vapour pressure is proportional to:

A) Mass fraction
B) Mole fraction
C) Molality
D) Molarity

Ans: B

12. Addition of a non-volatile solute to a solvent causes vapour pressure to:

A) Increase
B) Decrease
C) Remain constant
D) Become zero

Ans: B

13. An ideal solution obeys:

A) Henry’s law
B) Boyle’s law
C) Raoult’s law
D) Charles’ law

Ans: C

14. For an ideal solution:

A) ΔHmix = 0 and ΔVmix = 0
B) ΔHmix > 0
C) ΔVmix < 0
D) Vapour pressure is zero

Ans: A

15. Benzene and toluene form approximately:

A) Non-ideal solution
B) Ideal solution
C) Suspension
D) Azeotrope

Ans: B

16. Positive deviation from Raoult’s law occurs when:

A) A-B attraction is stronger
B) A-B attraction is weaker
C) Solute does not dissolve
D) Pressure decreases

Ans: B

17. Negative deviation occurs due to:

A) Weak solute-solvent interaction
B) Strong solute-solvent interaction
C) High temperature only
D) Low density

Ans: B

18. Minimum boiling azeotrope shows:

A) Negative deviation
B) Positive deviation
C) No deviation
D) Ideal behaviour

Ans: B

19. Ethanol-water mixture forms:

A) Maximum boiling azeotrope
B) Minimum boiling azeotrope
C) Ideal solution
D) Solid solution

Ans: B

20. Colligative properties depend on:

A) Nature of solute
B) Number of solute particles
C) Colour of solution
D) Density only

Ans: B

21. Elevation in boiling point is represented by:

A) ΔTf
B) ΔTb
C) π
D) KH

Ans: B

22. Depression in freezing point is:

A) ΔTb = Kbm
B) ΔTf = Kfm
C) P = KHx
D) π = CRT

Ans: B

23. Osmotic pressure is represented by:

A) P
B) π
C) Kf
D) Kb

Ans: B

24. The equation for osmotic pressure is:

A) π = CRT
B) P = KHx
C) P = xP°
D) ΔTf = Kfm

Ans: A

25. Solutions having same osmotic pressure are:

A) Hypertonic
B) Hypotonic
C) Isotonic
D) Ideal

Ans: C

26. Van’t Hoff factor is represented by:

A) K
B) i
C) x
D) R

Ans: B

27. Van’t Hoff factor for glucose solution is:

A) 0
B) 1
C) 2
D) 3

Ans: B

28. NaCl solution shows abnormal colligative properties due to:

A) Association
B) Dissociation
C) Evaporation
D) Condensation

Ans: B

29. Van’t Hoff factor for association is:

A) Greater than 1
B) Equal to 1
C) Less than 1
D) Infinite

Ans: C

30. Reverse osmosis is used for:

A) Gas purification
B) Water desalination
C) Increasing boiling point
D) Measuring vapour pressure

Ans: B

Section B: Assertion–Reason Questions (15 Important)

Choose the correct option:

A) Both Assertion and Reason are true, and Reason is the correct explanation.
B) Both are true, but Reason is not the correct explanation.
C) Assertion is true, Reason is false.
D) Assertion is false, Reason is true.

1.

Assertion: Molality is preferred over molarity for colligative property calculations.
Reason: Molality does not change with temperature.

Ans: A

2.

Assertion: Molarity changes with temperature.
Reason: Volume of solution changes with temperature.

Ans: A

3.

Assertion: Mole fraction has no unit.
Reason: It is the ratio of moles of one component to total moles.

Ans: A

4.

Assertion: Solubility of gases decreases with increase in temperature.
Reason: Dissolution of gases is generally an exothermic process.

Ans: A

5.

Assertion: Carbonated drinks are prepared under high pressure.
Reason: Gas solubility increases with increase in pressure.

Ans: A

6.

Assertion: Lower Henry’s constant means higher gas solubility.
Reason: KH is inversely related to gas solubility.

Ans: A

7.

Assertion: Ideal solutions obey Raoult’s law.
Reason: A-B interactions are nearly equal to A-A and B-B interactions.

Ans: A

8.

Assertion: Addition of non-volatile solute decreases vapour pressure.
Reason: Mole fraction of solvent decreases.

Ans: A

9.

Assertion: Positive deviation gives higher vapour pressure than expected.
Reason: Solute-solvent attraction is weaker.

