Chapter 1: Solutions (Complete Notes & Question Bank)

Chemistry
CBSE Class 12 Chemistry - Chapter 1: Solutions (Complete Notes & Question Bank)

📚 CBSE Class 12 Chemistry
Chapter 1: Solutions (Complete Notes & Question Bank)

1. Solutions and its Expression of Concentration, Solubility of Gases

Basic Concepts

Solution: A homogeneous mixture of two or more pure substances.

  • Solute: A substance that is dissolved in another substance in lesser amount.
  • Solvent: A substance in which another substance is dissolved in larger amount. Solvent determines the physical state of the solution.

Types of Solutions

Depending upon the physical states of solute and solvent, there are nine different types of solutions:

S. No.Types of SolutionsSoluteSolventExamples
1.Solid – SolidSolidSolidAlloys like brass, bronze, amalgam
2.Solid – LiquidSolidLiquidSolution of sugar, salt, urea in water
3.Solid – GasSolidGasDust or smoke particles in air
4.Liquid – SolidLiquidSolidHydrated salts, mercury in amalgamated zinc
5.Liquid – LiquidLiquidLiquidAlcohol in water, benzene in toluene
6.Liquid – GasLiquidGasAerosol, water vapour in air
7.Gas – SolidGasSolidHydrogen adsorbed in palladium
8.Gas – LiquidGasLiquidAerated drinks
9.Gas – GasGasGasMixture of gases (e.g., air)
  • Aqueous solution: A solution containing water as solvent (e.g., sugar solution).
  • Non-aqueous solution: A solution containing solvent other than water (e.g., iodine in alcohol).
  • Saturated solution: A solution in which no more solute can be dissolved at the same temperature.
  • Unsaturated solution: A solution in which more amount of solute can be dissolved at the same temperature.

Visual Summary: Types of Solutions

SOLUTIONS Solid Solvent Liquid Solvent Gaseous Solvent • Alloys (Solid/Solid) • Hydrated salts (Liquid/Solid) • H2 in Pd (Gas/Solid) • Sugar in water (Solid/Liquid) • Alcohol in water (Liquid/Liquid) • Aerated drinks (Gas/Liquid) • Smoke (Solid/Gas) • Fog (Liquid/Gas) • Air (Gas/Gas)

Methods of Expressing Concentration of Solution

(i) Mass percentage (w/w%)

$$\text{Mass\%} = \frac{\text{Mass of solute}}{\text{Total mass of solution}} \times 100$$

(ii) Volume percentage (v/v%)

$$\text{Volume\%} = \frac{\text{Volume of solute}}{\text{Total volume of solution}} \times 100$$

(iii) Mass by volume percentage (w/V%)

$$\text{Mass by volume\%} = \frac{\text{Mass of solute}}{\text{Volume of solution}} \times 100$$

Commonly used in medicine and pharmacy.

(iv) Parts per million (ppm)

$$\text{ppm} = \frac{\text{Number of parts of component}}{\text{Total number of parts of all components}} \times 10^6$$

Can be expressed as: (a) Mass to mass, (b) Volume to volume, (c) Mass to volume.

(v) Mole Fraction ($\chi$)

$$\chi_A = \frac{n_A}{n_A + n_B}$$ $$\chi_A + \chi_B = 1$$

(vi) Molarity (M)

$$M = \frac{\text{Moles of solute}}{\text{Volume of solution in L}} = \frac{W_B \times 1000}{M_B \times V_{\text{(mL)}}}$$

Unit: mol L⁻¹. Depends on temperature.

(vii) Molality (m)

$$m = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{W_B \times 1000}{M_B \times W_{A\text{(g)}}}$$

Unit: mol kg⁻¹. Independent of temperature.

(viii) Normality (N)

$$N = \frac{\text{Gram equivalents of solute}}{\text{Volume of solution in L}} = \frac{W_B \times 1000}{E_B \times V_{\text{(mL)}}}$$

Relationships Between Concentration Terms

$$\text{Molarity (M) and Molality (m): } m = \frac{1000 \times M}{1000 \times d - M \times M_B}$$ $$\text{Mole fraction } (\chi_B) \text{ and Molality (m): } m = \frac{1000 \times \chi_B}{(1 - \chi_B) M_A}$$

Solubility & Henry’s Law

Solubility: Maximum amount of solute that can be dissolved in 100 g of solvent to form a saturated solution at a given temperature.

