Solutions: CBSE Class 12 Chemistry NCERT Guide

Key Ways to Express the Concentration of Solutions

Molarity (M)

Number of moles of solute dissolved in one litre of the solution. Formula: M = (Moles of Solute) / (Volume of Solution in L). It is temperature-dependent because volume changes with temperature.

Molality (m)

Number of moles of solute dissolved per kilogram of the solvent. Formula: m = (Moles of Solute) / (Mass of Solvent in kg). It is temperature-independent, making it more reliable for scientific work.

Mole Fraction (χ)

The ratio of the number of moles of a particular component to the total number of moles of all components in the solution. For a binary solution with components A and B, χ_A = n_A / (n_A + n_B).

Mass Percentage (w/w)

The mass of the solute in grams present in 100 g of the solution. Formula: Mass % = [(Mass of Solute) / (Mass of Solution)] x 100.

Deep Dive: What Are Colligative Properties?

Colligative properties are a set of solution properties that depend on the concentration of solute particles, but not on their chemical identity. Essentially, it's the quantity, not the quality, of the solute that matters. Imagine adding salt to water. The properties we are about to discuss change simply because you've added something, not specifically because you've added NaCl. There are four main colligative properties:

1. Relative Lowering of Vapour Pressure: A non-volatile solute reduces the escaping tendency of solvent molecules, lowering the solution's vapour pressure compared to the pure solvent.
2. Elevation in Boiling Point (ΔT_b): Because the vapour pressure is lowered, a higher temperature is needed to make the vapour pressure equal to the atmospheric pressure. Thus, the boiling point of the solution is elevated.
3. Depression in Freezing Point (ΔT_f): The solute particles interfere with the formation of the solid solvent crystal lattice, so a lower temperature is required to freeze the solution.
4. Osmotic Pressure (Π): The pressure required to prevent the inward flow of solvent across a semipermeable membrane.

These properties are incredibly useful for determining the molar mass of unknown solutes.

Worked Examples: Applying Solution Formulas

• Question 1: Calculate the molality of 2.5 g of ethanoic acid (CH₃COOH) in 75 g of benzene.

  Step 1: Calculate the molar mass of ethanoic acid (CH₃COOH).
  Molar mass = (2 × 12.01) + (4 × 1.008) + (2 × 16.00) = 24.02 + 4.032 + 32.00 = 60.052 g/mol.

  Step 2: Calculate the number of moles of ethanoic acid.
  Moles = Mass / Molar Mass = 2.5 g / 60.052 g/mol = 0.0416 moles.

  Step 3: Convert the mass of the solvent (benzene) from grams to kilograms.
  Mass of solvent = 75 g = 75 / 1000 kg = 0.075 kg.

  Step 4: Calculate the molality (m).
  Molality (m) = Moles of solute / Mass of solvent in kg = 0.0416 mol / 0.075 kg = 0.5547 mol/kg.

  Final Answer: The molality of the solution is approximately 0.555 m.

• Question 2: 18 g of glucose (C₆H₁₂O₆) is dissolved in 1 kg of water in a saucepan. At what temperature will water boil at 1.013 bar? K_b for water is 0.52 K kg mol⁻¹.

  Step 1: Calculate the molar mass of glucose (C₆H₁₂O₆).
  Molar mass = (6 × 12.01) + (12 × 1.008) + (6 × 16.00) = 72.06 + 12.096 + 96.00 = 180.156 g/mol ≈ 180 g/mol.

  Step 2: Calculate the moles of glucose.
  Moles = Mass / Molar Mass = 18 g / 180 g/mol = 0.1 mol.

  Step 3: Calculate the molality (m) of the solution.
  Molality (m) = Moles of solute / Mass of solvent in kg = 0.1 mol / 1 kg = 0.1 m.

  Step 4: Calculate the elevation in boiling point (ΔT_b).
  The formula is ΔT_b = K_b × m.
  ΔT_b = 0.52 K kg mol⁻¹ × 0.1 mol kg⁻¹ = 0.052 K.

  Step 5: Determine the new boiling point.
  Water boils at 100 °C (373.15 K) at 1 atm (1.013 bar). The elevation is 0.052 K.
  New Boiling Point = Boiling point of pure water + ΔT_b = 373.15 K + 0.052 K = 373.202 K.

  Final Answer: The solution will boil at 373.202 K (or 100.052 °C).

