Class 12 — Solutions

Chapter Overview & Weightage

Solutions is one of the most reliably scored chapters in Class 12 Chemistry. CBSE pulls 5 to 7 marks from it every year, with predictable question patterns: a 2-mark concept ask, a 3-mark numerical, and occasionally a 5-mark colligative-properties problem. JEE Main and NEET also draw 1 to 2 questions per paper.

The chapter has three big themes: concentration units, Raoult’s law and vapour pressure, and colligative properties (the four classics — osmotic pressure, BP elevation, FP depression, and vapour pressure lowering).

Year	CBSE Weightage
2024	7 marks
2023	6 marks
2022	5 marks
2021	7 marks
2020	6 marks

Key Concepts You Must Know

Molality (m) is moles of solute per kg of solvent. Molarity (M) is moles per litre of solution. Molality is preferred for colligative properties because it does not change with temperature.
Raoult’s law for ideal solutions: vapour pressure of each component is proportional to its mole fraction.
Henry’s law governs gas solubility: $p = K_H x$ .
The four colligative properties (vapour pressure lowering, BP elevation, FP depression, osmotic pressure) depend only on the number of solute particles, not their identity.
For ionic solutes, use the van ‘t Hoff factor $i$ to account for dissociation.
Ideal vs non-ideal solutions: ideal solutions obey Raoult’s law at all concentrations. Non-ideal show positive or negative deviations.

Important Formulas

Mass percent: $\frac{m_{\text{solute}}}{m_{\text{solution}}} \times 100$

Molarity: $M = \frac{n_{\text{solute}}}{V_{\text{solution (L)}}}$

Molality: $m = \frac{n_{\text{solute}}}{m_{\text{solvent (kg)}}}$

Mole fraction: $x_A = \frac{n_A}{n_A + n_B}$

$P = x_A P_A^\circ + x_B P_B^\circ$

For a binary ideal solution. The vapour above is generally richer in the more volatile component.

$\frac{P_A^\circ - P_A}{P_A^\circ} = x_B = \frac{n_B}{n_A + n_B}$

For dilute solutions of non-volatile solute B in volatile solvent A.

$\Delta T_b = K_b \cdot m \cdot i$

$\Delta T_f = K_f \cdot m \cdot i$

where $K_b$ and $K_f$ are ebullioscopic and cryoscopic constants of the solvent. For water: $K_b = 0.52\,\text{K kg/mol}$ , $K_f = 1.86\,\text{K kg/mol}$ .

$\pi = c R T \cdot i = \frac{n}{V} R T \cdot i$

The most commonly tested colligative property in Board exams — used to find molar mass of unknown solutes.

$i = \frac{\text{observed colligative property}}{\text{calculated value (assuming no dissociation)}}$

For NaCl: $i \approx 2$ . For $CaCl_2$ : $i \approx 3$ . For sugar: $i = 1$ .

Solved Previous Year Questions

PYQ 1 (CBSE 2023, 3 marks)

A solution containing $5\,\text{g}$ of an unknown non-electrolyte in $100\,\text{g}$ of water freezes at $-1.0\,°\text{C}$ . Find the molar mass of the solute. ( $K_f = 1.86\,\text{K kg/mol}$ )

Solution:

$\Delta T_f = 0 - (-1.0) = 1.0\,\text{K}$ .

Molality: $m = \frac{\Delta T_f}{K_f} = \frac{1.0}{1.86} = 0.538\,\text{mol/kg}$ .

Moles of solute = $0.538 \times 0.1 = 0.0538\,\text{mol}$ .

Molar mass = $\frac{5}{0.0538} \approx 93\,\text{g/mol}$ .

PYQ 2 (CBSE 2022, 5 marks)

The osmotic pressure of a solution containing $1.5\,\text{g}$ of a polymer in $250\,\text{mL}$ of water is $2.45 \times 10^{-3}\,\text{atm}$ at $300\,\text{K}$ . Calculate the molar mass of the polymer. ( $R = 0.0821\,\text{L atm K}^{-1}\,\text{mol}^{-1}$ )

Solution:

$\pi V = nRT \implies n = \pi V / (RT)$ .

$n = \frac{2.45 \times 10^{-3} \times 0.250}{0.0821 \times 300} = \frac{6.125 \times 10^{-4}}{24.63} = 2.49 \times 10^{-5}\,\text{mol}$

Molar mass = $\frac{1.5}{2.49 \times 10^{-5}} \approx 60{,}300\,\text{g/mol}$ .

PYQ 3 (CBSE 2024, 2 marks)

Why is the relative lowering of vapour pressure used to determine the molar mass of a non-volatile solute?

Solution:

Relative lowering of vapour pressure depends only on the mole fraction of the solute. By measuring this ratio for a known mass of solute, the number of moles can be calculated, and hence the molar mass. The method works for non-volatile solutes because volatile solutes contribute their own vapour pressure and complicate the analysis.

Difficulty Distribution

Easy (concentration unit conversion): 2 marks
Medium (Raoult’s law / colligative numerical): 3 marks
Hard (osmotic pressure for molar mass + van ‘t Hoff): 5 marks

The chapter rewards practice — once you do 10 numericals across the four colligative properties, the patterns become predictable.

Expert Strategy

CBSE always asks at least one numerical involving osmotic pressure or freezing point depression. Master the substitution sequence: identify $\Delta T$ , find molality, find moles, find molar mass.

Use molality ( $m$ ) for $\Delta T_b$ and $\Delta T_f$ , and molarity ( $c$ ) for $\pi$ . They are easy to swap if you write the formulas wrong. Memorise: “B and F use small m; pi uses big M.”

For ionic solutes, multiply the calculated colligative property by the van ‘t Hoff factor $i$ . For NaCl $i \approx 2$ , $CaCl_2 \approx 3$ , $K_4[Fe(CN)_6] \approx 5$ . CBSE often tests whether you remember to include $i$ .

Common Traps

Trap 1 — Mixing molality and molarity. They are numerically close in dilute aqueous solutions but conceptually different. Always read the formula carefully.

Trap 2 — Forgetting the van ‘t Hoff factor for ionic solutes. A 0.1 m NaCl solution behaves like 0.2 m for colligative purposes.

Trap 3 — Using °C instead of K in osmotic pressure. Thermodynamic temperature must be in kelvin. Add 273 to °C.

Trap 4 — Forgetting unit conversion in molality. Solvent mass must be in kg, not grams. 100 g of water = 0.1 kg.

Trap 5 — Assuming all aqueous solutions are ideal. Concentrated solutions, especially of strong electrolytes, deviate significantly. CBSE rarely tests this but JEE does.