5: Colligative Properties

The usual expression for relative lowering of vapour pressure assumes that the solute is non-volatile, so only the solvent contributes appreciably to the vapour phase. If the solute is volatile, it also produces vapour and the total pressure depends on both solvent and solute partial pressures. Therefore, the simple direct relation for relative lowering of vapour pressure is not applicable without modification. The other listed effects can still be considered using appropriate conditions and equations.

In ordinary osmosis, solvent passes through a semipermeable membrane from the dilute solution toward the concentrated solution because of the osmotic pressure difference. In reverse osmosis, pressure greater than the osmotic pressure is applied to the concentrated side. This imposed pressure overcomes the natural osmotic tendency and forces solvent toward the dilute or pure-solvent side, allowing desalination and water purification.

For a solution containing a non-volatile solute, Raoult’s law gives p = xsolvent p°. Hence the lowering in vapour pressure is p° − p = p°(1 − xsolvent) = p°xsolute, because the mole fractions of solvent and solute add to unity. Dividing by p° gives (p° − p)/p° = xsolute. Thus the relative lowering equals the solute mole fraction, not its mass fraction or molality.

The freezing-point depression is ΔTf = iKf m. Since the molality and solvent are the same, the order depends only on the van’t Hoff factor. Glucose does not ionise, so i = 1. Sodium chloride ideally produces two ions, so i ≈ 2, while aluminium chloride produces four ions, so i ≈ 4. Therefore, aluminium chloride causes the greatest depression, followed by sodium chloride and glucose.

The ideal osmotic-pressure equation is π = iCRT. If the nature of the solute is unchanged, i remains constant, and the concentration C is also specified to remain constant. Therefore, osmotic pressure is directly proportional to absolute temperature: π₂/π₁ = T₂/T₁ = 600/300 = 2. Hence π₂ = 2π. The comparison must use kelvin temperatures, not Celsius values.

Freezing-point depression is given by ΔTf = iKf m. Urea is a non-electrolyte and does not dissociate appreciably, so iurea = 1. In the ideal case, calcium chloride dissociates completely as CaCl2 → Ca²⁺ + 2Cl⁻, producing three particles and iCaCl2 = 3. Since Kf and molality are the same in both solutions, the calcium chloride solution has three times the depression: 3x.

Freezing requires solvent molecules to arrange themselves into an ordered solid crystal lattice. When non-volatile solute particles are present, they occupy positions among solvent molecules and interfere with this regular arrangement. Therefore, the solution must be cooled to a temperature lower than the freezing point of the pure solvent before crystallisation can begin. The change depends mainly on the number of dissolved particles.

The van’t Hoff factor for complete dissociation is the total number of particles formed from one formula unit, assuming ideal behaviour. One formula unit of A₂B produces two A⁺ ions and one B²⁻ ion. Thus, the total number of particles is 2 + 1 = 3, so the ideal van’t Hoff factor is i = 3. This ignores ion pairing and other non-ideal effects.

Isotonic solutions have equal osmotic pressures at the same temperature. Osmotic pressure is given by π = iCRT, where iC represents the effective concentration of solute particles. Therefore, equal osmotic pressure means equal effective particle concentration, even when the solutions contain different solutes, different molarities, or different van’t Hoff factors. Equal molality is not necessary because electrolytes may produce more particles than nonelectrolytes.

Depression in freezing point follows ΔTf = iKf m. For nonelectrolytes, the van’t Hoff factor i is approximately 1, and Kf is fixed for the same solvent. Thus, two different non-volatile nonelectrolytes having equal molality produce equal depression. This demonstrates dependence on the number of dissolved particles rather than their chemical identity. Equal mass or equal molar mass does not guarantee equal particle concentration.

In reverse osmosis, pressure greater than the osmotic pressure is applied to seawater. This forces water through a semipermeable membrane in the direction opposite to natural osmosis. The membrane permits water molecules to pass more readily but rejects most dissolved ions and larger impurities. The process therefore reduces salinity and is widely used for desalination, although it requires energy and suitable membrane maintenance.

The magnitude of a colligative effect is proportional to the effective number of solute particles, represented by the van’t Hoff factor i. Complete dissociation would give an effective particle factor of 4, whereas the actual factor is only 3.2. Because 3.2 is smaller than 4, the observed freezing-point change, boiling-point change, osmotic pressure, or vapour-pressure effect will be less than the value for complete dissociation.