Ans: A

10.

Assertion: Negative deviation gives lower vapour pressure.
Reason: Solute-solvent attraction is stronger.

Ans: A

11.

Assertion: Azeotropes cannot be separated by fractional distillation.
Reason: Vapour and liquid phases have identical composition.

Ans: A

12.

Assertion: Boiling point increases after adding non-volatile solute.
Reason: Vapour pressure decreases.

Ans: A

13.

Assertion: Osmotic pressure is useful for finding molar mass of proteins.
Reason: Proteins may decompose on heating.

Ans: A

14.

Assertion: NaCl solution has higher colligative effect than glucose solution of same concentration.
Reason: NaCl produces more particles due to ionisation.

Ans: A

15.

Assertion: Van’t Hoff factor for associated solute is less than one.
Reason: Association decreases number of particles.

Ans: A

Section C: Fill in the Blanks (20 Important Only)

1. A solution is a ______ mixture.

Ans: homogeneous

2. The component present in larger amount in a solution is called ______.

Ans: solvent

3. The component present in smaller amount is called ______.

Ans: solute

4. Number of moles of solute per litre of solution is called ______.

Ans: molarity

5. Number of moles of solute per kg of solvent is called ______.

Ans: molality

6. The symbol for mole fraction is ______.

Ans: x

7. The sum of mole fractions of all components is ______.

Ans: 1

8. Molality is independent of ______.

Ans: temperature

9. Henry’s law is represented as ______.

Ans: P = KHx

10. Solubility of gases increases with increase in ______.

Ans: pressure

11. Raoult’s law is applicable to ______ solutions.

Ans: ideal

12. Addition of a non-volatile solute ______ vapour pressure.

Ans: decreases

13. The heat of mixing for an ideal solution is ______.

Ans: zero

14. Positive deviation occurs due to ______ solute-solvent attraction.

Ans: weaker

15. Negative deviation occurs due to ______ solute-solvent attraction.

Ans: stronger

16. Constant boiling mixtures are called ______.

Ans: azeotropes

17. Properties depending only on number of particles are called ______ properties.

Ans: colligative

18. Osmotic pressure is represented by the symbol ______.

Ans: π

19. Van’t Hoff factor is represented by ______.

Ans: i

20. Reverse osmosis is used for purification of ______.

Ans: water

Section D: Very Short Answer Questions (15 Important Only)

1. What is a solution?

Ans: A homogeneous mixture of two or more components.

2. Define solubility.

Ans: The maximum amount of solute that dissolves in a given amount of solvent at a specified temperature.

3. What is the difference between saturated and unsaturated solution?

Ans:

Saturated	Unsaturated
Contains maximum dissolved solute	Can dissolve more solute

4. Why is molality preferred over molarity in colligative properties?

Ans: Because molality does not change with temperature.

5. State Henry’s law.

Ans: At constant temperature, solubility of a gas is directly proportional to its partial pressure. $P=K_Hx$ P=KHx

6. Why are soft drinks stored under high pressure?

Ans: High pressure increases the solubility of CO₂ gas.

7. What happens to gas solubility on increasing temperature?

Ans: It decreases.

8. State Raoult’s law.

Ans: Partial vapour pressure of a component is proportional to its mole fraction. $P_A=X_AP_A^°$ PA=XAPA°

9. What is an ideal solution?

Ans: A solution that obeys Raoult’s law over the entire concentration range.

10. Give one example of an ideal solution.

Ans: Benzene–toluene mixture.

11. What are colligative properties?

Ans: Properties that depend only on the number of solute particles, not their nature.

12. Name the four colligative properties.

Ans:

Relative lowering of vapour pressure
Elevation of boiling point
Depression of freezing point
Osmotic pressure

13. Define osmotic pressure.

Ans: The minimum pressure required to stop osmosis.

14. What is an isotonic solution?

Ans: Solutions having the same osmotic pressure at the same temperature.

15. What is Van’t Hoff factor?

Ans: A factor used to explain abnormal colligative properties due to association or dissociation of solute.

