Class 12 Solutions — Every Colligative Property Formula
Colligative properties are the heart of the Class 12 Solutions chapter, and worth learning properly because the numericals are formula-driven and highly repeatable. There are four of them, one correction factor, and one recurring exam task — finding the molar mass of an unknown solute.
What "colligative" actually means
A colligative property depends only on the number of solute particles in the solution, not on what they are. One mole of glucose and one mole of urea in the same amount of water lower the freezing point by the same amount. But one mole of NaCl lowers it by nearly twice as much, because each formula unit gives two particles in solution. That single idea explains the entire chapter, including the van't Hoff factor.
1. Relative lowering of vapour pressure
A non-volatile solute reduces the vapour pressure of the solvent. Raoult's law for such a solution gives:
For a dilute solution: (p° − p) / p° ≈ n2 / n1 = (w2 / M2) × (M1 / w1)
Here p° is the vapour pressure of pure solvent, p that of the solution; subscript 1 is the solvent and 2 the solute; w is mass and M is molar mass.
Worked example 1. 5.0 g of a non-volatile solute (M = 100 g mol−1) is dissolved in 90 g of water. Pure water has p° = 760 mm Hg at 373 K. Find the vapour pressure of the solution.
n2 = 5.0 ÷ 100 = 0.050 mol
n1 = 90 ÷ 18 = 5.00 mol
x2 = 0.050 ÷ (5.00 + 0.050) = 0.050 ÷ 5.050 = 0.009901
p° − p = 0.009901 × 760 = 7.52 mm Hg
p = 760 − 7.52 = 752.48 mm Hg
2. Elevation of boiling point
M2 = (1000 × Kb × w2) / (ΔTb × w1) with w in grams
Kb is the molal elevation constant (ebullioscopic constant) of the solvent, in K kg mol−1.
Worked example 2. 1.80 g of glucose (M = 180 g mol−1) is dissolved in 100 g of water. Kb(water) = 0.512 K kg mol−1. Find the boiling point of the solution.
moles of glucose = 1.80 ÷ 180 = 0.0100 mol
mass of solvent = 100 g = 0.100 kg
m = 0.0100 ÷ 0.100 = 0.100 mol kg−1
ΔTb = 0.512 × 0.100 = 0.0512 K
Boiling point = 100 + 0.0512 = 100.05 °C
3. Depression of freezing point
M2 = (1000 × Kf × w2) / (ΔTf × w1)
This is the most common molar-mass question in the chapter, because Kf values are large and the temperature change is easy to measure in a laboratory.
Worked example 3. 1.00 g of a non-electrolyte dissolved in 50.0 g of benzene lowers the freezing point by 0.40 K. Kf(benzene) = 5.12 K kg mol−1. Find the molar mass of the solute.
M2 = (1000 × 5.12 × 1.00) ÷ (0.40 × 50.0)
Numerator = 5120
Denominator = 0.40 × 50.0 = 20.0
M2 = 5120 ÷ 20.0 = 256 g mol−1
Worth noticing: 256 is the molar mass of S8 (8 × 32) — this is how the molecular formula of sulphur in benzene was established historically.
4. Osmotic pressure
⇒ M2 = (w2 R T) / (π V)
C is molarity in mol L−1, V is the volume of solution in litres, T is in kelvin, and R = 0.0821 L atm K−1 mol−1 if you want π in atmospheres. Osmotic pressure is the preferred method for very large molecules such as proteins and polymers, because even a tiny concentration produces a measurable pressure.
Worked example 4. 3.0 g of urea (M = 60 g mol−1) is dissolved in water to make 250 mL of solution at 300 K. Find the osmotic pressure.
V = 250 mL = 0.250 L
π = (w2 R T) ÷ (M2 V) = (3.0 × 0.0821 × 300) ÷ (60 × 0.250)
Numerator: 3.0 × 0.0821 = 0.2463; 0.2463 × 300 = 73.89
Denominator: 60 × 0.250 = 15.0
π = 73.89 ÷ 15.0 = 4.93 atm
The van't Hoff factor — the correction for electrolytes
All four formulas above assume the solute neither splits up nor sticks together. When it does, multiply the right-hand side by i:
ΔTb = i Kb m · ΔTf = i Kf m · π = i C R T
Dissociation into n particles: α = (i − 1) / (n − 1)
Association of n molecules into one: α = (1 − i) / (1 − 1/n)
For complete dissociation, i = 2 for NaCl, 3 for CaCl2 and 2 for KCl. For carboxylic acids dimerising in benzene, i falls towards 0.5. A value of i between 1 and n means partial dissociation.
Worked example 5. A 0.100 molal aqueous NaCl solution freezes at −0.348 °C. Kf(water) = 1.86 K kg mol−1. Find i and the degree of dissociation.
Calculated ΔTf (no dissociation) = 1.86 × 0.100 = 0.186 K
Observed ΔTf = 0.348 K
i = 0.348 ÷ 0.186 = 1.871
NaCl gives n = 2 particles, so α = (i − 1) ÷ (n − 1) = (1.871 − 1) ÷ (2 − 1) = 0.871, i.e. the NaCl is about 87.1% dissociated.
Mistakes that cost marks
- Molality vs molarity. ΔTb and ΔTf use molality (per kg of solvent). Osmotic pressure uses molarity (per litre of solution). Swapping them is the single biggest error in this chapter.
- Using the mass of solution as w1. In the molar-mass formulas w1 is the mass of solvent only. If the question gives 100 g of solution containing 2 g of solute, then w1 = 98 g.
- Forgetting the 1000. It converts grams of solvent to kilograms inside the molar-mass formula. Drop it and your answer is off by a factor of 1000.
- Forgetting i for ionic solutes. Any salt, acid or base in water needs the van't Hoff factor. Glucose, urea and sucrose do not.
- Wrong R. Use R = 0.0821 L atm K−1 mol−1 for π in atm, or 8.314 J K−1 mol−1 for SI work. Do not mix them.
- Celsius in the osmotic pressure formula. T must be kelvin.
- Assuming Kb or Kf belongs to the solute. They are properties of the solvent.
Constants and formulas at a glance
| Property | Formula | Concentration used |
|---|---|---|
| Relative lowering of vapour pressure | (p° − p)/p° = x2 | Mole fraction |
| Elevation of boiling point | ΔTb = i Kb m | Molality |
| Depression of freezing point | ΔTf = i Kf m | Molality |
| Osmotic pressure | π = i C R T | Molarity |
Commonly tabulated solvent constants (K kg mol−1). Values differ slightly between textbook editions and data tables, so use the value printed in your own question paper if one is supplied:
| Solvent | Kb | Kf |
|---|---|---|
| Water | 0.512 | 1.86 |
| Benzene | 2.53 | 5.12 |
| Chloroform | 3.63 | 4.79 |
| Acetic acid | 2.93 | 3.90 |
| Camphor | 5.95 | 39.7 |
Check your i values and colligative numericals instantly. The free van't Hoff Factor calculator relates observed and calculated colligative properties, so you can confirm a degree of dissociation or association in one step after solving it yourself.
Open the van't Hoff Factor Calculator →ABC Chemistry teaches Class 11–12 chemistry at its Gurugram centre and online across India, with worked numerical practice every week — see abcchemistry.in.