Factors affecting solubility:

  • Nature of Solute and Solvent: "Like dissolves like" (polar dissolves polar, non-polar dissolves non-polar).
  • Temperature: Increases for endothermic reactions, decreases for exothermic reactions.
  • Pressure: Does not affect solids/liquids significantly, but greatly affects gases.

Henry’s Law

$$p = K_H \cdot \chi$$

The partial pressure of the gas is proportional to its mole fraction in the solution.

Applications: (i) Sealing soda bottles under high pressure. (ii) Scuba diving tanks diluted with He to avoid N₂ toxicity. (iii) Anoxia at high altitudes.

Limitations: Applicable only when pressure is not too high, temperature is not too low, and gas doesn't undergo chemical change, association, or dissociation.

Example 1 [Board 2023]

If N₂ gas is bubbled through water at 293 K, how many millimoles of N₂ gas would dissolve in 1 litre of water? Assume that N₂ exerts a partial pressure of 0.987 bar. Given that Henry’s law constant for N₂ at 293 K is 76.48 k bar.

Solution:

$$\begin{aligned} \chi &= \frac{P(\text{nitrogen})}{K_H} \\ &= \frac{0.987 \text{ bar}}{76480 \text{ bar}} \\ &= 1.29 \times 10^{-5} \end{aligned}$$ $$\text{Since } 1 \text{ mol} = 1000 \text{ mmol}$$ $$\chi = 1.29 \times 10^{-5} \times 1000 = 0.0129 \text{ mmol/L of water}$$

2. Raoult’s Law, Ideal and Non-Ideal Solutions

Vapour Pressure & Raoult's Law

Vapour pressure: Pressure exerted by vapours over a liquid at equilibrium at constant temperature. Depends on nature of liquid and temperature.

Raoult’s law for volatile liquids:

$$p_A = p_A^\circ \cdot \chi_A \quad \text{and} \quad p_B = p_B^\circ \cdot \chi_B$$ $$p_{\text{total}} = p_A + p_B = p_A^\circ \chi_A + p_B^\circ \chi_B$$

Raoult’s law as a special case of Henry’s law:

Comparing $p_A = p_A^\circ \chi_A$ with Henry's law $p_A = K_H \chi_A$, we get $p_A^\circ = K_H$.

Raoult’s law for non-volatile solute:

$$\frac{p_A^\circ - p_A}{p_A^\circ} = \chi_B$$

Relative lowering of vapour pressure is equal to the mole fraction of solute.

Ideal and Non-Ideal Solutions

PropertyIdeal SolutionNon-Ideal Solution
Raoult's LawObeys at all concentrationsDoes not obey
ΔmixH= 0≠ 0
ΔmixV= 0≠ 0
InteractionsA-A ≈ B-B ≈ A-BA-A ≠ A-B
Examplesn-hexane + n-heptane, Benzene + TolueneWater + Ethanol, Chloroform + Acetone

Visual Summary: Deviations from Raoult's Law

Positive Deviation A-B interactions weaker than A-A/B-B Mole Fraction ($\chi$) VP Forms Minimum Boiling Azeotrope Negative Deviation A-B interactions stronger than A-A/B-B Mole Fraction ($\chi$) VP Forms Maximum Boiling Azeotrope

Azeotropes

Azeotropes: Liquid mixtures that distil over without change in composition (constant boiling mixtures).

  • Minimum boiling azeotropes: Formed by solutions showing large positive deviation (e.g., water + benzene, chloroform + methanol).
  • Maximum boiling azeotropes: Formed by solutions showing large negative deviation (e.g., HNO₃ + H₂O).

Example 2 [Board 2023]

Suppose a solution is prepared by mixing two volatile liquids, A and B. Let $\chi_A$ and $\chi_B$ respectively be their mole fractions, and let $p_A$ and $p_B$ be their partial vapour pressures, respectively, in the solution at a particular temperature. Calculate the composition of the vapour phase in equilibrium with the solution.