Compact Revision Table

No.	Question Topic	Key Answer
1	Solution	Homogeneous mixture
2	Solubility	Maximum dissolved solute
3	Saturated	Maximum solute
4	Molality	Temperature independent
5	Henry’s law	P = KHx
6	Soft drinks	High pressure CO₂
7	Gas solubility	Decreases with temperature
8	Raoult’s law	P = xP°
9	Ideal solution	Obeys Raoult’s law
10	Example	Benzene–toluene
11	Colligative	Depends on particles
12	Four properties	RLVP, ΔTb, ΔTf, π
13	Osmotic pressure	Stops osmosis
14	Isotonic	Same osmotic pressure
15	Van’t Hoff factor	Corrects abnormality

Section E: 2–3 Mark Board Questions (20 Important Only)

Part 1 (Q1–10)

1. Differentiate between molarity and molality.

Ans:

Molarity	Molality
Moles of solute per litre of solution	Moles of solute per kg of solvent
Unit: mol L⁻¹	Unit: mol kg⁻¹
Temperature dependent	Temperature independent

2. Explain why molality is preferred for colligative properties.

Ans:
Molality depends on mass of solvent, which does not change with temperature. Therefore, it gives more accurate colligative property calculations.

3. Define mole fraction and write its formula.

Ans:
Mole fraction is the ratio of moles of a component to total moles of all components. $X_A=\frac{n_A}{n_A+n_B}$ XA=nA+nBnA

4. State Henry’s law and mention two applications.

Ans:
At constant temperature, solubility of a gas is directly proportional to its partial pressure. $P=K_Hx$ P=KHx

Applications:

Carbonation of soft drinks.
Scuba diving gas mixtures.

5. Why does solubility of gases decrease with increase in temperature?

Ans:
Dissolution of gases is generally exothermic. Increasing temperature shifts equilibrium towards escaping gas molecules, decreasing solubility.

6. State Raoult’s law for a solution containing volatile components.

Ans:
The partial vapour pressure of each volatile component is directly proportional to its mole fraction. $P_A=X_AP_A^\circ$ PA=XAPA∘

7. Why does addition of non-volatile solute lower vapour pressure?

Ans:
The solute reduces the mole fraction of solvent, decreasing the number of solvent molecules escaping into vapour phase.

8. Define ideal solution. Write its characteristics.

Ans:
An ideal solution obeys Raoult’s law throughout the concentration range.

Characteristics:

ΔHmix = 0
ΔVmix = 0
Similar intermolecular forces

9. Explain positive deviation from Raoult’s law with example.

Ans:
When vapour pressure is higher than expected, it shows positive deviation.

Reason: A-B interactions are weaker.

Example: Ethanol + acetone.

10. Explain negative deviation from Raoult’s law with example.

Ans:
When vapour pressure is lower than expected, it shows negative deviation.

Reason: A-B interactions are stronger.

Example: Chloroform + acetone.

Quick Revision Table

Q	Topic	Key Point
1	Molarity vs Molality	Volume vs mass
2	Molality	Temperature independent
3	Mole fraction	n component/total n
4	Henry’s law	P = KHx
5	Gas solubility	↓ with temperature
6	Raoult’s law	P = xP°
7	Non-volatile solute	↓ vapour pressure
8	Ideal solution	ΔH=0, ΔV=0
9	Positive deviation	Weak A-B force
10	Negative deviation	Strong A-B force

Section E: 2–3 Mark Board Questions (20 Important Only)

Part 2 (Q11–20)

11. What are azeotropes? Why cannot they be separated by fractional distillation?

Ans:
Azeotropes are constant boiling mixtures having the same composition in liquid and vapour phases.

They cannot be separated by fractional distillation because vapour and liquid phases have identical composition.

12. Differentiate between minimum and maximum boiling azeotropes.

Minimum boiling azeotrope	Maximum boiling azeotrope
Shows positive deviation	Shows negative deviation
Higher vapour pressure	Lower vapour pressure
Example: Ethanol-water	Example: HNO₃-water

13. Define colligative properties. Name them.

Ans:
Properties that depend only on the number of solute particles and not on their nature are called colligative properties.

They are:

Relative lowering of vapour pressure
Elevation of boiling point
Depression of freezing point
Osmotic pressure

14. Explain elevation in boiling point.

Ans:
Addition of a non-volatile solute lowers vapour pressure of solvent. More heating is required to make vapour pressure equal to atmospheric pressure, causing an increase in boiling point. $\Delta T_b=K_bm$ ΔTb=Kbm

15. Explain depression in freezing point.

Ans:
Solute particles interfere with formation of solvent crystals, causing lowering of freezing point. $\Delta T_f=K_fm$ ΔTf=Kfm

16. Define osmosis and osmotic pressure.

Ans:

Osmosis: Movement of solvent molecules through a semipermeable membrane from dilute solution to concentrated solution.

