Colligative Properties — Boiling Point, Freezing Point, Osmosis

Dissolve salt in water and it boils at a higher temperature. Add antifreeze to a car radiator and the liquid won’t freeze in winter. Blood doesn’t mix freely with pure water because of osmotic pressure. All of these are colligative properties — and they all have one thing in common: they depend on how many solute particles are present, not on what those particles are.

The word “colligative” comes from the Latin colligare — to bind together. These properties are bound together by a single principle: adding solute particles lowers the chemical potential (escaping tendency) of the solvent, changing its physical behaviour.

In CBSE Class 12 and JEE, colligative properties appear in the Solutions chapter. Numericals on elevation of boiling point, depression of freezing point, and osmotic pressure are standard 3–5 mark questions.

Key Terms & Definitions

Colligative Property: A property of a solution that depends on the number of solute particles (not their identity) per mole of solvent.

Mole Fraction ( $\chi$ ): Fraction of moles of a component to total moles in solution.

Molality ( $m$ ): Moles of solute per kilogram of solvent. (Not affected by temperature changes — preferred for colligative properties.)

Molarity ( $M$ ): Moles of solute per litre of solution. (Affected by temperature.)

Van’t Hoff Factor ( $i$ ): Accounts for dissociation or association of solute. $i = \dfrac{\text{actual particles after dissolution}}{\text{formula units dissolved}}$ .

For non-electrolytes (like glucose): $i = 1$ . For NaCl: $i \approx 2$ (Na⁺ + Cl⁻). For $\text{CaCl}_2$ : $i \approx 3$ (Ca²⁺ + 2Cl⁻). For associating solutes (e.g., acetic acid in benzene, forms dimers): $i < 1$ .

The Four Colligative Properties

1. Relative Lowering of Vapour Pressure (RLVP)

When a non-volatile solute is dissolved, the vapour pressure of the solvent decreases. Solute particles occupy surface area, reducing the rate of evaporation.

Raoult’s Law:

p = p^* \chi_{\text{solvent}} = p^*(1 - \chi_{\text{solute}})

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

Where $p^*$ = vapour pressure of pure solvent, $p$ = vapour pressure of solution.

The relative lowering of vapour pressure equals the mole fraction of solute — a purely colligative relationship.

With Van’t Hoff factor:

\frac{p^* - p}{p^*} = i \cdot \chi_{\text{solute}}

2. Elevation of Boiling Point ( $\Delta T_b$ )

Lower vapour pressure means higher temperature needed to reach atmospheric pressure → higher boiling point.

\Delta T_b = K_b \cdot m \cdot i

Where:

$K_b$ = ebullioscopic constant (boiling point elevation constant) in K·kg/mol
$m$ = molality of solution
$i$ = Van’t Hoff factor

For water: $K_b = 0.52$ K·kg/mol. The new boiling point: $T_b = 373 + \Delta T_b$ (in K).

3. Depression of Freezing Point ( $\Delta T_f$ )

Solute disrupts the crystal lattice formation of the solvent → lower temperature needed to freeze → lower freezing point.

\Delta T_f = K_f \cdot m \cdot i

Where $K_f$ = cryoscopic constant (freezing point depression constant) in K·kg/mol.

For water: $K_f = 1.86$ K·kg/mol. The new freezing point: $T_f = 273 - \Delta T_f$ (in K).

$K_f > K_b$ for water (1.86 vs 0.52). Depression of freezing point is more sensitive than elevation of boiling point for the same solution. This is why freezing point depression is preferred for molecular weight determination.

4. Osmotic Pressure ( $\pi$ )

Osmosis: the flow of solvent from lower concentration to higher concentration across a semipermeable membrane. Osmotic pressure is the pressure needed to just stop this flow.

\pi = iMRT = \frac{inRT}{V}

Where:

$M$ = molarity
$R = 8.314$ J/(mol·K) or $0.0821$ L·atm/(mol·K)
$T$ = temperature in Kelvin

Isotonic solutions: Same osmotic pressure — no net osmosis.

Hypertonic: Higher solute concentration (higher $\pi$ ). Water leaves cells placed in it (cell shrinks — plasmolysis).

Hypotonic: Lower solute concentration. Water enters cells (cell swells — haemolysis in RBCs).

Important Formulas

\Delta T_b = K_b \times m \times i

For molecular weight determination: $M_2 = \dfrac{K_b \times w_2 \times 1000}{\Delta T_b \times w_1}$

where $w_2$ = mass of solute, $w_1$ = mass of solvent in grams.

\Delta T_f = K_f \times m \times i

For molecular weight determination: $M_2 = \dfrac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1}$

\pi = iMRT = \frac{inRT}{V}

At STP: $\pi V = nRT$ (analogous to ideal gas law — this is why dilute solutions are called “ideal”).

