Solutions — Concepts, Reactions & Solved Examples

Solutions: The Chemistry of Mixing Things Right

When salt dissolves in water, something interesting happens — the sodium and chloride ions scatter uniformly throughout the water, and no matter which drop you pick, you get the same concentration. That uniformity is the defining feature of a solution: a homogeneous mixture of two or more substances.

We call the substance present in larger quantity the solvent, and the one dissolved in it the solute. Water is the most common solvent — so common that it’s called the “universal solvent.” But solutions aren’t always liquid. Air is a gaseous solution, brass is a solid solution, and your cold drink is a liquid solution. The chapter covers all of these, but Class 12 boards and JEE focus almost entirely on liquid solutions.

This is one of the most scoring chapters in Class 12 Chemistry. CBSE allocates around 5-7 marks directly, and JEE Main has asked at least one question from colligative properties in nearly every session since 2019. If you understand the “why” behind each concept — not just the formulas — this chapter becomes straightforward.

Key Terms and Definitions

Molarity (M): Moles of solute per litre of solution.

M = \frac{\text{moles of solute}}{\text{volume of solution in litres}}

Molarity changes with temperature because volume changes with temperature.

Molality (m): Moles of solute per kilogram of solvent.

m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}

Molality is temperature-independent. Use it for colligative property calculations.

Mole Fraction (χ): Fraction of moles of one component in total moles of mixture.

\chi_A = \frac{n_A}{n_A + n_B}

Note that $\chi_A + \chi_B = 1$ always.

Parts Per Million (ppm): For very dilute solutions — water pollutants, for instance.

\text{ppm} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6

Mass by Volume Percentage (w/v %): Grams of solute per 100 mL of solution. Common in lab work and pharmacy.

Henry’s Law: Gas-in-Liquid Solutions

When a gas dissolves in a liquid, Henry’s Law tells us how much dissolves:

p = K_H \cdot \chi

Where $p$ is the partial pressure of the gas, $K_H$ is Henry’s Law constant, and $\chi$ is the mole fraction of gas in solution.

Higher $K_H$ means the gas is less soluble. This is the key relationship to remember.

Why does oxygen concentration in water decrease as temperature rises? As temperature increases, $K_H$ increases, so for the same partial pressure, the mole fraction (solubility) decreases. This is why fish struggle in warm, polluted water — lower dissolved oxygen.

Henry’s Law appeared in JEE Main 2024 (Session 1) as a direct application: given the partial pressure of CO₂ above a soda bottle and $K_H$ , find the mole fraction of CO₂. Know the units of $K_H$ — it’s in bar (or atm).

Raoult’s Law and Vapour Pressure

For an ideal solution, the partial vapour pressure of each component is proportional to its mole fraction:

p_A = \chi_A \cdot p_A^*

p_{total} = \chi_A \cdot p_A^* + \chi_B \cdot p_B^*

Where $p_A^*$ is the vapour pressure of pure component A.

Ideal vs Non-Ideal Solutions

An ideal solution obeys Raoult’s Law across all concentrations. The condition: solute-solvent interactions ≈ solvent-solvent interactions. Benzene-toluene is the textbook example.

Positive deviation from Raoult’s Law: solute-solvent interactions are weaker than solvent-solvent. Vapour pressure is higher than expected. Ethanol-acetone is a classic example. $\Delta H_{mix} > 0$ , $\Delta V_{mix} > 0$ .

Negative deviation: solute-solvent interactions are stronger. Vapour pressure is lower than expected. Chloroform-acetone forms H-bonds between the two. $\Delta H_{mix} < 0$ , $\Delta V_{mix} < 0$ .

To remember which deviation is which: think about escaping. If molecules “want to escape” more (weaker attractive forces in the mixture), pressure goes up — positive deviation.

Azeotropes

When positive or negative deviation is extreme, we get azeotropes — mixtures that boil at a constant temperature and can’t be separated by simple distillation.

Minimum boiling azeotrope: maximum positive deviation (ethanol-water at 95.5% ethanol, boils at 351.1 K). This is why we can’t get 100% ethanol by distillation alone.
Maximum boiling azeotrope: maximum negative deviation (HNO₃-water at 68% HNO₃, boils at 393.5 K).

Colligative Properties

These four properties depend only on the number of solute particles, not their identity. That’s the central idea — write it on your palm before the exam.