Osmotic pressure: Minimum pressure required to stop osmosis.

17. Why is osmotic pressure preferred for finding molar mass of biomolecules?

Ans:

Biomolecules have very high molar masses.
They may decompose on heating.
Osmotic pressure can be measured at room temperature.

18. What are isotonic, hypertonic and hypotonic solutions?

Ans:

Type	Meaning
Isotonic	Same osmotic pressure
Hypertonic	Higher osmotic pressure
Hypotonic	Lower osmotic pressure

19. Explain Van’t Hoff factor and abnormal molar mass.

Ans:
Van’t Hoff factor (i) accounts for abnormal colligative properties due to association or dissociation. $i=\frac{\text{Observed colligative property}}{\text{Calculated colligative property}}$ i=Calculated colligative propertyObserved colligative property

Dissociation → i > 1
Association → i < 1

20. Calculate Van’t Hoff factor for NaCl and glucose solutions.

Ans:

Glucose: $C_6H_{12}O_6 \rightarrow \text{no ions}$ C6H12O6→no ions $i=1$ i=1

NaCl: $NaCl \rightarrow Na^+ + Cl^-$ NaCl→Na++Cl−

Number of particles = 2 $i\approx2$ i≈2

Compact Revision Table

Q	Topic	Key Point
11	Azeotrope	Constant boiling mixture
12	Azeotrope types	Positive/negative deviation
13	Colligative properties	Particle number dependent
14	ΔTb	Boiling point increases
15	ΔTf	Freezing point decreases
16	Osmosis	Solvent movement
17	Biomolecules	Use osmotic pressure
18	Tonicity	Relative osmotic pressure
19	Van’t Hoff factor	Corrects abnormality
20	NaCl vs glucose	i = 2 vs 1

Section F: Important Numericals (20 Only)

Part 1 (Q1–10)

All major formulas covered

1. Calculate molarity of a solution containing 5.85 g NaCl in 500 mL solution.

(Molar mass NaCl = 58.5 g mol⁻¹)

Solution:

Moles of NaCl: $=\frac{5.85}{58.5}=0.1$ =58.55.85=0.1

Volume = 0.5 L $M=\frac{0.1}{0.5}$ M=0.50.1

Ans = 0.2 M

2. Calculate molality of a solution containing 18 g glucose in 180 g water.

(Molar mass glucose = 180 g mol⁻¹)

Moles glucose: $=\frac{18}{180}=0.1$ =18018=0.1

Mass solvent = 0.18 kg $m=\frac{0.1}{0.18}$ m=0.180.1

Ans = 0.56 m

3. Calculate mole fraction of ethanol in a mixture of 46 g ethanol and 54 g water.

Moles ethanol: $=\frac{46}{46}=1$ =4646=1

Moles water: $=\frac{54}{18}=3$ =1854=3 $X_{ethanol}=\frac{1}{1+3}$ Xethanol=1+31

Ans = 0.25

4. Calculate mole fraction of water in the above solution.

$X_{water}=\frac{3}{4}$ Xwater=43

Ans = 0.75

5. Calculate molarity of 10% (w/v) glucose solution.

(Molar mass glucose = 180 g mol⁻¹)

10% w/v means:

10 g glucose in 100 mL solution

In 1 L:

Glucose = 100 g

Moles: $=\frac{100}{180}=0.556$ =180100=0.556

Ans = 0.556 M

6. Calculate molality of 20 g urea dissolved in 500 g water.

(Molar mass urea = 60 g mol⁻¹)

Moles: $=\frac{20}{60}=0.333$ =6020=0.333

Solvent = 0.5 kg $m=\frac{0.333}{0.5}$ m=0.50.333

Ans = 0.67 m

7. Calculate molarity of 4.9 g H₂SO₄ in 250 mL solution.

(Molar mass = 98 g mol⁻¹)

Moles: $=\frac{4.9}{98}=0.05$ =984.9=0.05

Volume = 0.25 L $M=\frac{0.05}{0.25}$ M=0.250.05

Ans = 0.2 M

8. Find mole fraction of solute when 2 mol solute is mixed with 8 mol solvent.

Total moles: $=2+8=10$ =2+8=10 $X_{solute}=\frac{2}{10}$ Xsolute=102

Ans = 0.2

9. Calculate molarity of 5 g NaOH in 250 mL solution.

(Molar mass NaOH = 40 g mol⁻¹)

