Ch. 2 | Solutions0/13

Colligative Properties

What they are, why they depend only on particle count, and relative lowering of vapour pressure

Logical ReasoningLevel 2 · Application

You add 10 g of table salt (NaCl) to boiling water while cooking pasta. A friend says this "raises the boiling point and cooks pasta faster." Is this chemically correct? Does 10 g of NaCl in 1 litre of water produce a meaningful boiling point elevation?

✦Real Life Hook

The four colligative properties were discovered and quantified between 1870 and 1900 — and they revolutionised chemistry. Before them, chemists had no reliable method for determining molar masses of large molecules. After them, you could dissolve a mysterious protein in water, measure how much the freezing point dropped, and calculate its molar mass. François-Marie Raoult and Jacobus van't Hoff (Nobel Prize 1901) developed this toolkit. Every molar mass in 1890s chemistry was determined this way.

What Are Colligative Properties?

Colligative properties are solution properties that depend only on the number of solute particles (moles), not on the identity, size, charge, or chemical nature of those particles.

The four colligative properties:

  1. Relative lowering of vapour pressure (RLVP)
  2. Elevation of boiling point (ΔTb\Delta T_b)
  3. Depression of freezing point (ΔTf\Delta T_f)
  4. Osmotic pressure (π\pi)

Why particle count only? These properties arise from how solute particles dilute the solvent — replacing solvent molecules at the liquid surface or membrane interface. A large molecule and a small molecule, if equally numerous, have the same effect.

Relative Lowering of Vapour Pressure (RLVP)

From Raoult's law for non-volatile solute:
RLVP=p1∘−psolutionp1∘=x2=n2n1+n2\text{RLVP} = \frac{p_1^\circ - p_{\text{solution}}}{p_1^\circ} = x_2 = \frac{n_2}{n_1 + n_2}

For dilute solutions (n2≪n1n_2 \ll n_1):
p1∘−psp1∘≈n2n1=w2⋅M1w1⋅M2\frac{p_1^\circ - p_s}{p_1^\circ} \approx \frac{n_2}{n_1} = \frac{w_2 \cdot M_1}{w_1 \cdot M_2}

This equation is the basis of molar mass determination from vapour pressure measurements.

📖NCERT 2.7NCERT Intext

Problem

The vapour pressure of water is 12.3 kPa at 300 K. Calculate the vapour pressure of 1 molal solution of a non-volatile solute at this temperature. (Molar mass of water = 18 g/mol)

✦JEE / NEET Exam InsightJEE / NEET
◆Colligative property identification: "Depends on number of particles, not identity" → colligative. If a question says "0.1 m glucose" vs "0.1 m NaCl," they do NOT give the same colligative effect — NaCl gives 0.2 m effective particles (i = 2).
◆RLVP for molar mass: Rearranged: M2=w2M1p1∘w1(p1∘−ps)M_2 = \frac{w_2 M_1 p_1^\circ}{w_1 (p_1^\circ - p_s)}
◆Key numbers to remember:
    ◆KbK_b (water) = 0.52 K·kg/mol
    ◆KfK_f (water) = 1.86 K·kg/mol
    ◆KbK_b(benzene) = 2.53, KfK_f(benzene) = 5.12
    ◆KfK_f(camphor) = 37.7 — very large, useful for molar mass of polymers
Quick Check

Q1.Which of the following is NOT a colligative property?