Solved Examples

Example 1 — Easy (CBSE Level)

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

Solution:

m = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{1.8/180}{100/1000} = \frac{0.01}{0.1} = 0.1 \text{ mol/kg}

Glucose is non-electrolyte, so $i = 1$ :

\Delta T_f = 1.86 \times 0.1 \times 1 = \mathbf{0.186 \text{ K}}

New freezing point = $0 - 0.186 = -0.186°\text{C}$ .

Example 2 — Medium (JEE Level)

0.585 g of NaCl ( $M = 58.5$ g/mol) is dissolved in 200 g of water. Find the elevation in boiling point. Assume complete dissociation. ( $K_b = 0.52$ K·kg/mol)

Solution:

m = \frac{0.585/58.5}{200/1000} = \frac{0.01}{0.2} = 0.05 \text{ mol/kg}

NaCl dissociates: Na⁺ + Cl⁻, so $i = 2$ :

\Delta T_b = 0.52 \times 0.05 \times 2 = \mathbf{0.052 \text{ K}}

New boiling point = $100 + 0.052 = 100.052°\text{C}$ .

Example 3 — Hard (JEE Main Level)

The osmotic pressure of a 5% solution of urea ( $M = 60$ g/mol) at 27°C is:

Solution: 5% w/v means 5 g in 100 mL = 50 g in 1 L.

M = \frac{50}{60} = 0.833 \text{ mol/L}

$T = 27 + 273 = 300$ K, $i = 1$ (non-electrolyte):

\pi = 0.833 \times 0.0821 \times 300 = \mathbf{20.5 \text{ atm}}

Example 4 — Molecular Weight Determination

An unknown compound: 2 g dissolved in 100 g of benzene raises boiling point by 0.52°C. ( $K_b \text{ for benzene} = 2.6$ K·kg/mol). Find molecular weight.

Solution:

M_2 = \frac{K_b \times w_2 \times 1000}{\Delta T_b \times w_1} = \frac{2.6 \times 2 \times 1000}{0.52 \times 100} = \frac{5200}{52} = \mathbf{100 \text{ g/mol}}

Exam-Specific Tips

JEE Main favourite: Questions on Van’t Hoff factor with degree of dissociation. If NaCl has degree of dissociation $\alpha$ : $i = 1 + \alpha$ (for binary electrolytes). In general: $i = 1 + (n-1)\alpha$ where $n$ = number of ions formed. For $\text{CaCl}_2$ ( $n=3$ ): $i = 1 + 2\alpha$ .

CBSE Class 12: Numerical on freezing point depression and molecular weight determination appear almost every year. The formula $M_2 = K_f \times w_2 \times 1000 / (\Delta T_f \times w_1)$ must be memorised with correct units.

Common Mistakes to Avoid

Mistake 1: Using molarity instead of molality. Colligative properties use molality ( $m$ ) not molarity ( $M$ ). Molality = moles / kg solvent. Molarity = moles / L solution. Using molarity gives wrong answers.

Mistake 2: Forgetting Van’t Hoff factor for electrolytes. For NaCl, $\Delta T_f = K_f \times m \times 2$ , not $K_f \times m$ . Forgetting $i$ halves your answer for NaCl problems.

Mistake 3: Wrong osmotic pressure units. Use $R = 0.0821$ L·atm/(mol·K) when molarity is in mol/L and $T$ in K. Or use $R = 8.314$ J/(mol·K) for SI units. Mixing unit systems gives nonsensical answers.

Mistake 4: Saying osmotic pressure is “just a pressure.” Osmotic pressure is a colligative property — it depends on solute concentration, not on what the solute is. Two solutions with the same molar concentration have the same osmotic pressure, regardless of the chemical identity.

Practice Questions

Q1. 5.85 g of NaCl dissolved in 500 g water. Find molality. ( $M_{\text{NaCl}} = 58.5$ )

$m = \frac{5.85/58.5}{500/1000} = \frac{0.1}{0.5} = 0.2$ mol/kg

Q2. What is the freezing point of 0.1 m glucose solution? ( $K_f = 1.86$ )

$\Delta T_f = 1.86 \times 0.1 \times 1 = 0.186$ K. Freezing point = $0 - 0.186 = -0.186°C$

Q3. A 0.2 m NaCl solution: find elevation of boiling point. ( $K_b = 0.52$ , assume $i = 2$ )

$\Delta T_b = 0.52 \times 0.2 \times 2 = 0.208$ K. Boiling point = $100.208°C$

Q4. Calculate osmotic pressure of 0.01 M NaCl at 298 K. ( $R = 0.0821$ )

$i = 2$ for NaCl. $\pi = 2 \times 0.01 \times 0.0821 \times 298 = 0.489$ atm

Additional Worked Examples

Example 5 — Degree of Dissociation from Freezing Point

The freezing point of 0.01 m NaCl solution is found to be $-0.0361°\text{C}$ . Calculate the degree of dissociation of NaCl. ( $K_f = 1.86$ K·kg/mol)

\Delta T_f = 0.0361 \text{ K}

Without dissociation: $\Delta T_f^{\text{calc}} = K_f \times m = 1.86 \times 0.01 = 0.0186 \text{ K}$

i = \frac{\Delta T_f^{\text{obs}}}{\Delta T_f^{\text{calc}}} = \frac{0.0361}{0.0186} = 1.94

For NaCl ( $n = 2$ ions): $i = 1 + (n-1)\alpha = 1 + \alpha$

1.94 = 1 + \alpha \implies \alpha = 0.94

NaCl is 94% dissociated — close to complete, as expected for a strong electrolyte.