Moles: $=\frac{5}{40}=0.125$ =405=0.125

Volume = 0.25 L $M=\frac{0.125}{0.25}$ M=0.250.125

Ans = 0.5 M

10. Calculate molality of 9.8 g H₂SO₄ dissolved in 100 g water.

Moles: $=\frac{9.8}{98}=0.1$ =989.8=0.1

Solvent = 0.1 kg $m=\frac{0.1}{0.1}$ m=0.10.1

Ans = 1 m

Formula Revision

Quantity	Formula
Molarity	moles/L solution
Molality	moles/kg solvent
Mole fraction	moles of component/total moles
Moles	given mass/molar mass

Section F: Important Numericals (20 Only)

Part 2 (Q11–20)

Henry’s Law + Raoult’s Law + Colligative Properties

11. A gas has mole fraction 0.02 in a solution. Henry’s constant is 4×1044 \times 10^44×104 bar. Find pressure.

Given: $x=0.02,\quad K_H=4\times10^4$ x=0.02,KH=4×104

Henry’s law: $P=K_Hx$ P=KHx $P=4\times10^4 \times0.02$ P=4×104×0.02

Ans = 800 bar

12. Calculate mole fraction of dissolved gas if pressure is 50 bar and KH=2.5×104K_H=2.5\times10^4KH=2.5×104 bar.

$x=\frac{P}{K_H}$ x=KHP $x=\frac{50}{25000}$ x=2500050

Ans = 0.002

13. Vapour pressure of pure benzene is 200 mm Hg. Mole fraction of benzene is 0.6. Find partial vapour pressure.

Raoult’s law: $P_A=X_AP_A^\circ$ PA=XAPA∘ $P=0.6\times200$ P=0.6×200

Ans = 120 mm Hg

14. A solution contains 2 mol benzene and 3 mol toluene. Vapour pressure of benzene is 150 mm Hg. Find partial pressure of benzene.

$X_{benzene}=\frac{2}{2+3}=0.4$ Xbenzene=2+32=0.4 $P=0.4\times150$ P=0.4×150

Ans = 60 mm Hg

15. Pure solvent vapour pressure is 100 mm Hg and solution vapour pressure is 95 mm Hg. Calculate relative lowering of vapour pressure.

Formula: $\frac{P^o-P}{P^o}$ PoPo−P $=\frac{100-95}{100}$ =100100−95

Ans = 0.05

16. Calculate elevation in boiling point of 1 m glucose solution.

Given: $K_b=0.52\,K\,kg\,mol^{-1}$ Kb=0.52Kkgmol−1 $\Delta T_b=K_bm$ ΔTb=Kbm $=0.52\times1$ =0.52×1

Ans = 0.52 K

17. Calculate depression in freezing point of 0.5 m urea solution.

Given: $K_f=1.86$ Kf=1.86 $\Delta T_f=K_fm$ ΔTf=Kfm $=1.86\times0.5$ =1.86×0.5

Ans = 0.93 K

18. Calculate osmotic pressure of 0.1 M glucose solution at 300 K.

Given: $R=0.082\,L\,atm\,mol^{-1}K^{-1}$ R=0.082Latmmol−1K−1 $\pi=CRT$ π=CRT $=0.1\times0.082\times300$ =0.1×0.082×300

Ans = 2.46 atm

19. Calculate osmotic pressure of 0.05 M solution at 27°C.

$T=273+27=300K$ T=273+27=300K $\pi=0.05\times0.082\times300$ π=0.05×0.082×300

Ans = 1.23 atm

20. Calculate Van’t Hoff factor for 80% ionised NaCl solution.

NaCl: $NaCl\rightarrow Na^++Cl^-$ NaCl→Na++Cl−

Number of ions: $n=2$ n=2

Formula: $i=1+\alpha(n-1)$ i=1+α(n−1) $i=1+0.8(1)$ i=1+0.8(1)

Ans = 1.8

Final Formula Sheet

Concept	Formula
Henry’s law	P = KHx
Raoult’s law	P = xP°
Relative lowering	(P°−P)/P°
Boiling point elevation	ΔTb = Kbm
Freezing point depression	ΔTf = Kfm
Osmotic pressure	π = CRT
Van’t Hoff factor	i = 1 + α(n−1)

Section G: Case-Based Questions (5 CBSE Pattern Sets Only)

Case Study 1: Carbonated Drinks and Henry’s Law

Carbon dioxide is dissolved in cold drinks under high pressure. When the bottle is opened, pressure decreases and CO₂ escapes.

