JEE Chemistry — Solutions Complete Chapter Guide

Chapter Overview & Weightage

Solutions is a reliable 4-5% chapter in JEE Main — you can almost always count on 1-2 questions per paper. In JEE Advanced, it appears less frequently but tests deeper conceptual understanding, especially around ideal vs. non-ideal behavior and van’t Hoff factor edge cases.

In JEE Main 2024 (both sessions combined), Solutions contributed 2 questions across Shift 1 and Shift 2 — one on colligative properties calculation and one on vapour pressure. The chapter has been consistently present every year since 2013.

Year	JEE Main Questions	Topics Covered
2024	2	Vapour pressure, osmotic pressure
2023	1-2	Elevation in boiling point, van’t Hoff factor
2022	2	Raoult’s law, depression in freezing point
2021	1-2	Henry’s law, mole fraction
2020	2	Osmotic pressure, molality
2019	1	Colligative properties with electrolytes

The pattern is clear: colligative properties dominate — depression in freezing point ( $\Delta T_f$ ) and osmotic pressure ( $\pi$ ) are the two most-tested sub-topics. Henry’s law and Raoult’s law usually appear as conceptual or data-based questions rather than heavy calculations.

Key Concepts You Must Know

Prioritised by how often they appear in actual JEE papers:

Tier 1 — Always prepare these:

Colligative properties: $\Delta T_b$ , $\Delta T_f$ , osmotic pressure, relative lowering of vapour pressure. Know all four formulas and when to apply each.
Van’t Hoff factor ( $i$ ): For electrolytes that dissociate or associate. The relation between $i$ , degree of dissociation ( $\alpha$ ), and number of ions is the most-tested idea in this chapter.
Raoult’s law for volatile solutes: Both components in a binary liquid mixture. Know the total vapour pressure formula and the composition of vapour above the solution.

Tier 2 — High probability in any given paper:

Henry’s law: $p = K_H \cdot x$ — used for dissolved gases. Know that $K_H$ increases with temperature (gases become less soluble on heating).
Ideal vs. non-ideal solutions: Positive deviation (A-B interactions weaker than A-A, B-B) vs. negative deviation. Azeotropes are the conceptual endpoint here.
Molality vs. molarity: Molality is temperature-independent; molarity changes with temperature. Examiners test whether you pick the right concentration unit.

Tier 3 — Conceptual, low calculation:

Solubility and temperature trends for solids and gases
Osmosis vs. reverse osmosis vs. isotonic solutions
Abnormal molar masses for associating/dissociating solutes

Important Formulas

p_A = x_A \cdot p_A^*

p_{total} = x_A p_A^* + x_B p_B^*

When to use: Binary liquid mixtures where both components are volatile (e.g., benzene-toluene). $x_A$ is mole fraction of A in the liquid phase.

Note: Composition of vapour ≠ composition of liquid. Vapour is richer in the more volatile component.

\frac{p^* - p_s}{p^*} = x_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}}

When to use: Non-volatile solute in a volatile solvent. This is a colligative property — only mole fraction matters, not the identity of the solute.

\Delta T_b = K_b \cdot m \qquad \Delta T_f = K_f \cdot m

When to use: $m$ is molality (mol solute per kg solvent). $K_b$ and $K_f$ are solvent constants — memorise for water: $K_b = 0.52\ \text{K·kg/mol}$ , $K_f = 1.86\ \text{K·kg/mol}$ .

For electrolytes, multiply by van’t Hoff factor: $\Delta T_f = i \cdot K_f \cdot m$

\pi = iCRT

When to use: $C$ is molar concentration (mol/L), $R = 0.0821\ \text{L·atm/mol·K}$ , $T$ in Kelvin. This is the most formula-heavy colligative property question type — unit consistency is critical.

i = 1 + (n-1)\alpha \quad \text{(dissociation)}

i = 1 - \left(1 - \frac{1}{n}\right)\alpha \quad \text{(association)}

When to use: $n$ = number of particles formed per formula unit, $\alpha$ = degree of dissociation/association. For complete dissociation of $\text{Na}_2\text{SO}_4$ : $n=3$ , $\alpha=1$ , so $i=3$ .

p = K_H \cdot x

When to use: Dissolution of a gas in a liquid. $K_H$ has units of pressure. Higher $K_H$ → lower solubility at the same partial pressure. $K_H$ increases with temperature.

Solved Previous Year Questions

PYQ 1 — JEE Main 2023 (Depression in Freezing Point)

Question: 1.8 g of glucose ( $M = 180$ g/mol) is dissolved in 100 g of water. What is the depression in freezing point? ( $K_f$ for water = 1.86 K·kg/mol)

Solution:

First, find moles of glucose:

n = \frac{1.8}{180} = 0.01\ \text{mol}

Molality of solution:

m = \frac{0.01\ \text{mol}}{0.100\ \text{kg}} = 0.1\ \text{mol/kg}

Glucose doesn’t dissociate, so $i = 1$ :

\Delta T_f = K_f \cdot m = 1.86 \times 0.1 = 0.186\ \text{K}

Glucose is non-electrolyte — always $i = 1$ . The trap in this question is using 100 g (not converting to kg for molality). Write units at every step.

