5: Colligative Properties

For AB₂, complete dissociation produces three particles: A and two B ions. If α is the degree of dissociation, the van’t Hoff factor is i = 1 + α(n − 1), where n = 3. Therefore, 2.5 = 1 + α(3 − 1) = 1 + 2α. Hence α = 1.5/2 = 0.75, or 75%. Thus, three-fourths of the AB₂ molecules dissociate into ions.

For freezing-point depression, ΔTf = iKf m. Therefore, i = ΔTf/(Kf m) = 0.186/(1.86 × 0.2) = 0.186/0.372 = 0.5. Since the value is less than unity, the number of solute particles has decreased compared with the undissociated state. This decrease is explained by association, such as dimerisation, rather than dissociation.

A non-volatile solute lowers the vapour pressure of the solution compared with the pure solvent. Boiling begins only when the vapour pressure becomes equal to the external pressure. Since the solution starts with a lower vapour pressure, it must be heated to a higher temperature to reach that condition. This increase in boiling temperature is called elevation of boiling point, so the two effects are directly related.

For a dilute solution, osmotic pressure is given by π = iCRT. At the same temperature and concentration, C, R, and T are identical for both solutions, so the comparison depends on i. Glucose is a non-electrolyte and has i approximately equal to 1. Sodium sulfate dissociates ideally as Na₂SO₄ → 2Na⁺ + SO₄²⁻, giving i approximately equal to 3. Therefore, sodium sulfate produces the higher osmotic pressure.

The elevation in boiling point is calculated using ΔTb = iKb m. Because the solute is non-dissociating, i = 1. Substituting the values gives 0.52 K = (1)(0.52 K kg mol⁻¹)m. Therefore, m = 0.52/0.52 = 1.00 mol kg⁻¹. The unit is molality, meaning moles of solute per kilogram of solvent, not per kilogram of solution.

Boiling-point elevation follows the relation ΔTb = iKb m. When the solvent is the same, its ebullioscopic constant Kb is fixed. Equal molality makes m identical as well. Thus, the only relevant changing quantity in the comparison is i, the van’t Hoff factor, which represents the effective number of dissolved particles. A solute producing more particles gives a larger boiling-point elevation.

The osmotic pressure is the minimum external pressure that must be applied to stop the natural flow of solvent through a semipermeable membrane. At exactly 5 atm, osmosis is merely stopped and there is no net reverse flow. To force solvent from the solution side toward the pure-solvent side, the applied pressure must exceed the osmotic pressure. Therefore, reverse osmosis requires a pressure greater than 5 atm.

For elevation of boiling point, the relation is ΔTb = iKb m. In the first solution, the solute does not dissociate, so i = 1 and the elevation is 0.26 K. The second solution has the same solvent and the same molality, but its i value is 3. Thus its elevation is three times as large: 3 × 0.26 = 0.78 K. Hence option C is correct.

For freezing-point depression, the relation is ΔT_f = iK_fm, where i is the van’t Hoff factor. Rearranging gives i = ΔT_f/(K_fm). Substituting the data, i = 0.372/(1.86 × 0.100) = 0.372/0.186 = 2.00. Therefore, the solution produces twice the ideal number of solute particles, usually because the solute dissociates into ions.

NaCl dissociates into two ions, Na⁺ and Cl⁻. For a solute that produces n particles on complete dissociation, the van’t Hoff factor is i = 1 + α(n − 1), where α is the degree of dissociation. Here α = 0.75 and n = 2, so i = 1 + 0.75(2 − 1) = 1.75. Thus, option C is correct. The value lies between 1 and 2 because dissociation is incomplete.

Molality is calculated from moles of solute divided by kilograms of solvent: m = 0.1/2 = 0.05 mol kg⁻¹. For a solute that dissociates or otherwise changes the number of particles, the effective molality is i multiplied by the ordinary molality. Therefore, effective molality = 3 × 0.05 = 0.15 m. Hence option C is correct.

MgCl₂ dissociates according to MgCl₂ → Mg²⁺ + 2Cl⁻, producing a total of n = 3 ions when dissociation is complete. For a degree of dissociation α, the van’t Hoff factor is i = 1 + α(n − 1). Substituting α = 0.50 and n = 3 gives i = 1 + 0.50(3 − 1) = 1 + 1 = 2. Therefore, the correct answer is 2.0, option B. This factor indicates that the number of solute particles has doubled relative to the original undissociated formula units under these conditions.