1. Relative Lowering of Vapour Pressure

\frac{p_A^* - p_A}{p_A^*} = \chi_B

For dilute solutions: $\approx \dfrac{n_B}{n_A} = \dfrac{w_B \cdot M_A}{w_A \cdot M_B}$

This formula lets us find the molar mass of an unknown solute.

2. Elevation of Boiling Point

\Delta T_b = K_b \cdot m

$K_b$ is the ebullioscopic constant (molal elevation constant), unique to each solvent. For water, $K_b = 0.512$ K·kg/mol.

3. Depression of Freezing Point

\Delta T_f = K_f \cdot m

$K_f$ is the cryoscopic constant. For water, $K_f = 1.86$ K·kg/mol. This is why we add salt to roads in winter.

4. Osmotic Pressure

\pi = CRT = \frac{n}{V}RT

Where $C$ is molar concentration (mol/L), $R = 0.0821$ L·atm·mol⁻¹·K⁻¹, $T$ in Kelvin.

Osmotic pressure is the most sensitive colligative property — useful for finding molar masses of large molecules like proteins.

Van’t Hoff Factor (i)

For electrolytes, particles dissociate. The van’t Hoff factor accounts for this:

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

NaCl → Na⁺ + Cl⁻, so $i = 2$ (theoretically)
K₂SO₄ → 2K⁺ + SO₄²⁻, so $i = 3$
Glucose doesn’t dissociate, $i = 1$

Modified colligative property equations become:

\Delta T_b = i \cdot K_b \cdot m \qquad \Delta T_f = i \cdot K_f \cdot m \qquad \pi = i \cdot CRT

Association (like acetic acid in benzene, which dimerises) gives $i < 1$ .

Solved Examples

Example 1 — Easy (CBSE Level)

Q: Calculate the molality of a solution containing 18 g of glucose (M = 180 g/mol) in 500 g of water.

Solution:

Moles of glucose = $\frac{18}{180} = 0.1$ mol

Mass of water = 500 g = 0.5 kg

m = \frac{0.1}{0.5} = 0.2 \text{ mol/kg}

Example 2 — Medium (JEE Main Level)

Q: The vapour pressure of pure water at 25°C is 23.8 mmHg. When 10 g of a non-volatile solute is dissolved in 180 g of water, the vapour pressure drops to 23.4 mmHg. Find the molar mass of the solute.

Solution:

Using the relative lowering formula:

\frac{p^* - p_s}{p^*} = \chi_{solute}

\frac{23.8 - 23.4}{23.8} = \frac{0.4}{23.8} = 0.01681

Moles of water = $\frac{180}{18} = 10$ mol

Let moles of solute = $n$

\chi_{solute} = \frac{n}{n + 10} = 0.01681

n = 0.01681(n + 10) \implies n(1 - 0.01681) = 0.1681

n = \frac{0.1681}{0.9832} = 0.171 \text{ mol}

Molar mass = $\frac{10}{0.171} \approx 58.5$ g/mol

(This is close to NaCl, but since the formula gives molar mass without dissociation correction, this would be the “apparent” molar mass if it were an electrolyte — check context.)

Example 3 — Hard (JEE Advanced Level)

Q: A 5% solution (w/v) of cane sugar ( $M = 342$ g/mol) is isotonic with a 1% solution (w/v) of a substance X. What is the molar mass of X?

Solution:

Two solutions are isotonic when their osmotic pressures are equal: $\pi_1 = \pi_2$ .

Since $\pi = CRT$ , isotonic means $C_1 = C_2$ .

C_{sugar} = \frac{5/342}{0.1 \text{ L}} = \frac{50}{342} = 0.1462 \text{ mol/L}

(5 g in 100 mL = 50 g in 1 L)

C_X = \frac{1/M_X}{0.1 \text{ L}} = \frac{10}{M_X} \text{ mol/L}

Setting equal:

\frac{10}{M_X} = 0.1462

M_X = \frac{10}{0.1462} = 68.4 \text{ g/mol}

Exam-Specific Tips

CBSE Board (5-7 marks): CBSE loves numericals on all four colligative properties — expect one on $\Delta T_f$ or osmotic pressure. Write the formula, substitute with units, and show intermediate steps. Definition-based questions on Henry’s Law and azeotropes appear in 1-2 mark questions. The 3-mark question often combines van’t Hoff factor with $\Delta T_b$ or $\Delta T_f$ .

