Class 12 Chemistry – Solutions Notes

1. Introduction

A solution is a homogeneous mixture of two or more substances.

• Solvent: Component present in larger amount
• Solute: Component present in smaller amount

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1.1 Types of Solutions

Solutions are classified on the basis of the physical state of solute and solvent.

Solvent	Solute	Example
Gas	Gas	Air
Liquid	Gas	CO₂ in water
Liquid	Liquid	Alcohol in water
Liquid	Solid	Sugar in water
Solid	Solid	Alloys (brass)

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1.2 Expressing Concentration of Solutions

1. Mass Percentage (w/w)

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

2. Volume Percentage (v/v)

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

3. Mass by Volume Percentage (w/v)

\text{w/v %} = \frac{\text{Mass of solute (g)}}{\text{Volume of solution (mL)}} \times 100

4. Mole Fraction

$$X_A = \frac{n_A}{n_A + n_B}$$XA​=nA​+nB​nA​​

• Sum of mole fractions = 1

5. Molarity (M)

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

6. Molality (m)

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

✔ Independent of temperature

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1.3 Solubility

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

Factors Affecting Solubility

1. Nature of solute and solvent
2. Temperature
3. Pressure (only for gases)

Henry’s Law

The solubility of a gas in a liquid is directly proportional to the pressure of the gas.$$p = k_H x$$p=kH​x

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1.4 Vapour Pressure of Liquid Solutions

Vapour pressure is the pressure exerted by vapours in equilibrium with a liquid.

Raoult’s Law

$$p_A = x_A p_A^0$$pA​=xA​pA0​

Total vapour pressure:$$p = p_A + p_B$$p=pA​+pB​

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1.5 Ideal and Non-Ideal Solutions

Ideal Solutions

• Obey Raoult’s law
• No heat change on mixing
• No volume change

Example: Benzene and toluene

Non-Ideal Solutions

• Do not obey Raoult’s law
• Show heat or volume change

Positive Deviation

• Weaker interactions
• Higher vapour pressure
  Example: Ethanol + acetone

Negative Deviation

• Stronger interactions
• Lower vapour pressure
  Example: Chloroform + acetone

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1.6 Colligative Properties and Determination of Molar Mass

Colligative properties depend only on the number of solute particles.

Types

1. Relative lowering of vapour pressure
2. Elevation of boiling point

$$\Delta T_b = K_b m$$ΔTb​=Kb​m

3. Depression of freezing point

$$\Delta T_f = K_f m$$ΔTf​=Kf​m

4. Osmotic pressure

$$\pi = CRT$$π=CRT

These properties help in determining the molar mass of solute.

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1.7 Abnormal Molar Masses

Observed molar mass may differ due to:

1. Association

• Solute particles combine
• Example: Acetic acid in benzene

2. Dissociation

• Solute particles break into ions
• Example: NaCl in water

van’t Hoff Factor

$$i = \frac{\text{Observed colligative property}}{\text{Theoretical value}}$$i=Theoretical valueObserved colligative property​