PYQ 2 — JEE Main 2022 (Van’t Hoff Factor with Association)

Question: Acetic acid dimerises in benzene. A solution of 1.2 g of acetic acid ( $M = 60$ g/mol) in 100 g benzene shows a freezing point depression of 0.45 K. Find the van’t Hoff factor. ( $K_f$ for benzene = 5.12 K·kg/mol)

Solution:

First, what $\Delta T_f$ should be without association (i.e., $i = 1$ ):

m = \frac{1.2/60}{0.1} = \frac{0.02}{0.1} = 0.2\ \text{mol/kg}

\Delta T_f^{\text{theoretical}} = 5.12 \times 0.2 = 1.024\ \text{K}

Now use actual $\Delta T_f$ to find $i$ :

i = \frac{\Delta T_f^{\text{observed}}}{\Delta T_f^{\text{theoretical}}} = \frac{0.45}{1.024} \approx 0.44

This confirms association ( $i < 1$ always signals association, $i > 1$ signals dissociation).

Students often confuse the formula for $i$ in association vs. dissociation. Association gives $i < 1$ because two molecules become one. Dissociation gives $i > 1$ because one molecule becomes many ions.

PYQ 3 — JEE Main 2024 Shift 1 (Osmotic Pressure)

Question: At 27°C, the osmotic pressure of a solution containing 6 g of a non-electrolyte in 2 litres of solution is 1.23 atm. What is the molar mass of the solute? ( $R = 0.0821\ \text{L·atm/mol·K}$ )

Solution:

Convert temperature: $T = 300\ \text{K}$

From $\pi = CRT$ , molar concentration is:

C = \frac{\pi}{RT} = \frac{1.23}{0.0821 \times 300} = \frac{1.23}{24.63} = 0.05\ \text{mol/L}

Moles in 2 litres: $n = 0.05 \times 2 = 0.1\ \text{mol}$

Molar mass:

M = \frac{6\ \text{g}}{0.1\ \text{mol}} = 60\ \text{g/mol}

Osmotic pressure problems almost always give you 27°C (= 300 K) or 0°C (= 273 K). The moment you see “27°C”, add 273 first — then check units of $R$ to match the pressure unit given.

Difficulty Distribution

For JEE Main, Solutions questions break down roughly like this:

Difficulty	Proportion	What It Tests
Easy	~40%	Direct formula application — $\Delta T_f$ , osmotic pressure with non-electrolytes
Medium	~45%	Van’t Hoff factor with incomplete dissociation/association; Raoult’s law with vapour composition
Hard	~15%	Multi-step: find molar mass from colligative data, then identify compound; or abnormal molar mass combined with structure reasoning

For JEE Advanced, the difficulty shifts — most questions are Medium-Hard, testing the reasoning behind ideal/non-ideal deviations or combining Solutions with electrochemistry concepts.

Expert Strategy

How toppers approach this chapter:

The smartest approach is to master colligative properties first and lock in all four formulas as a unit. They are different expressions of the same underlying idea — solute particles disrupting the solvent’s behaviour. If you understand why adding solute lowers vapour pressure, the other three colligative properties make sense automatically.

Solve 15-20 PYQs on $\Delta T_f$ and osmotic pressure back-to-back before moving to Raoult’s law. The calculation pattern repeats — you’ll start seeing it as almost mechanical. Then spend time on van’t Hoff factor, which is where marks actually get differentiated.

For Raoult’s law, the key insight toppers use is distinguishing between liquid phase composition (what you mix) and vapour phase composition (what rises above). Questions on vapour composition are medium-difficulty but feel hard if you haven’t practised this distinction specifically.

Time allocation in exam: Solutions questions usually take 2.5–3 minutes each. If a question involves molar mass calculation from colligative properties (a common 5-step problem), budget 3.5 minutes.

Common Traps

Trap 1 — Molality vs. Molarity confusion. The colligative property formulas use molality ( $m$ , mol per kg solvent). Osmotic pressure uses molarity ( $C$ , mol per litre solution). Students accidentally use the wrong one. Read the unit carefully before plugging in.

Trap 2 — Van’t Hoff factor for partial electrolytes. JEE frequently gives you a “30% ionised” or “40% associated” scenario. You must apply the $i$ formula with $\alpha = 0.3$ or $0.4$ — not assume complete dissociation. $\text{NaCl}$ fully dissociating gives $i = 2$ , but $\text{NaCl}$ at 60% dissociation gives $i = 1 + (2-1)(0.6) = 1.6$ .

Trap 3 — Mole fraction of solute vs. solvent. In the relative lowering of vapour pressure formula, the right-hand side is $x_{solute}$ (mole fraction of solute). Some students write $x_{solvent}$ by mistake. The formula says: the fraction by which vapour pressure drops equals the mole fraction of the solute.

Trap 4 — Temperature in Henry’s law questions. Increasing temperature decreases the solubility of gases (think about opening a cold vs. warm cola bottle). But in numerical questions, all you need is $p = K_H \cdot x$ . The temperature trend is tested conceptually, usually as a true/false-style assertion-reason question.

One last pattern to watch: JEE sometimes gives an “abnormal molar mass” from freezing point data and asks you to find degree of dissociation. The logic chain is: observed $\Delta T_f$ → observed $i$ → use $i = 1 + (n-1)\alpha$ to back-calculate $\alpha$ . This appeared in JEE Main 2023 Session 2 and is a reliable medium-difficulty question type.