For a solution containing an electrolyte or any particle-number correction, the depression in freezing point is ΔTf = iKf m. Substituting the given values gives ΔTf = 2.5 × 1.86 × 0.2 K. First, 1.86 × 0.2 = 0.372; multiplying by 2.5 gives 0.930 K. Therefore, option C is correct. Omitting i would incorrectly give 0.372 K.

The van’t Hoff factor compares the actual number of particles in solution with the number expected for an undissociated solute. When i = 1, there is no net change. A value below 1, such as 0.75, means that particles have combined to form larger associated units. Therefore, the correct answer is particle association, not dissociation or complete ionisation.

Molality is the number of moles of solute per kilogram of solvent. Here, 500 g of solvent equals 0.5 kg, so the ordinary molality is 1 ÷ 0.5 = 2 mol kg⁻¹. Colligative effects depend on the effective particle concentration, which is i times the molality. Therefore effective molality = 2 × 2 = 4 mol kg⁻¹, giving option C.

For a solution containing a non-volatile solute, Raoult’s law gives relative lowering of vapour pressure as the mole fraction of the solute: (P° − P)/P° = Xsolute. Therefore, Xsolute = moles of solute / total moles = 0.25. With total moles equal to 4, moles of solute = 0.25 × 4 = 1.0 mole. Hence, option B is correct.

The electrolyte XY dissociates into two particles, X and Y. For a substance producing two ions, the van’t Hoff factor is i = 1 + α, where α is the fractional degree of dissociation. Thirty percent dissociation means α = 0.30. Therefore i = 1 + 0.30 = 1.30. The factor is greater than one because dissociation increases the number of particles.

In complete trimerisation, every group of three original solute molecules combines to produce one associated particle. Thus, if the initial number of particles is N, the final number is N/3. The van’t Hoff factor is the ratio of actual effective particles to the original particles, so i = (N/3)/N = 1/3. Association reduces the particle count, making i less than one.

A smaller-than-expected relative lowering means that fewer effective solute particles are present than the formula assumes. Association combines two or more solute molecules into larger units, decreasing particle number and producing a van’t Hoff factor below one, \(i<1\). Dissociation would instead increase the effect and give \(i>1\).

The van’t Hoff factor represents the ratio of the actual number of dissolved particles to the number expected if no association or dissociation occurred. During association, two or more solute particles combine into one larger species, reducing the number of independent particles. Consequently, i becomes less than 1. Dissociation produces additional particles and gives i greater than 1, while no change gives i approximately equal to 1.

For a solution containing a non-volatile solute, adding more solute lowers the mole fraction of the solvent. By Raoult’s law, the partial vapour pressure of the solvent decreases. The solution must then be heated to a higher temperature before its vapour pressure equals the external pressure, so its boiling point rises. These are connected colligative effects: lowering of vapour pressure and elevation of boiling point.

At equal molality, the depression in freezing point is ΔTf = iKf m, so the deciding factor is the van’t Hoff factor. Complete dissociation gives approximately i = 2 for NaCl because it forms Na+ and Cl−, while CaCl2 gives approximately i = 3 because it forms Ca2+ and two Cl− ions. Therefore, under complete dissociation, CaCl2 produces more dissolved particles and a greater freezing-point depression.

Osmotic pressure is given by π = iCRT. Urea is a non-electrolyte, so one mole gives one particle and i = 1. Completely dissociated K₂SO₄ gives two K⁺ ions and one SO₄²⁻ ion, so i = 3. Since temperature and molar concentration are equal, π(urea):π(K₂SO₄) = 1:3.

For a dilute solution, boiling-point elevation is ΔT_b = iK_bm. Dissociation converts one solute unit into two or more particles, increasing the van’t Hoff factor i above its ideal nonelectrolyte value. At the same molality and with the same solvent, this produces a larger elevation than expected. Association would instead reduce the number of particles.

Freezing-point depression is ΔT_f = iK_fm. Both solutions have the same solvent and molality, so their depressions are proportional only to i. Completely dissociated BaCl₂ produces Ba²⁺ and two Cl⁻ ions, giving i = 3. Glucose does not ionise and gives i = 1. Hence the ratio is 3:1.