Solution:

$$\begin{aligned} \text{According to Raoult's law:} \\ p_A &= p_A^\circ \chi_A \\ p_B &= p_B^\circ \chi_B \\ \text{By Dalton's law of partial pressures:} \\ p_{\text{total}} &= p_A + p_B \\ &= \chi_A p_A^\circ + \chi_B p_B^\circ \\ &= (1 - \chi_B) p_A^\circ + \chi_B p_B^\circ \\ &= p_A^\circ + (p_B^\circ - p_A^\circ)\chi_B \end{aligned}$$ $$\begin{aligned} \text{Composition of vapour phase } (\chi_A', \chi_B'): \\ p_A &= \chi_A' p_{\text{total}} \implies \chi_A' = \frac{p_A}{p_{\text{total}}} \\ p_B &= \chi_B' p_{\text{total}} \implies \chi_B' = \frac{p_B}{p_{\text{total}}} \end{aligned}$$

3. Colligative Properties, Determination of Molecular Mass, Abnormal Molecular Mass, Van’t Hoff Factor

Colligative Properties

Properties of solutions that depend only on the number of particles of solute and not on the nature of the solute.

Visual Summary: Colligative Properties

COLLIGATIVE PROPERTIES Relative Lowering of Vapour Pressure Elevation of Boiling Point Depression of Freezing Point Osmotic Pressure Depend ONLY on number of solute particles

(i) Relative Lowering of Vapour Pressure

$$\frac{\Delta p}{p_A^\circ} = \frac{p_A^\circ - p_A}{p_A^\circ} = \chi_B = \frac{n}{n + N}$$

(ii) Elevation of Boiling Point

$$\Delta T_b = K_b \cdot m = \frac{K_b \times w_2 \times 1000}{M_2 \times w_1}$$

$K_b$ = Molal elevation constant (Ebullioscopic constant).

(iii) Depression of Freezing Point

$$\Delta T_f = K_f \cdot m = \frac{K_f \times w_2 \times 1000}{M_2 \times w_1}$$

$K_f$ = Molal depression constant (Cryoscopic constant).

(iv) Osmotic Pressure

$$\pi = C R T = \frac{n_B}{V} R T$$

Osmosis: Net flow of solvent to the solution through a semipermeable membrane.

Reverse Osmosis: When applied pressure > osmotic pressure, pure solvent flows out of solution. Used in water purification.

Abnormal Molecular Mass & Van't Hoff Factor

Abnormal molecular mass: When molecular mass calculated from colligative properties differs from theoretical value due to association or dissociation of solute.

$$i = \frac{\text{Observed colligative property}}{\text{Normal colligative property}} = \frac{\text{Normal molecular mass}}{\text{Observed molecular mass}}$$

Modified colligative property equations:

$$\Delta T_b = i \cdot K_b \cdot m \quad ; \quad \Delta T_f = i \cdot K_f \cdot m \quad ; \quad \pi = i \cdot C R T$$

Example 3

The freezing point is reduced from 5.51 to 5.03°C when 0.721 g of a compound is added to 75 mL of benzene. (density of benzene = 0.879 g/mL, $K_f$ for benzene = 5.12 K kg mol⁻¹). Calculate the molecular mass of the compound.

Solution:

$$\begin{aligned} \text{Mass of benzene } (w_1) &= 75 \text{ mL} \times 0.879 \text{ g/mL} \\ &= 65.925 \text{ g} = 0.06593 \text{ kg} \\ \Delta T_f &= 5.51 - 5.03 = 0.48 \text{ K} \\ \text{We know, } \Delta T_f &= \frac{K_f \times w_2 \times 1000}{M_2 \times w_1} \\ 0.48 &= \frac{5.12 \times 0.721 \times 1000}{M_2 \times 65.925} \\ M_2 &= \frac{5.12 \times 0.721 \times 1000}{0.48 \times 65.925} \\ M_2 &= 116.65 \text{ g/mol} \end{aligned}$$

End of Chapter 1: Solutions

Complete Revision Notes & Question Bank for CBSE Class 12 Chemistry

Chemistry Chemistry 12