Solutions and colligative properties for NEET

Solutions is one of the most dependable scoring chapters in NEET physical chemistry. Almost every question is one of four kinds: rank some solutions by boiling or freezing point, calculate a molar mass from ΔTb, ΔTf or osmotic pressure, find the van 't Hoff factor for an electrolyte, or classify a pair of liquids as showing positive or negative deviation. Each kind has a fixed method.

The chapter rests on one idea. Colligative properties depend on the number of solute particles in a given amount of solvent, not on what those particles are. Hold on to that and the formulas make sense.

Concentration terms

Molality is used in the boiling and freezing point formulas because it does not change with temperature. For more practice on these, see concentration terms.

Gases in liquids: Henry's law

At constant temperature, the partial pressure of a gas above a solution is proportional to its mole fraction in the solution: p = K_H × x.

• A larger K_H means a lower solubility at the same pressure.
• Gas solubility falls as temperature rises, so K_H increases with temperature.

Everyday applications: soft drinks are sealed under high CO₂ pressure. Scuba divers risk "the bends" when nitrogen dissolved in the blood at depth forms bubbles as they rise. At high altitude, low oxygen pressure means less dissolved oxygen, which causes weakness and slow thinking (anoxia).

Raoult's law, ideal and non-ideal solutions

For a solution of two volatile liquids, the partial vapour pressure of each is its mole fraction times its pure vapour pressure:

p₁ = x₁p₁°, p₂ = x₂p₂°, and p_total = p₁ + p₂

When the solute is non-volatile, only the solvent contributes, and the relative lowering of vapour pressure equals the mole fraction of the solute:

(p₁° − p₁)/p₁° = x₂

An ideal solution obeys Raoult's law at all compositions. This happens when the A–B attractions are about the same as the A–A and B–B attractions. Then ΔmixH = 0 and ΔmixV = 0. Examples: n-hexane + n-heptane, benzene + toluene, bromoethane + chloroethane.

Why chloroform + acetone deviates negatively: chloroform's H forms a hydrogen bond with acetone's C=O oxygen, an attraction neither liquid had on its own. Why ethanol + acetone deviates positively: acetone molecules get between ethanol molecules and break some of their hydrogen bonds.

An azeotrope boils at a constant temperature with the same composition in liquid and vapour, so it cannot be separated by fractional distillation.

The four colligative properties

For water, Kb = 0.52 K kg mol⁻¹ and Kf = 1.86 K kg mol⁻¹. Kb and Kf depend only on the solvent.

Molar mass from colligative data. Rearranging with w₂ g of solute in w₁ g of solvent:

• M₂ = (Kb × w₂ × 1000)/(ΔTb × w₁)
• M₂ = (Kf × w₂ × 1000)/(ΔTf × w₁)
• M₂ = (w₂ × R × T)/(π × V)

Osmotic pressure is preferred for proteins and polymers. Their molar masses are large, so ΔTb and ΔTf would be too small to measure, but π is still measurable at room temperature.

Osmosis: solvent flows through a semipermeable membrane from the dilute side to the concentrated side. Osmotic pressure is the extra pressure needed to stop that flow.

• Isotonic solutions have the same osmotic pressure. A 0.9% (mass/volume) NaCl solution is isotonic with blood, which is why it is used as a drip.
• A cell placed in a hypertonic solution shrinks. In a hypotonic solution, it swells.
• Reverse osmosis: applying a pressure larger than π pushes solvent the other way. It is used to desalinate sea water.

Abnormal molar mass and the van 't Hoff factor

Electrolytes dissociate, making more particles than the formula suggests. Some solutes associate, making fewer. The van 't Hoff factor corrects for this:

i = normal molar mass/observed molar mass = observed colligative property/calculated colligative property

The corrected formulas become ΔTb = iKb m, ΔTf = iKf m and π = iCRT.

With partial dissociation into n ions and degree of dissociation α: i = 1 + (n − 1)α.

With association of n molecules into one, and degree of association α: i = 1 − (1 − 1/n)α.