Exam Traps and Key Points to Remember

Pay close attention to these common pitfalls in your exams:

• Molarity (M) vs. Molality (m): Do not use them interchangeably. Molarity is moles/Litre of solution, while Molality is moles/kg of solvent. Molality is independent of temperature, which is a key difference.
• Units: Always convert mass to kg for molality and volume to L for molarity. A common mistake is using grams directly in the formula.
• van 't Hoff Factor (i): For colligative properties, remember to include the van 't Hoff factor 'i' for electrolytes (salts, acids, bases) that dissociate in solution. For NaCl, i = 2. For non-electrolytes like glucose or urea, i = 1.
• Solvent vs. Solution: Read the question carefully. Is the mass/volume given for the solvent or the entire solution? This is crucial for calculating molality correctly.
• Raoult's Law: It applies only to the volatile component. For a solution of a non-volatile solute, the vapour pressure of the solution is solely due to the solvent (P_solution = P°_solvent × χ_solvent).

Practice Questions with Solutions

• Q: Vapour pressure of chloroform (CHCl₃) and dichloromethane (CH₂Cl₂) at 298 K are 200 mm Hg and 415 mm Hg respectively. Calculate the vapour pressure of the solution prepared by mixing 25.5 g of CHCl₃ and 40 g of CH₂Cl₂ at 298 K.
  A: Step 1: Calculate moles of each component.
  Molar mass of CHCl₃ = 119.5 g/mol. Moles CHCl₃ = 25.5 g / 119.5 g/mol = 0.213 mol.
  Molar mass of CH₂Cl₂ = 85 g/mol. Moles CH₂Cl₂ = 40 g / 85 g/mol = 0.47 mol.
  Total moles = 0.213 + 0.47 = 0.683 mol.
  Step 2: Calculate the mole fraction (χ) of each component.
  χ_CHCl₃ = 0.213 / 0.683 = 0.312
  χ_CH₂Cl₂ = 0.47 / 0.683 = 0.688
  Step 3: Apply Raoult's Law: P_total = P°_A χ_A + P°_B χ_B
  P_total = (200 mm Hg 0.312) + (415 mm Hg 0.688)
  P_total = 62.4 + 285.52 = 347.92 mm Hg.
  Final answer: The vapour pressure of the solution is 347.92 mm Hg.

• Q: 45 g of ethylene glycol (C₂H₆O₂) is mixed with 600 g of water. Calculate the freezing point depression.
  (K_f for water = 1.86 K kg mol⁻¹)
  A: Step 1: Calculate moles of ethylene glycol.
  Molar mass of C₂H₆O₂ = (212) + (61) + (2*16) = 24 + 6 + 32 = 62 g/mol.
  Moles = 45 g / 62 g/mol = 0.726 mol.
  Step 2: Calculate the molality (m) of the solution.
  Mass of solvent (water) = 600 g = 0.6 kg.
  Molality (m) = 0.726 mol / 0.6 kg = 1.21 mol/kg.
  Step 3: Calculate freezing point depression (ΔT_f).
  ΔT_f = K_f m = 1.86 K kg mol⁻¹ 1.21 mol kg⁻¹ = 2.25 K.
  Final answer: The freezing point depression is 2.25 K.

• Q: Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of a polymer of molar mass 185,000 in 450 mL of water at 37 °C.
  A: Step 1: Convert all units to SI units for the formula Π = (n/V)RT.
  n = 1.0 g / 185,000 g/mol = 5.405 x 10⁻⁶ mol.
  V = 450 mL = 0.450 L = 0.450 x 10⁻³ m³.
  T = 37 °C = 37 + 273 = 310 K.
  R = 8.314 Pa m³ K⁻¹ mol⁻¹.
  Step 2: Calculate the concentration (n/V).
  C = 5.405 x 10⁻⁶ mol / (0.450 x 10⁻³ m³) = 0.01201 mol m⁻³.
  Step 3: Calculate Osmotic Pressure (Π).
  Π = C R T = 0.01201 mol m⁻³ 8.314 Pa m³ K⁻¹ mol⁻¹ 310 K.
  Π = 30.96 Pa.
  Final answer: The osmotic pressure is approximately 30.96 Pa.

• Q: Why does a solution of ethanol and acetone show a positive deviation from Raoult's Law?
  A: Step 1: Analyze the intermolecular forces in the pure components. In pure ethanol, molecules are held together by strong hydrogen bonds. In pure acetone, there are weaker dipole-dipole interactions.
  Step 2: Analyze the forces in the mixture. When ethanol and acetone are mixed, the acetone molecules get in between the ethanol molecules, breaking some of the hydrogen bonds. The new interactions (ethanol-acetone) are weaker than the original ethanol-ethanol hydrogen bonds.
  Step 3: Relate forces to vapour pressure. Because the intermolecular forces of attraction in the solution are weaker, the molecules can escape into the vapour phase more easily. This results in a higher vapour pressure for the solution than predicted by Raoult's Law, which is defined as a positive deviation.
  Final answer: The solution shows positive deviation because the solute-solvent interactions are weaker than the solute-solute and solvent-solvent interactions, leading to an increased vapour pressure.

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