Exhaustive Guide: Concentration Terms (Solutions) | CBSE Class 12 Chemistry | ChemCA

Expressing Concentration of Solutions

Comprehensive NCERT-Based Module for CBSE Class 12 Chemistry

1. Introduction to Solutions

In chemistry, a solution is a homogeneous mixture of two or more chemically non-reacting substances whose composition can be varied within certain limits. The substance present in the largest quantity is usually called the solvent, and the substance present in a lesser quantity is called the solute.

To accurately study physical and chemical properties, it is essential to quantify how much solute is dissolved in the solvent. This quantitative representation is called the concentration of a solution.

2. Methods of Expressing Concentration

According to the NCERT Class 12 syllabus, the concentration of a solution can be expressed in several ways, broadly categorized into temperature-dependent (volume-based) and temperature-independent (mass-based) terms.

2.1 Mass Percentage (w/w)

Definition: The mass of the solute in grams present in 100 grams of the solution.

Mass % of a component = (Mass of the component in solution / Total mass of the solution) × 100

Example: A 10% solution of sodium chloride in water by mass means that 10 g of NaCl is dissolved in 90 g of water to make 100 g of the solution.

NCERT Real-World Application: Commercial bleaching solution contains 3.62% by mass of sodium hypochlorite (NaClO) in water.

2.2 Volume Percentage (v/v)

Definition: The volume of a given component present in 100 parts by volume of the solution. It is commonly used for liquid-liquid solutions.

Volume % of a component = (Volume of the component / Total volume of solution) × 100

Example: A 10% ethanol solution in water means 10 mL of ethanol is dissolved in water such that the total volume of the solution is 100 mL.

NCERT Real-World Application: A 35% (v/v) solution of ethylene glycol is used as an antifreeze in cars. It lowers the freezing point of water to 255.4 K (-17.6°C).

2.3 Mass by Volume Percentage (w/v)

Definition: The mass of solute dissolved in 100 mL of the solution. This unit is heavily used in the pharmaceutical industry and medicine.

2.4 Parts Per Million (ppm)

Definition: When a solute is present in trace quantities (very small amounts), it is convenient to express concentration in parts per million.

ppm = (Number of parts of the component / Total number of parts of all components of the solution) × 10⁶

Concentration in ppm can be expressed as mass to mass, volume to volume, or mass to volume. It is standard for expressing atmospheric pollution (e.g., SO₂ in air) or dissolved ions in water.

NCERT Insight on Fluoride: 1 ppm of fluoride ions in water prevents tooth decay. However, 1.5 ppm causes mottling of teeth, and high concentrations (like in sodium fluoride) are used as rat poison.

2.5 Mole Fraction (x)

Definition: The ratio of the number of moles of one component to the total number of moles of all the components present in the solution.

Mole fraction of component A (x_A) = n_A / (n_A + n_B)

If a solution contains 'i' number of components, we have: x₁ + x₂ + ... + x_i = 1.
Note: Mole fraction is a dimensionless quantity (it has no units) and is independent of temperature.

2.6 Molarity (M)

Definition: The number of moles of solute dissolved per litre (or cubic decimetre) of the solution. It is the most common method of expressing concentration in laboratories.

Molarity (M) = Moles of solute / Volume of solution in litres (L)

Units: mol L^-1 or mol dm^-3, simply denoted as 'M'.

Calculation Tip: If volume is given in mL, M = (Mass of solute / Molar Mass of solute) × (1000 / Volume in mL).

2.7 Molality (m)

Definition: The number of moles of solute dissolved in 1 kg (1000 g) of the solvent.

Molality (m) = Moles of solute / Mass of solvent in kg

Units: mol kg^-1, simply denoted as 'm'.

3. Effect of Temperature on Concentration Terms

This is a highly tested concept in CBSE Board Exams. Concentration terms fall into two distinct categories based on whether they involve the measurement of mass or volume.

Temperature Independent (Mass-based)	Temperature Dependent (Volume-based)
Mass Percentage (w/w)	Volume Percentage (v/v)
Mole Fraction (x)	Mass by Volume Percentage (w/v)
Molality (m)	Molarity (M)
Parts per million (w/w)	Normality (N)

Why? Volumes of liquids expand or contract with changes in temperature due to thermal expansion. Therefore, Molarity decreases with an increase in temperature (Volume increases). Mass, however, remains constant regardless of temperature changes. This is why molality is considered a better way to express concentration than molarity.

4. NCERT Solved Examples (Step-by-Step)

NCERT Example 2.1: Calculate the mole fraction of ethylene glycol (C₂H₆O₂) in a solution containing 20% of C₂H₆O₂ by mass.

Solution:
Assume 100 g of solution.
Mass of ethylene glycol (W_B) = 20 g
Mass of water (W_A) = 100 - 20 = 80 g
Molar mass of C₂H₆O₂ = (2×12) + (6×1) + (2×16) = 62 g mol^-1
Molar mass of water (H₂O) = 18 g mol^-1

Moles of ethylene glycol (n_B) = 20 / 62 = 0.322 mol
Moles of water (n_A) = 80 / 18 = 4.444 mol

Mole fraction of glycol (x_B) = n_B / (n_A + n_B) = 0.322 / (4.444 + 0.322) = 0.068
Mole fraction of water (x_A) = 1 - 0.068 = 0.932

NCERT Example 2.2: Calculate the molarity of a solution containing 5 g of NaOH in 450 mL solution.

