Colligative properties — which formula to use for which property

medium CBSE JEE-MAIN NEET 3 min read

Question

A solution contains 6 g of urea (molecular mass = 60) dissolved in 500 g of water. Calculate the boiling point elevation, freezing point depression, and relative lowering of vapour pressure. Given: Kb=0.52 K kg mol1K_b = 0.52\text{ K kg mol}^{-1}, Kf=1.86 K kg mol1K_f = 1.86\text{ K kg mol}^{-1}.

How do we decide which colligative formula applies to which situation?

(CBSE 12 board + JEE Main pattern)


Solution — Step by Step

Moles of urea =660=0.1 mol= \frac{6}{60} = 0.1\text{ mol}

Mass of solvent =500 g=0.5 kg= 500\text{ g} = 0.5\text{ kg}

m=0.10.5=0.2 mol kg1m = \frac{0.1}{0.5} = 0.2\text{ mol kg}^{-1}

Molality is the starting point for boiling point elevation and freezing point depression. Both ΔTb\Delta T_b and ΔTf\Delta T_f depend on the number of solute particles per kg of solvent.

ΔTb=Kb×m=0.52×0.2=0.104 K\Delta T_b = K_b \times m = 0.52 \times 0.2 = 0.104\text{ K}

Use this formula when: the question asks about boiling point change, or gives KbK_b.

ΔTf=Kf×m=1.86×0.2=0.372 K\Delta T_f = K_f \times m = 1.86 \times 0.2 = 0.372\text{ K}

Use this formula when: the question asks about freezing point change, or gives KfK_f.

ΔPP0=nsolutensolute+nsolvent\frac{\Delta P}{P_0} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}}

Moles of water =50018=27.78 mol= \frac{500}{18} = 27.78\text{ mol}

ΔPP0=0.10.1+27.78=0.127.88=0.00359\frac{\Delta P}{P_0} = \frac{0.1}{0.1 + 27.78} = \frac{0.1}{27.88} = 0.00359

Use Raoult’s law when: the question mentions vapour pressure or gives P0P_0.


Colligative Property Formula Selection Flowchart

flowchart TD
    A["Colligative property problem"] --> B{"What is asked?"}
    B -->|"Boiling point"| C["Use ΔTb = Kb × m"]
    B -->|"Freezing point"| D["Use ΔTf = Kf × m"]
    B -->|"Vapour pressure"| E["Use Raoult's law: ΔP/P₀ = x_solute"]
    B -->|"Osmotic pressure"| F["Use π = CRT"]
    C --> G{"Electrolyte?"}
    D --> G
    E --> G
    F --> G
    G -->|"Yes"| H["Multiply by van't Hoff factor i"]
    G -->|"No"| I["Use formula directly"]

Why This Works

All four colligative properties depend on the number of solute particles, not their identity. That is what makes them “colligative” — they depend on quantity, not nature.

The key decision is: which formula maps to which physical quantity? Boiling point and freezing point use molality (because they depend on solvent mass). Osmotic pressure uses molarity (because it depends on solution volume). Vapour pressure uses mole fraction (Raoult’s law).

For electrolytes like NaCl or CaCl2\text{CaCl}_2, multiply by the van’t Hoff factor ii (number of ions the solute dissociates into). For NaCl, i=2i = 2. For non-electrolytes like urea or glucose, i=1i = 1.


Common Mistake

Students often confuse molality and molarity. ΔTb\Delta T_b and ΔTf\Delta T_f use molality (moles per kg of solvent), while osmotic pressure π=CRT\pi = CRT uses molarity (moles per litre of solution). Mixing these up gives wrong answers even when the concept is correct. If the question gives solution volume, think osmotic pressure. If it gives solvent mass, think boiling/freezing point.

Quick memory aid: Boiling and Freezing use molality (m), Osmotic pressure uses molarity (C), Vapour pressure uses mole fraction (xx). The formulas are simple — the real skill is picking the right one.

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