5: Colligative Properties

For each non-electrolyte, i is approximately 1. In the same solvent, Kb is also the same, and equal molality gives the same value of m. Substitution in ΔTb = iKb m therefore gives equal boiling-point elevations, assuming dilute ideal behaviour. Molar mass, colour or the label ‘heavy’ does not independently determine the colligative effect.

A non-volatile solute lowers the vapour pressure of the solvent at a given temperature. Boiling occurs when vapour pressure equals the external pressure, so the solution must be heated to a higher temperature to reach that condition. The increase is therefore linked to vapour-pressure lowering, not to colour, loss of solvent mass or escape of the solute.

Adding a solute lowers the chemical potential and escaping tendency of the solvent in the liquid solution. As a result, the liquid and solid phases reach equilibrium only at a lower temperature than for the pure solvent. The effect does not require the solute to produce heat, remove all particles or transform the solvent into another substance.

For a fixed mass, n = mass/molar mass, so a lower molar mass produces a larger number of moles. A non-electrolyte contributes approximately one particle per molecule, giving more effective particles in solution. Since colligative effects depend on particle number, the boiling-point elevation or freezing-point depression is larger. Colour and physical state are not the governing reasons.

In colligative-property equations such as ΔTb = iKb m and ΔTf = iKf m, the van’t Hoff factor multiplies the particle-based effect. An ideal non-electrolyte has i = 1, whereas i = 2 represents twice the effective particle number at the same concentration. Therefore the predicted colligative effect is approximately twice as large, assuming all other quantities are equal.

Freezing-point depression is ΔTf = iKf m. At equal molality and in the same solvent, Kf and m are the same, so the solution with the larger van’t Hoff factor has the greater depression. A solute that dissociates into more effective ions generally has a larger i, subject to the actual degree of dissociation. Colour and container size are irrelevant.

For a non-electrolyte solution, boiling-point elevation is given by ΔT_b = K_b m. To find molality, divide both sides by K_b: m = ΔT_b / K_b. Substitution gives m = 2 / 1 = 2. The value 0.5 would result from reversing the division, and adding the quantities would not follow the governing relation.

For a dilute non-electrolyte solution, the freezing-point relation is \(\Delta T_f = K_fm\). Rearranging gives \(m = \Delta T_f/K_f\). Substitution yields \(m = 3.6/1.8 = 2\). Therefore, the molality is 2, in the compatible molality unit implied by the given values. Option B would result from an incorrect division, C from multiplication, and D from reversing the ratio.

Boiling-point elevation is given by ΔTb = iKb m. The two solutions have the same solvent and molality, so Kb and m are identical. Glucose does not ionise and has i = 1, whereas completely ionised NaCl produces Na+ and Cl−, giving i approximately 2. Therefore NaCl produces nearly twice the elevation under the stated ideal assumption.

For a dilute solution of a non-electrolyte, elevation in boiling point is ΔTb = Kb m, where Kb is the solvent’s ebullioscopic constant and m is molality. If dissociation or association occurs, the more general form includes the van’t Hoff factor i. Option B describes freezing-point depression, while C describes osmotic pressure and D describes relative vapour-pressure lowering.

For freezing-point depression, the relation is ΔTf = iKf m. Here, both solutions use the same solvent, so Kf is the same; their molalities m are also equal; and non-dissociating solutes have van’t Hoff factor i = 1. Therefore, each solution has the same value of ΔTf. The depression is not zero, because dissolved particles lower the solvent’s freezing point. Options B and C incorrectly assume one solution has more effect, while D ignores the presence of solute.

Equal moles alone do not guarantee equal particle numbers. Glucose remains largely as one molecular species, so i is about 1. An ionic salt dissociates into ions, giving a larger i and more effective particles under the stated ideal assumption. Therefore the salt solution has the greater colligative effect.