For a solution containing a non-volatile solute, Raoult’s law gives the relative lowering of vapour pressure as (p° − p)/p° = Xsolute, where Xsolute is the mole fraction of the solute. Therefore, increasing the solute mole fraction directly increases the relative lowering. It does not become zero or show a turning trend.

Molarity is defined as M = n/V, where n is the amount of solute in moles and V is the volume of the solution in litres. Rearranging this equation gives n = M × V. Therefore, after molarity has been obtained from osmotic pressure, multiply it by the solution volume expressed in litres. For example, if M = 0.4 mol L⁻¹ and V = 0.25 L, then n = 0.10 mol. Millilitres must first be converted to litres.

For a dilute solution containing a non-volatile solute, Raoult's law gives (p° − p)/p° = x₂, where p° is the vapour pressure of the pure solvent, p is the vapour pressure of the solution, and x₂ is the solute mole fraction. Because the solution is dilute, this relation is an excellent approximation and is used to determine molar mass. Therefore, option A is correct.

Colligative properties depend on the number of solute particles actually present in solution. A dilute solution containing a pure solute that has dissolved completely is closest to the assumptions used in the ideal equations and minimizes complications such as association, incomplete dissolution, and non-ideal interactions. A concentrated, impure, or incompletely dissolved system gives less reliable results. Hence, option A is correct.

For equal masses, the number of moles is n = mass/molar mass. Therefore, the solute with the lower molar mass provides more moles and hence more solute particles. Since the solutes are non-dissociating, each mole remains the same number of particles, so no additional van’t Hoff factor changes the comparison. The greater number of particles produces the greater colligative effect. Thus, the first solute is correct.

A lower-than-expected apparent molar mass means that the observed freezing-point depression is greater than the value predicted for undissociated solute molecules. An electrolyte can dissociate into ions in solution, increasing the number of effective particles. The increased particle number enhances the freezing-point depression and makes the calculated molar mass appear smaller. Therefore, dissociation is the suitable explanation. Association would instead reduce the particle number and raise the apparent molar mass.

For complete dissociation into three ions, the van’t Hoff factor i equals 3 because the number of particles becomes three times the number of formula units. The relation between observed and normal molar mass is Mobserved = Mnormal/i. Therefore, Mobserved = Mnormal/3, or one-third of the normal molar mass. Dissociation lowers the observed molar mass calculated from a colligative property.

Osmotic pressure is most suitable for proteins and other macromolecules because it can be measured accurately in very dilute solutions at or near room temperature. The changes in boiling point or freezing point are often extremely small for high-molar-mass solutes. Using π = CRT, the molar mass can be calculated without heating or cooling the sample strongly. Hence, option D is correct.

For a high-molar-mass solute, the changes in vapour pressure, boiling point, and freezing point may be too small for reliable measurement. Osmotic pressure can be measured in very dilute solutions at room temperature, so the heat-sensitive solute is not exposed to heating. The relation π = CRT permits molar-mass calculation. Therefore, option A is correct.

Freezing-point depression is a colligative property and depends on the number of dissolved particles: \(\Delta T_f=iK_fm\). An electrolyte dissociates into ions, increasing the number of effective particles and making \(i>1\). Consequently, the observed depression is greater than the value calculated by treating the electrolyte as undissociated. Therefore, option A is correct.

Elevation of boiling point depends on the number of solute particles according to \(\Delta T_b=iK_bm\). Some acids, such as ethanoic acid, associate through hydrogen bonding and form dimers in suitable nonpolar organic solvents. The number of effective particles decreases, so \(i<1\), the boiling-point elevation becomes smaller, and the calculated molar mass appears higher. Option A is correct.

For freezing-point depression, ΔTf = iKf m, and the molar mass calculated by ignoring the van’t Hoff factor is inversely related to i. Dissociation increases the number of solute particles, so i becomes greater than 1 and the observed colligative effect becomes larger. Consequently, the calculated molar mass is smaller than the true value. Therefore, dissociation is the correct reason.

The ordinary molality of the solution is given as 0.3 m. In colligative-property equations, the particle effect is represented by the product i m, called effective molality. Therefore, effective molality = i × m = 2 × 0.3 = 0.60 m. The value doubles because the solute produces twice the number of effective particles compared with a non-dissociating solute at the same analytical molality.