Questions:

1. The law explaining gas solubility in liquids is:
A) Raoult’s law
B) Henry’s law
C) Boyle’s law
D) Dalton’s law

Ans: B

2. According to Henry’s law, gas solubility increases with:
A) Increase in pressure
B) Increase in temperature
C) Decrease in pressure
D) Decrease in gas amount

Ans: A

3. Gas solubility decreases when:
A) Pressure increases
B) Temperature increases
C) Temperature decreases
D) Gas is compressed

Ans: B

4. Henry’s law equation is:
A) P = xP°
B) π = CRT
C) P = KHx
D) ΔTf = Kfm

Ans: C

Case Study 2: Antifreeze Solution

Ethylene glycol is added to car radiators because it lowers the freezing point of water.

Questions:

5. This phenomenon is called:
A) Elevation of boiling point
B) Depression of freezing point
C) Osmosis
D) Vapour pressure increase

Ans: B

6. Freezing point depression depends on:
A) Number of solute particles
B) Colour of solution
C) Shape of container
D) Density only

Ans: A

7. Formula for depression in freezing point is:

A) ΔTb = Kbm
B) ΔTf = Kfm
C) π = CRT
D) P = KHx

Ans: B

8. The constant used is:
A) Kb
B) Kf
C) KH
D) R

Ans: B

Case Study 3: Osmosis in Biological Systems

A semipermeable membrane allows only solvent molecules to pass through it. Osmosis is important in biological processes.

Questions:

9. Movement of solvent through a semipermeable membrane is called:

A) Diffusion
B) Osmosis
C) Sublimation
D) Evaporation

Ans: B

10. Pressure required to stop osmosis is:

A) Atmospheric pressure
B) Osmotic pressure
C) Vapour pressure
D) Gas pressure

Ans: B

11. Osmotic pressure equation is:

A) π = CRT
B) P = KHx
C) P = xP°
D) ΔTb = Kbm

Ans: A

12. Solutions having same osmotic pressure are:

A) Hypertonic
B) Hypotonic
C) Isotonic
D) Ideal

Ans: C

Case Study 4: Ideal and Non-Ideal Solutions

An ideal solution obeys Raoult’s law because intermolecular forces between different molecules are similar.

Questions:

13. An ideal solution obeys:

A) Henry’s law only
B) Raoult’s law
C) Boyle’s law
D) Charles’ law

Ans: B

14. For an ideal solution:

A) ΔHmix = 0
B) ΔHmix > 0
C) ΔVmix ≠ 0
D) Vapour pressure is zero

Ans: A

15. Positive deviation occurs due to:

A) Strong A-B attraction
B) Weak A-B attraction
C) No solute
D) Low temperature only

Ans: B

16. Negative deviation occurs due to:

A) Weak attraction
B) Strong attraction
C) High pressure only
D) Low density

Ans: B

Case Study 5: Van’t Hoff Factor

NaCl dissociates into ions, while some molecules like acetic acid associate to form dimers.

Questions:

17. NaCl shows abnormal colligative properties due to:

A) Association
B) Dissociation
C) Evaporation
D) Condensation

Ans: B

18. Van’t Hoff factor for glucose is:

A) 0.5
B) 1
C) 2
D) 3

Ans: B

19. Dissociation causes Van’t Hoff factor to be:

A) Less than 1
B) Equal to 1
C) Greater than 1
D) Zero

Ans: C

20. Association causes Van’t Hoff factor to be:

A) Greater than 1
B) Less than 1
C) Equal to 2
D) Infinite

Ans: B

Section H: One Page Final Revision Sheet

Class 12 Chemistry (CBSE)

1. Important Definitions

Term	Meaning
Solution	Homogeneous mixture of two or more components
Solute	Component present in smaller amount
Solvent	Component present in larger amount
Solubility	Maximum amount of solute dissolved at given temperature
Ideal solution	Solution obeying Raoult’s law completely
Azeotrope	Constant boiling mixture