JEE Main (1-2 questions per session): Focus areas: (1) Calculating molar mass from colligative property data, (2) Henry’s Law application, (3) Positive vs negative deviation identification from given interactions, (4) Osmotic pressure calculations. JEE Main 2023 Session 2 had a question on the mole fraction of a gas using Henry’s Law — straightforward if you know $\chi = p/K_H$ .

NEET (1-2 questions): NEET focuses on conceptual clarity over heavy numericals. Know: what happens to solubility of a gas when temperature increases, the biological significance of osmosis (dialysis, IV fluids being isotonic with blood), and definitions of all concentration terms. Osmotic pressure formula $\pi V = nRT$ in NEET is often tested in context of RBC behavior in hypertonic/hypotonic solutions.

Common Mistakes to Avoid

Mistake 1 — Using molarity instead of molality in colligative property formulas. $\Delta T_b = K_b \cdot m$ uses molality, not molarity. Molarity changes with temperature; molality doesn’t. If the question gives you mass of solvent, you’re computing molality. If it gives volume of solution, that’s molarity — and you may need to convert.

Mistake 2 — Confusing $p_A^*$ (pure solvent) with $p_A$ (in solution). $p_A = \chi_A \cdot p_A^*$ means the partial pressure in solution is always less than the pure solvent’s vapour pressure (for a non-volatile solute). Students flip this and get $p_A > p_A^*$ .

Mistake 3 — Applying Henry’s Law to Raoult’s Law situations. Henry’s Law is for gases dissolved in liquids (dilute range). Raoult’s Law is for both components of a liquid-liquid mixture. Don’t use $p = K_H \cdot \chi$ when both components are liquids.

Mistake 4 — Forgetting the van’t Hoff factor for electrolytes. In any colligative property calculation involving NaCl, CaCl₂, K₂SO₄, or similar salts, multiply by $i$ . Students doing $\Delta T_f = K_f \cdot m$ for NaCl are losing marks unnecessarily. The answer will be half of what it should be.

Mistake 5 — Swapping $\Delta H_{mix}$ signs for ideal vs non-ideal solutions. Ideal solution: $\Delta H_{mix} = 0$ . Positive deviation: $\Delta H_{mix} > 0$ (energy needed to break stronger solvent-solvent bonds). Negative deviation: $\Delta H_{mix} < 0$ (new, stronger bonds form). Writing these backwards in a CBSE 3-mark question drops you to 1 mark.

Practice Questions

Q1. What is the mole fraction of ethanol in a solution containing 46 g of ethanol (M = 46) and 36 g of water?

Moles of ethanol = 46/46 = 1 mol. Moles of water = 36/18 = 2 mol. Total moles = 3. χ(ethanol) = 1/3 ≈ 0.333.

Q2. A solution is prepared by dissolving 4 g of NaOH (M = 40) in 500 mL of solution. Find its molarity.

Moles of NaOH = 4/40 = 0.1 mol. Volume = 500 mL = 0.5 L. Molarity = 0.1/0.5 = 0.2 M.

Q3. The freezing point of 0.1 m NaCl solution is −0.372°C. Calculate the van’t Hoff factor. ( $K_f$ for water = 1.86 K·kg/mol)

Expected $\Delta T_f$ (without dissociation) = 1.86 × 0.1 = 0.186°C. Observed $\Delta T_f$ = 0.372°C. $i = 0.372/0.186 = 2$ . This matches the expected dissociation of NaCl into 2 ions.

Q4. At 100°C, water has a vapour pressure of 760 mmHg. What is the vapour pressure of a solution containing 0.1 mol of glucose in 18 g of water?

Moles of water = 18/18 = 1 mol. χ(water) = 1/(1 + 0.1) = 1/1.1 = 0.909. Vapour pressure = 0.909 × 760 = 690.9 mmHg.

Q5. Which aqueous solution has the lowest freezing point: 0.1 m glucose, 0.1 m NaCl, or 0.1 m CaCl₂?

