Class 12 Chemistry MCQ Test – Solutions (Chapter 1) | Full Chapter Test

Class 12 Chemistry – Chapter: Solutions (Detailed Notes)

These notes are designed for CBSE, GSEB, JEE and NEET students. It includes detailed theory, concepts, formulas and exam-oriented explanations.

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1. What is a Solution?

A solution is a homogeneous mixture of two or more substances whose composition is uniform throughout the system. This means that any small sample taken from the solution will have the same composition and properties.

A solution consists of:

• Solvent: The component present in larger amount and determines the physical state.
• Solute: The component present in smaller amount and dissolved in solvent.

Example: In salt water, water is solvent and salt is solute.

Types of Solutions

Solutions can exist in different physical states such as gaseous, liquid and solid depending on the nature of solute and solvent.

• Gas in Gas → Air
• Gas in Liquid → Oxygen in water
• Liquid in Liquid → Alcohol in water
• Solid in Liquid → Sugar in water
• Solid in Solid → Alloys (Brass)

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2. Methods of Expressing Concentration

Concentration tells how much solute is dissolved in a given amount of solvent or solution. It is very important in chemistry as it helps in quantitative analysis.

1. Mass Percentage

It is defined as the mass of solute present in 100 g of solution.

\( \text{Mass \%} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100 \)

Used in industrial solutions and medicines.

2. Molarity (M)

Molarity is defined as the number of moles of solute dissolved in one litre of solution.

\( M = \frac{n}{V(\text{L})} \)

Important: Molarity depends on temperature because volume changes with temperature.

3. Molality (m)

Molality is defined as the number of moles of solute per kilogram of solvent.

\( m = \frac{n}{W(\text{kg})} \)

Molality is temperature independent because mass does not change.

4. Mole Fraction

It is the ratio of moles of one component to total moles in solution.

\( x_A = \frac{n_A}{n_A + n_B} \)

Total mole fraction is always:

\( x_A + x_B = 1 \)

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3. Solubility of Gases – Henry’s Law

The solubility of a gas in a liquid depends on pressure and temperature.

Henry’s Law: At constant temperature, the solubility of a gas is directly proportional to its partial pressure.

\( p = K_H \cdot x \)

Where:

• \( p \) = Partial pressure of gas
• \( x \) = Mole fraction of gas
• \( K_H \) = Henry’s constant

Higher value of \( K_H \) means lower solubility.

Applications:

• Scuba diving (avoiding bends)
• Carbonated drinks

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4. Vapour Pressure & Raoult’s Law

Vapour pressure is the pressure exerted by vapour in equilibrium with its liquid.

Raoult’s Law

The partial vapour pressure of each component is proportional to its mole fraction.

\( p_1 = p_1^\circ x_1 \)

\( p_2 = p_2^\circ x_2 \)

Total Pressure:

\( P = p_1^\circ x_1 + p_2^\circ x_2 \)

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5. Ideal and Non-Ideal Solutions

Ideal Solutions

• Obey Raoult’s law completely
• \( \Delta H = 0 \)
• \( \Delta V = 0 \)
• Forces between molecules are similar

Non-Ideal Solutions

• Do not obey Raoult’s law
• \( \Delta H \neq 0 \)
• Show deviation

Types of deviation:

• Positive: weaker forces → higher vapour pressure
• Negative: stronger forces → lower vapour pressure

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6. Colligative Properties

Colligative properties depend only on number of particles, not nature of solute.

1. Relative Lowering of Vapour Pressure

\( \frac{p^\circ - p}{p^\circ} = x_2 \)

2. Elevation of Boiling Point

\( \Delta T_b = K_b m \)

3. Depression of Freezing Point

\( \Delta T_f = K_f m \)

4. Osmotic Pressure

\( \pi = CRT \)

Highly important for determining molar mass of biomolecules.

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7. Van’t Hoff Factor

Sometimes solutes dissociate or associate, changing number of particles.