Solution:
Mass of NaOH (W_B) = 5 g
Molar mass of NaOH = 23 + 16 + 1 = 40 g mol^-1
Volume of solution = 450 mL = 0.450 L

Molarity (M) = (Mass of Solute / Molar Mass) × (1000 / Vol in mL)
M = (5 / 40) × (1000 / 450) = 0.278 M or mol L^-1

NCERT Example 2.3: Calculate molality of 2.5 g of ethanoic acid (CH₃COOH) in 75 g of benzene.

Solution:
Mass of solute (CH₃COOH) = 2.5 g
Molar mass of CH₃COOH = 12 + (3×1) + 12 + 16 + 16 + 1 = 60 g mol^-1
Mass of solvent (benzene) = 75 g = 0.075 kg

Moles of ethanoic acid = 2.5 / 60 = 0.0417 mol
Molality (m) = Moles of solute / Mass of solvent in kg = 0.0417 / 0.075 = 0.556 m or mol kg^-1

5. Previous Year Questions (PYQs) & Exhaustive Question Bank

Part A: Conceptual & Assertion-Reason (1 Mark)

[CBSE 2019]

Q1. Why is molality of a solution preferred over molarity for expressing the concentration of a solution?

Answer: Molality involves masses of the solute and solvent, which are independent of temperature. Molarity involves the volume of the solution, which changes with temperature. Hence, molality is a more precise and reliable unit across varying thermal conditions.

[CBSE 2023 Sample Paper]

Q2. Assertion (A): Molarity of a solution in liquid state changes with temperature.
Reason (R): The volume of a solution changes with a change in temperature.

Answer: Both Assertion and Reason are correct, and Reason is the correct explanation for Assertion. Since Molarity = Moles / Volume(L), and volume expands upon heating, molarity decreases as temperature increases.

[CBSE 2016]

Q3. Define the term mole fraction.

Answer: Mole fraction of a component in a solution is the ratio of the number of moles of that component to the total number of moles of all the components present in the solution.

Part B: Short Answer Type (2-3 Marks)

[CBSE 2020]

Q4. A solution of glucose (molar mass = 180 g mol^-1) in water is labelled as 10% (by mass). What would be the molality and mole fraction of each component in the solution?

Answer:
10% by mass means 10 g of glucose is present in 90 g of water.
Moles of glucose (n_B) = 10 / 180 = 0.0555 mol.
Mass of solvent (water) = 90 g = 0.090 kg.
Molality (m) = 0.0555 / 0.090 = 0.617 m.

Moles of water (n_A) = 90 / 18 = 5 mol.
Total moles = 5 + 0.0555 = 5.0555 mol.
Mole fraction of glucose = 0.0555 / 5.0555 = 0.011.
Mole fraction of water = 1 - 0.011 = 0.989.

[CBSE 2015, 2018]

Q5. Calculate the molarity of 9.8% (w/w) solution of H₂SO₄ if the density of the solution is 1.02 g mL^-1. (Molar mass of H₂SO₄ = 98 g mol^-1)

Answer:
Let the total mass of the solution be 100 g.
Mass of H₂SO₄ = 9.8 g.
Volume of solution = Mass / Density = 100 g / 1.02 g mL^-1 = 98.04 mL = 0.09804 L.
Moles of H₂SO₄ = 9.8 / 98 = 0.1 mol.
Molarity (M) = Moles / Volume (L) = 0.1 / 0.09804 = 1.02 M.
Shortcut Formula: M = ( %w/w × density × 10 ) / Molar Mass = (9.8 × 1.02 × 10) / 98 = 1.02 M.

Part C: Advanced Interconversion Questions (High Order Thinking Skills)

Q6. An antifreeze solution is prepared from 222.6 g of ethylene glycol (C₂H₆O₂) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL^-1, what will be the molarity of the solution? (NCERT Exercise 2.11)

Answer:
Step 1: Calculate Molality
Mass of solute (C₂H₆O₂) = 222.6 g
Molar mass of ethylene glycol = (2×12) + (6×1) + (2×16) = 62 g mol^-1
Moles of ethylene glycol = 222.6 / 62 = 3.59 mol
Mass of solvent (water) = 200 g = 0.200 kg
Molality (m) = 3.59 / 0.200 = 17.95 m

Step 2: Calculate Molarity
Total mass of solution = Mass of solute + Mass of solvent = 222.6 g + 200 g = 422.6 g
Density of solution = 1.072 g mL^-1
Volume of solution = Total mass / Density = 422.6 / 1.072 = 394.2 mL = 0.3942 L
Molarity (M) = Moles of solute / Volume of solution in L = 3.59 / 0.3942 = 9.11 M

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Exhaustive Guide: Concentration Terms (Solutions) | CBSE Class 12 Chemistry