$\Delta T_f = i \cdot K_f \cdot m$ . Glucose: i = 1, $\Delta T_f$ = 0.186°C. NaCl: i = 2, $\Delta T_f$ = 0.372°C. CaCl₂: i = 3, $\Delta T_f$ = 0.558°C. CaCl₂ solution has the lowest freezing point (−0.558°C).

Q6. The osmotic pressure of 0.01 M solution of a compound at 25°C is 0.25 atm. Is this compound an electrolyte? ( $R = 0.0821$ L·atm·mol⁻¹·K⁻¹)

Expected $\pi$ (non-electrolyte) = CRT = 0.01 × 0.0821 × 298 = 0.245 atm ≈ 0.25 atm. The observed and expected values match, so the compound is a non-electrolyte (i ≈ 1).

Q7. The boiling point of a solution of 2 g of a non-volatile solute in 50 g of benzene is 80.94°C. The boiling point of pure benzene is 80.10°C. Find the molar mass of the solute. ( $K_b$ for benzene = 2.53 K·kg/mol)

$\Delta T_b$ = 80.94 − 80.10 = 0.84°C. $m = \Delta T_b / K_b = 0.84/2.53 = 0.332$ mol/kg. Mass of benzene = 50 g = 0.05 kg. Moles of solute = 0.332 × 0.05 = 0.0166 mol. Molar mass = 2/0.0166 ≈ 120.5 g/mol.

Q8. Two liquids A and B form an ideal solution. At 300 K, the vapour pressure of pure A is 400 mmHg and pure B is 600 mmHg. What is the composition of the vapour above an equimolar mixture?

χ(A) = χ(B) = 0.5. p(A) = 0.5 × 400 = 200 mmHg. p(B) = 0.5 × 600 = 300 mmHg. p(total) = 500 mmHg. Mole fraction of A in vapour = 200/500 = 0.4. Mole fraction of B in vapour = 300/500 = 0.6. The more volatile component (B) is enriched in the vapour phase — this is the principle behind distillation.

Frequently Asked Questions

Why is molality preferred over molarity for colligative properties?

Molarity depends on the volume of solution, which changes with temperature. Molality uses mass of solvent, which doesn’t change with temperature. Since colligative properties are measured at different temperatures (boiling point, freezing point), molality gives consistent results.

What is the difference between osmosis and reverse osmosis?

Osmosis is the spontaneous flow of solvent from a dilute solution to a concentrated solution through a semipermeable membrane. Reverse osmosis applies external pressure greater than the osmotic pressure to force solvent from the concentrated side to the dilute side. This is how seawater desalination plants work, and it’s conceptually tested in both CBSE and NEET.

Why do we add antifreeze (ethylene glycol) to car radiators?

Ethylene glycol lowers the freezing point of water (depression of freezing point). Even in −30°C winters, the coolant stays liquid. The same principle applies to adding salt on icy roads — $\Delta T_f = K_f \cdot m$ , so the road surface requires a lower temperature to freeze.

Can a solution have more than one solute?

Yes. Blood plasma, seawater, and most real-world solutions have multiple solutes. Colligative properties depend on the total number of solute particles, so we add up the contributions of each solute. This is why saline drips use NaCl at just the right concentration to match blood’s osmotic pressure (~7.4 atm).

What does “ideal” mean in the context of solutions, and which real pairs come closest?

An ideal solution satisfies three conditions: obeys Raoult’s Law, $\Delta H_{mix} = 0$ , and $\Delta V_{mix} = 0$ . Real pairs that come closest are those with very similar molecules: benzene-toluene (similar structure, similar size), hexane-heptane. When molecules are chemically similar, the intermolecular forces in the mixture are essentially the same as in the pure components.

Why does the solubility of gases decrease with temperature while solids generally increase?

For gases, dissolving is an exothermic process — energy is released when a gas molecule is captured by solvent. By Le Chatelier’s principle, increasing temperature shifts equilibrium toward the reverse (gas coming out of solution). For most solids, dissolving is endothermic, so higher temperature favours dissolution.

How is Henry’s Law related to scuba diving “the bends”?

At depth, a diver breathes air at high pressure. By Henry’s Law, more nitrogen dissolves in the blood at higher pressure. If the diver ascends too fast, pressure drops rapidly — like opening a soda bottle — and nitrogen bubbles form in the bloodstream. Slow, staged ascent gives time for nitrogen to leave the blood safely.