\( i = \frac{\text{Observed property}}{\text{Calculated property}} \)

• \( i = 1 \) → normal
• \( i > 1 \) → dissociation
• \( i < 1 \) → association

Modified Equations

\( \Delta T_b = i K_b m \)

\( \Delta T_f = i K_f m \)

\( \pi = i CRT \)

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⬇️ Attempt the test given below after revising these notes

Class 12 Chemistry

Chapter 1 : Solutions | Full Chapter MCQ Test

Time Left: 20:00

Student Name:

1. The vapour pressure of a solution containing non-volatile solute is:

greater than solvent equal to solvent less than pure solvent independent of solute

2. Raoult’s law is applicable to:

ideal solutions electrolytes only strong electrolytes colloidal solutions

3. Henry’s law constant increases with:

decrease in temperature increase in temperature pressure decrease solvent density

4. Osmotic pressure depends on:

nature of solute number of solute particles solvent type viscosity

5. Which property is NOT colligative?

Osmotic pressure Elevation of boiling point Vapour pressure lowering Density

6. Van’t Hoff factor is defined as:

observed colligative property / calculated calculated / observed temperature ratio mole fraction ratio

7. Depression in freezing point depends on:

molality molarity mole fraction normality

8. Unit of molality is:

mol L⁻¹ mol kg⁻¹ kg mol⁻¹ g mol⁻¹

9. Ideal solutions obey:

Raoult’s law Henry’s law Dalton law Boyle law

10. Positive deviation from Raoult’s law means:

weaker intermolecular forces stronger intermolecular forces no interaction ionic bonding

11. Osmotic pressure equation is:

π = CRT π = MRT π = nRT π = PV

12. Colligative properties depend on:

nature of solvent number of particles volume of solution pressure

13. If i > 1, the solute undergoes:

association dissociation polymerization neutralization

14. Example of ideal solution:

Benzene + Toluene Ethanol + Water Acetone + Water HCl + Water

15. Unit of Henry’s law constant:

atm atm⁻¹ mol K

16. Osmosis occurs through:

semi-permeable membrane filter paper glass plate metal sheet

17. Elevation in boiling point is proportional to:

molality molarity mole fraction density

18. If solute associates, Van’t Hoff factor becomes:

>1 <1 label=""> =1 0

19. Raoult's law formula:

P = X₁P₁° PV=nRT π = MRT Kb m

20. Colligative properties are useful to determine:

molecular mass density viscosity pressure

Vedant Classes | Class 12 Chemistry Test Series

Class 12 Chemistry Chapter 1: Solutions

This chapter is one of the most important topics in Class 12 Chemistry. It builds a strong foundation for understanding concentration terms, colligative properties, and their applications in real life. Mastering this chapter is essential for board exams as well as competitive exams like NEET and JEE.

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Learning Outcomes

After completing this chapter, students will be able to:

• Understand the concept of solutions and types of solutions (solid, liquid, gaseous).
• Differentiate between solute and solvent in a mixture.
• Express concentration using various methods like molarity, molality, mass percentage, and mole fraction.
• Understand and apply Henry’s Law and Raoult’s Law.
• Explain colligative properties such as:
  • Relative lowering of vapour pressure
  • Elevation in boiling point
  • Depression in freezing point
  • Osmotic pressure
• Calculate molar mass using colligative properties.
• Identify abnormal molecular masses using van’t Hoff factor.

Exam Tip: Numerical problems based on molality and colligative properties are frequently asked in CBSE exams.

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Real Life Applications

Concepts from this chapter are widely used in daily life and industries:

• Refrigeration: Antifreeze solutions are used in car radiators to lower the freezing point.
• Food Preservation: High salt or sugar concentration prevents microbial growth.
• Medical Field: IV fluids are isotonic to blood to prevent damage to cells.
• Desalination: Reverse osmosis is used to purify water.
• Soft Drinks: CO₂ dissolves in beverages under high pressure following Henry’s Law.

Important Concept: Osmotic pressure plays a crucial role in biological systems and is frequently asked in case-based questions.

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Common Mistakes Students Make

• Confusing molarity with molality.
• Ignoring temperature dependence of molarity.
• Wrong unit conversions in numerical problems.
• Not using van’t Hoff factor correctly in calculations.
• Mixing up formulas of colligative properties.
• Forgetting to convert mass into kg in molality.

Pro Tip: Always write units at each step while solving numericals to avoid silly mistakes.

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Conclusion

The chapter “Solutions” is highly scoring if practiced properly. Focus on understanding concepts and solving numerical problems regularly. Avoid common mistakes and revise formulas consistently to achieve excellent marks in exams.

