Solutions: Concentration, Raoult's Law and Colligative Properties Made Simple
A friendly, NCERT-aligned walkthrough of the Solutions chapter: how to measure concentration (molarity, molality, mole fraction), what Raoult's Law says about vapour pressure, and how the four colligative properties let us find molar masses, with two fully worked numericals.
By the PadhoDost Team · 📖 8 min read · Updated 4 August 2026
Part of Class 12 (CBSE) prep🧠 Start with a glass of nimbu paani
Think about making nimbu paani. A little sugar and a squeeze of lemon dissolved in a full glass of water tastes light; the same sugar in half a glass tastes much sweeter. Nothing changed about the sugar itself, only how crowded it is in the water. That single idea, how much stuff is dissolved in how much liquid, is the heart of this whole chapter. Once you can measure that crowding precisely, you can predict how the solution boils, freezes and behaves. Padho, dost, it is simpler than it looks!
Solute, solvent and solution: the basic words
A solution is a homogeneous mixture of two or more substances. The component present in the larger amount is the solvent (usually water), and the component dissolved in it is the solute (like sugar or salt). In nimbu paani, water is the solvent and sugar is the solute. To do any real chemistry we must state exactly how much solute is present, and that measurement is called concentration.
Three precise ways to express concentration
Vague words like 'strong' or 'dilute' are not enough for Class 12. NCERT expects you to use three exact measures: molarity, molality and mole fraction. Each answers 'how much solute?' but divides it by a different reference, so read the units carefully every single time.
| Measure | Formula | Unit | Changes with temperature? |
|---|---|---|---|
| Molarity (M) | moles of solute / volume of solution in litres | mol/L | Yes, volume expands or shrinks with temperature |
| Molality (m) | moles of solute / mass of solvent in kilograms | mol/kg | No, mass never changes with temperature |
| Mole fraction (x) | moles of a component / total moles of all components | no unit (a ratio) | No |
📝 Worked example: molality and mole fraction
Problem: Dissolve 45 g of glucose (molar mass 180 g/mol) in 500 g of water. Find the molality and the mole fraction of glucose.
Step 1: Moles of glucose = 45 / 180 = 0.25 mol.
Step 2: Mass of solvent (water) = 500 g = 0.5 kg.
Step 3: Molality = 0.25 mol / 0.5 kg = 0.5 mol/kg (0.5 m).
Step 4: Moles of water = 500 / 18 = 27.78 mol.
Step 5: Mole fraction of glucose = 0.25 / (0.25 + 27.78) = 0.25 / 28.03 = 0.0089.
Answer: molality = 0.5 m and mole fraction of glucose = 0.0089 (so mole fraction of water = 1 - 0.0089 = 0.9911).
Raoult's Law: the pressure of escaping molecules
Every liquid has a vapour pressure, the push of molecules escaping into the space above it. Raoult's Law says that for a solution of volatile liquids, the partial vapour pressure of each component is proportional to its mole fraction in the solution. For a solution of two volatile liquids A and B, each contributes its own share. When the solute is non-volatile (like sugar, which does not evaporate), only the solvent contributes vapour, and its vapour pressure drops below that of the pure solvent because solute particles occupy part of the surface and reduce the escaping tendency.
Colligative properties: it is the count that matters
Colligative properties depend only on the NUMBER of solute particles dissolved, not on their chemical nature. One mole of glucose and one mole of urea, both non-electrolytes, produce the same effect even though they are different substances. NCERT lists four colligative properties, and all four flow from the same 'crowding lowers vapour pressure' idea you just met.
In plain words: dissolving a non-volatile solute makes a liquid harder to boil (boiling point rises) and harder to freeze (freezing point falls). This is why we add salt to icy roads and antifreeze to car radiators. Osmotic pressure is the pressure needed to just stop solvent from flowing through a semipermeable membrane into the solution, and because it is measured at room temperature and gives large, easily measured values, it is the best method for finding the molar mass of delicate large molecules like proteins.
📝 Worked example: freezing point depression to find molar mass
Problem: 18 g of a non-volatile, non-electrolyte solute is dissolved in 200 g of water. The solution freezes at -0.93 °C. Find the molar mass. (Kf for water = 1.86 K kg/mol.)
Step 1: Pure water freezes at 0 °C, so ΔTf = 0 - (-0.93) = 0.93 K.
Step 2: Start from ΔTf = Kf * m, where molality m = (w2 * 1000)/(M2 * w1). Combining and rearranging for M2 gives M2 = (Kf * w2 * 1000)/(ΔTf * w1).
Step 3: Put in values (w2 = 18 g, w1 = 200 g): M2 = (1.86 * 18 * 1000) / (0.93 * 200).
Step 4: Numerator = 1.86 * 18 * 1000 = 33480; denominator = 0.93 * 200 = 186; M2 = 33480 / 186 = 180 g/mol.
Answer: the molar mass is 180 g/mol, which matches glucose.
Quick revision before the exam
- ✓Molarity (mol/L) depends on temperature; molality (mol/kg) and mole fraction (no unit) do not.
- ✓Raoult's Law: the partial vapour pressure of a volatile component is proportional to its mole fraction.
- ✓Relative lowering of vapour pressure equals the mole fraction of the solute.
- ✓Four colligative properties: relative lowering of vapour pressure, ΔTb = Kb·m, ΔTf = Kf·m, and Π = CRT.
- ✓For water, Kb = 0.52 and Kf = 1.86 K kg/mol; remember R = 0.0821 L atm K⁻¹ mol⁻¹.
- ✓Colligative properties count particles, so multiply by the van't Hoff factor i for electrolytes.
⚡ Quick check
Three separate 1 molal (1 m) aqueous solutions are made using glucose, NaCl and CaCl2. Which one shows the GREATEST depression of freezing point?
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