Use ΔTf = iKf m. At equal molality in the same solvent, the solution with the largest i has the greatest depression and therefore the lowest freezing point. CaCl₂ ideally gives three ions, Ca²⁺ and two Cl⁻, so i = 3. Glucose, urea, and sucrose are non-electrolytes with i about 1.

For a dilute solution, ΔTb = iKb m. With the same solvent, Kb is common to both solutions, and a non-dissociating solute has i = 1. Since the molalities are also equal, both products iKb m are equal, so the boiling-point elevations are equal. They are not zero because dissolved solute particles are present; colour and total mass do not determine this colligative effect.

For a dilute solution, osmotic pressure follows the ideal dilute-solution relation πV = nRT, or π = nRT/V. Here n is the amount of dissolved solute particles, V is the solution volume, T is absolute temperature, and R is the gas constant. The other equations mix unrelated quantities and do not represent the osmotic-pressure relation.

From πV = nRT, rearrangement gives π = nRT/V. If n, V, and R remain constant, osmotic pressure is directly proportional to the absolute temperature T. Thus increasing temperature increases π. The pressure does not become zero or negative merely because temperature rises; a decrease would contradict the direct proportionality in the stated dilute-solution model.

For a dilute solution of a non-volatile solute, the relative lowering of vapour pressure is (p° − p)/p° = x_solute, where p° is the vapour pressure of the pure solvent and p is that of the solution. Thus it depends on the solute mole fraction. A larger number of dissolved particles generally gives a larger lowering, while colour, shape and ice hardness are unrelated.

Colligative properties depend on the number of dissolved particles. Sugar and urea are treated as non-electrolytes, so one mole of either gives approximately one mole of solute particles. At equal molality in the same solvent, their ideal boiling-point elevation, freezing-point depression, or osmotic effect is therefore nearly the same, despite their different chemical identities.

For a non-electrolyte, boiling point elevation follows ΔT_b = K_bm. To find molality, divide both sides by K_b: m = ΔT_b/K_b. Thus m = 1.5/0.5 = 3. The calculation assumes i = 1 because no dissociation or association is stated. Multiplying the two values would not isolate m and would give the wrong result.

For a non-electrolyte solution, ΔT_f = K_fm. Rearranging gives m = ΔT_f/K_f. Substituting the data, m = 0.9/1.8 = 0.5 mol kg⁻¹. The answer is not 2 because that reverses the division, and 1.62 is the product 0.9 × 1.8 rather than the required quotient. No van't Hoff correction is needed because none is specified.

In dilute ideal solutions, colligative effects are proportional to the effective number of dissolved particles. Thus, if the solvent and temperature remain the same and the particle number doubles, the relative vapour-pressure lowering, boiling-point elevation, freezing-point depression, or osmotic pressure is approximately doubled. The word ‘nearly’ allows for non-ideal interactions at higher concentration.

After calculating the magnitude of the change, the sign must be applied correctly. For boiling-point elevation, T_b(solution) = T_b° + ΔT_b; for freezing-point depression, T_f(solution) = T_f° − ΔT_f. Forgetting this distinction gives a numerically plausible but physically wrong answer. The other choices are not calculation rules.

For a non-electrolyte, ΔTb = Kb m. Equal molality makes m the same, but Kb is a characteristic constant of the solvent and can differ from one solvent to another. Therefore the two elevations need not be equal. The non-electrolyte condition only sets i = 1; it does not make different solvents have the same Kb.

Freezing-point depression follows ΔTf = iKf m, so a larger depression can indicate a larger product of effective particle factor and molality. Under a comparison where the solvent and molality basis are appropriate, it is consistent with more effective solute particles, often due to a higher concentration or dissociation. The observation alone does not identify the exact cause, so molality and i should also be checked.

For a solution in which association or dissociation may occur, the freezing-point relation is ΔTf = iKf m. Rearranging gives i = ΔTf/(Kf m). Substitution gives i = 1.86/(1.86 × 0.5) = 1/0.5 = 2. Thus, the solution produces twice the ideal number of effective particles, and option C is correct.