Class 12 Chemistry Chapter 1: Solutions

10 Detailed Solved Examples

Example 1: Molarity Calculation

Calculate the molarity of a solution prepared by dissolving 11.7 g of sodium chloride (NaCl) in enough water to make 500 mL of solution. (Molar mass of NaCl = 58.5 g/mol)

Solution:
Step 1: Calculate number of moles of NaCl
Moles = Mass / Molar Mass = 11.7 / 58.5 = 0.2 mol

Step 2: Convert volume into litres
Volume = 500 mL = 0.5 L

Step 3: Apply formula
Molarity (M) = Moles / Volume = 0.2 / 0.5 = 0.4 M

Example 2: Molality Calculation

Calculate the molality of a solution containing 10 g of glucose dissolved in 200 g of water. (Molar mass of glucose = 180 g/mol)

Solution:
Moles of glucose = 10 / 180 = 0.0556 mol
Mass of solvent = 200 g = 0.2 kg

Molality (m) = 0.0556 / 0.2 = 0.278 m

Example 3: Mole Fraction

A solution contains 3 moles of ethanol and 7 moles of water. Calculate mole fraction of ethanol.

Solution:
Total moles = 3 + 7 = 10
Mole fraction = 3 / 10 = 0.3

Example 4: Raoult’s Law

The vapour pressure of pure benzene is 120 mmHg. If the mole fraction of benzene in solution is 0.75, calculate the vapour pressure of solution.

Solution:
P = P° × X = 120 × 0.75 = 90 mmHg

Example 5: Elevation in Boiling Point

Calculate the increase in boiling point when 1 mol of solute is dissolved in 1 kg of solvent. (Kb = 0.52 K kg mol⁻¹)

Solution:
ΔTb = Kb × m = 0.52 × 1 = 0.52 K

Example 6: Depression in Freezing Point

Calculate the depression in freezing point for a solution with molality 2 m. (Kf = 1.86 K kg mol⁻¹)

Solution:
ΔTf = 1.86 × 2 = 3.72 K

Example 7: Osmotic Pressure

Calculate osmotic pressure of a solution at 300 K having molarity 0.5 M. (R = 0.0821 L atm mol⁻¹ K⁻¹)

Solution:
π = MRT = 0.5 × 0.0821 × 300 = 12.315 atm

Example 8: Mass Percentage

Calculate mass percentage of a solution prepared by dissolving 20 g solute in 80 g solvent.

Solution:
Total mass = 20 + 80 = 100 g
% = (20/100) × 100 = 20%

Example 9: Henry’s Law

The Henry’s law constant for a gas is 2000 atm. Calculate mole fraction when pressure is 4 atm.

Solution:
X = P / kH = 4 / 2000 = 0.002

Example 10: van’t Hoff Factor

The observed molar mass of a substance is 40 g/mol, while the actual molar mass is 80 g/mol. Calculate van’t Hoff factor.

Solution:
i = Actual / Observed = 80 / 40 = 2

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10 Detailed FAQs

Q1: What is a solution and how is it different from a mixture?
A solution is a homogeneous mixture where composition is uniform throughout. Unlike heterogeneous mixtures, solutions do not show visible boundaries between components.

Q2: Why is molality preferred over molarity in colligative properties?
Molality is independent of temperature because it depends on mass, while molarity depends on volume which changes with temperature.

Q3: What is the significance of Raoult’s Law?
It helps in determining vapour pressure and understanding ideal and non-ideal solutions.

Q4: What are colligative properties and why are they important?
These properties depend only on number of solute particles and help determine molar mass.

Q5: What is osmotic pressure and its importance?
It is pressure required to stop osmosis and is used in biological and medical applications.

Q6: What is van’t Hoff factor and why is it used?
It corrects calculations when solute dissociates or associates.

Q7: What causes abnormal molar mass?
Dissociation or association of molecules leads to abnormal molar mass.

Q8: What is Henry’s Law?
It states that solubility of gas is directly proportional to pressure.

Q9: What is an ideal solution?
A solution that obeys Raoult’s Law at all concentrations.

Q10: Why are colligative properties important in daily life?
They are used in antifreeze, food preservation, and medical saline solutions.