Worked NEET-style questions

Example 1: freezing points

Find the freezing points of 0.1 m aqueous glucose and 0.1 m aqueous NaCl (assume complete dissociation). Kf = 1.86 K kg mol⁻¹.

• Glucose: ΔTf = 1 × 1.86 × 0.1 = 0.186 K, so it freezes at −0.186 °C.
• NaCl: ΔTf = 2 × 1.86 × 0.1 = 0.372 K, so it freezes at −0.372 °C.

Example 2: molar mass from boiling point

1.8 g of a non-electrolyte dissolved in 100 g of water raises the boiling point by 0.052 K. Kb = 0.52 K kg mol⁻¹. Find its molar mass.

• M₂ = (0.52 × 1.8 × 1000)/(0.052 × 100) = 936/5.2 = 180 g mol⁻¹. The solute could be glucose.

Example 3: relative lowering of vapour pressure

18 g of glucose is dissolved in 178.2 g of water. The vapour pressure of pure water is 23.8 mm Hg. Find the vapour pressure of the solution.

• Moles of glucose = 18/180 = 0.1. Moles of water = 178.2/18 = 9.9.
• x₂ = 0.1/(0.1 + 9.9) = 0.01.
• Δp = 0.01 × 23.8 = 0.238 mm Hg, so p = 23.8 − 0.238 = 23.562 mm Hg.

Example 4: van 't Hoff factor with partial dissociation

K₂SO₄ is 80% dissociated in a 0.1 m aqueous solution. Find i and ΔTf. Kf = 1.86 K kg mol⁻¹.

• K₂SO₄ gives 3 ions, so n = 3 and α = 0.8.
• i = 1 + (3 − 1) × 0.8 = 2.6.
• ΔTf = 2.6 × 1.86 × 0.1 ≈ 0.484 K.

Practice MCQs

1. Which 0.1 m aqueous solution has the highest boiling point? (a) glucose (b) urea (c) NaCl (d) BaCl₂
2. 4 g of NaOH is dissolved in 500 g of water. The molality is: (a) 0.1 m (b) 0.2 m (c) 0.8 m (d) 8 m
3. A mixture of chloroform and acetone shows: (a) ideal behaviour (b) positive deviation (c) negative deviation (d) no deviation at any composition
4. If ethanoic acid dimerises completely in benzene, its i is: (a) 2 (b) 1 (c) 0.5 (d) 0.25
5. Which concentration term is independent of temperature? (a) molarity (b) volume percentage (c) molality (d) mass/volume percentage
6. Assuming complete dissociation, which is isotonic with 0.2 M glucose? (a) 0.2 M NaCl (b) 0.1 M NaCl (c) 0.1 M CaCl₂ (d) 0.4 M urea
7. At the same temperature, the gas with the higher Henry's law constant is: (a) more soluble (b) less soluble (c) equally soluble (d) insoluble
8. The osmotic pressure of 0.2 M urea at 300 K (R = 0.0821 L atm K⁻¹ mol⁻¹) is about: (a) 0.49 atm (b) 2.46 atm (c) 4.93 atm (d) 49.3 atm

Answers

1. (d) BaCl₂ has i = 3, the largest i × m.
2. (b) 4/40 = 0.1 mol in 0.5 kg, so 0.2 m.
3. (c) Hydrogen bonding between them makes A–B attraction stronger.
4. (c) i = 1 − (1 − ½) × 1 = 0.5.
5. (c) Molality uses mass of solvent, which does not change with temperature.
6. (b) 0.1 × 2 = 0.2 M of particles.
7. (b) p = K_H × x, so for a given p, a larger K_H means a smaller x.
8. (c) π = 0.2 × 0.0821 × 300 ≈ 4.93 atm.

What to do next

• Copy the colligative property table and the van 't Hoff factor table onto one page and revise them daily for a week.
• Solve ten molar-mass questions, three from each of ΔTb, ΔTf and π, writing units at every step.
• Learn three examples each of positive and negative deviation, with the reason for each.
• Take Henry's law back to chemical equilibrium, where the syllabus lists it, then move on to electrochemistry.