2. Concentration Terms

Quantity	Formula	Unit
Mass %	$\frac{Mass\ of\ solute}{Mass\ of\ solution}\times100$ Mass of solutionMass of solute×100	%
Volume %	$\frac{Volume\ of\ solute}{Volume\ of\ solution}\times100$ Volume of solutionVolume of solute×100	%
Molarity (M)	$\frac{Moles\ of\ solute}{Volume\ of\ solution(L)}$ Volume of solution(L)Moles of solute	mol L⁻¹
Molality (m)	$\frac{Moles\ of\ solute}{Mass\ of\ solvent(kg)}$ Mass of solvent(kg)Moles of solute	mol kg⁻¹
Mole fraction	$\frac{n_A}{n_A+n_B}$ nA+nBnA	No unit

3. Henry’s Law

Statement:
Solubility of a gas is directly proportional to its partial pressure. $P=K_Hx$ P=KHx

Important Points:

Pressure ↑ → Gas solubility ↑
Temperature ↑ → Gas solubility ↓

Applications:

Carbonated drinks
Scuba diving
High altitude breathing

4. Raoult’s Law

For volatile component: $P_A=X_AP_A^\circ$ PA=XAPA∘

Where:

$P_A$ PA = partial vapour pressure
$X_A$ XA = mole fraction
$P_A^\circ$ PA∘ = vapour pressure of pure component

5. Ideal Solution

Conditions:

$\Delta H_{mix}=0$ ΔHmix=0 $\Delta V_{mix}=0$ ΔVmix=0

Example:

Benzene + Toluene

6. Deviations from Raoult’s Law

Positive Deviation	Negative Deviation
Vapour pressure higher	Vapour pressure lower
Weak A-B attraction	Strong A-B attraction
Endothermic mixing	Exothermic mixing
Example: Ethanol + acetone	Example: Chloroform + acetone

7. Azeotropes

Minimum Boiling Azeotrope:

Positive deviation
Higher vapour pressure
Example: Ethanol-water

Maximum Boiling Azeotrope:

Negative deviation
Lower vapour pressure
Example: HNO₃-water

8. Colligative Properties

Depend only on number of solute particles.

Four Properties:

Relative lowering of vapour pressure

$\frac{P^\circ-P}{P^\circ}$ P∘P∘−P

Elevation in boiling point

$\Delta T_b=K_bm$ ΔTb=Kbm

Depression in freezing point

$\Delta T_f=K_fm$ ΔTf=Kfm

Osmotic pressure

$\pi=CRT$ π=CRT

9. Osmosis

Osmosis:
Movement of solvent through semipermeable membrane from dilute to concentrated solution.

Osmotic pressure:
Pressure required to stop osmosis.

Types:

Solution	Osmotic Pressure
Isotonic	Same
Hypertonic	Higher
Hypotonic	Lower

10. Van’t Hoff Factor (i)

Used for abnormal molar mass. $i=\frac{Observed\ colligative\ property}{Calculated\ colligative\ property}$ i=Calculated colligative propertyObserved colligative property

Dissociation:

$i>1$ i>1

Example: $NaCl \rightarrow Na^+ + Cl^-$ NaCl→Na++Cl−

Association:

$i<1$ i<1

Example: $CH_3COOH \rightarrow (CH_3COOH)_2$ CH3COOH→(CH3COOH)2

11. Important Formula Box

Concept	Formula
Moles	$\frac{Given\ mass}{Molar\ mass}$ Molar massGiven mass
Molarity	$\frac{n}{V}$ Vn
Molality	$\frac{n}{kg\ solvent}$ kg solventn
Henry’s law	$P=K_Hx$ P=KHx
Raoult’s law	$P=xP^\circ$ P=xP∘
Boiling elevation	$\Delta T_b=K_bm$ ΔTb=Kbm
Freezing depression	$\Delta T_f=K_fm$ ΔTf=Kfm
Osmotic pressure	$\pi=iCRT$ π=iCRT

12. Last Minute Exam Traps

✅ Molarity → solution volume
✅ Molality → solvent mass
✅ Mole fraction has no unit
✅ Colligative properties depend on particles, not nature
✅ Dissociation → i increases
✅ Association → i decreases
✅ Ideal solution → ΔH = 0 and ΔV = 0
✅ Gas solubility increases with pressure
✅ Gas solubility decreases with temperature