Example 6 — Reverse Osmosis Pressure

Sea water contains approximately 3.5% NaCl by weight. Estimate the minimum pressure needed for reverse osmosis desalination at 25°C.

3.5 g NaCl in 100 mL (approximately). $M = \frac{3.5/58.5}{0.1} \approx 0.6$ mol/L.

$i \approx 2$ for NaCl. $\pi = iMRT = 2 \times 0.6 \times 0.0821 \times 298 \approx 29.3$ atm.

For reverse osmosis, the applied pressure must exceed the osmotic pressure. In practice, RO plants operate at 50–80 atm to achieve reasonable flow rates.

i = 1 + (n - 1)\alpha

where $n$ = number of ions the solute dissociates into and $\alpha$ = degree of dissociation.

For association (e.g., acetic acid dimerising in benzene):

i = 1 - \alpha + \frac{\alpha}{n}

where $n$ = number of molecules that associate together.

Mistake 5: Applying osmotic pressure formula with molality instead of molarity. The formula $\pi = iMRT$ uses molarity (M), not molality (m). This is because osmotic pressure depends on the number of particles per unit volume of solution. Mixing up M and m changes the answer, especially for concentrated solutions.

Q5. A 0.05 m solution of $K_2SO_4$ has a freezing point of $-0.256°\text{C}$ . Find $i$ and the degree of dissociation.

$\Delta T_f = 0.256$ K. Without dissociation: $\Delta T_f = 1.86 \times 0.05 = 0.093$ K.

$i = 0.256/0.093 = 2.75$ . For $K_2SO_4$ ( $n = 3$ ): $i = 1 + 2\alpha$ .

$2.75 = 1 + 2\alpha \implies \alpha = 0.875$ (87.5% dissociation).

Q6. Which has a higher boiling point: 0.1 m glucose or 0.1 m NaCl? Explain.

0.1 m NaCl has a higher boiling point. NaCl dissociates into Na⁺ and Cl⁻ ( $i \approx 2$ ), effectively doubling the number of solute particles. Since $\Delta T_b = K_b \times m \times i$ , NaCl gives $\Delta T_b \approx 2 \times$ that of glucose ( $i = 1$ ).

FAQs

Why do colligative properties depend on number of particles, not their type?

All colligative properties originate from the lowering of chemical potential (or vapour pressure) of the solvent when solute is added. Each solute particle contributes equally to this reduction, regardless of its nature. The entropy of mixing depends only on the number of particles.

Can colligative properties be used to find molecular weight?

Yes — this is one of the main applications. By measuring $\Delta T_b$ or $\Delta T_f$ experimentally and using the formula, we can calculate $M_2$ (molar mass of solute). This is especially useful for polymers and large biological molecules.

Why is $K_f$ larger than $K_b$ for water?

The origin is thermodynamic: $K_f = RT_f^2 M_1/(\Delta H_{fus} \times 1000)$ and $K_b = RT_b^2 M_1/(\Delta H_{vap} \times 1000)$ . For water, $\Delta H_{vap} >> \Delta H_{fus}$ (vaporisation requires more energy than melting), so $K_b < K_f$ .

What is abnormal molar mass?

When a solute ionises ( $i > 1$ ) or associates ( $i < 1$ ), the calculated molar mass from colligative properties differs from the theoretical value. This “abnormal” value is explained by the Van’t Hoff factor. It’s how we experimentally determine degree of dissociation.

Key Terms & Definitions

The Four Colligative Properties

1. Relative Lowering of Vapour Pressure (RLVP)

2. Elevation of Boiling Point (ΔTb\Delta T_bΔTb​)

3. Depression of Freezing Point (ΔTf\Delta T_fΔTf​)

4. Osmotic Pressure (π\piπ)

Important Formulas

Solved Examples

Example 1 — Easy (CBSE Level)

Example 2 — Medium (JEE Level)

Example 3 — Hard (JEE Main Level)

Example 4 — Molecular Weight Determination

Exam-Specific Tips

Common Mistakes to Avoid

Practice Questions

Additional Worked Examples

Example 5 — Degree of Dissociation from Freezing Point

Example 6 — Reverse Osmosis Pressure

FAQs

Practice Questions

2. Elevation of Boiling Point ( $\Delta T_b$ )

3. Depression of Freezing Point ( $\Delta T_f$ )

4. Osmotic Pressure ( $\pi$ )