5: Colligative Properties

Correct answer: A. Actual number of particles due to ionisation or association

Explanation: Colligative properties depend on the number of solute particles, not simply on the formula units initially added. A solute may dissociate into more particles, as an electrolyte does, or associate into fewer particles. The van’t Hoff factor i adjusts the calculated colligative effect to reflect this actual effective particle number. It does not correct apparatus shape, colour, or thermometer length.

Correct answer: A. One

Explanation: A non-electrolyte generally remains as individual molecules in solution and neither dissociates into ions nor associates appreciably. Thus the actual number of particles is approximately the number expected from the dissolved formula units. The ratio is therefore i = 1. A value of two would indicate additional particles, not ordinary non-electrolyte behaviour.

Correct answer: A. It increases

Explanation: Dissociation changes one solute formula unit into two or more ions, increasing the effective number of particles. Since colligative effects are proportional to particle number, the observed effect is generally larger than for an undissociated solute at the same formal concentration. The exact increase depends on the degree of dissociation, represented by i.

Correct answer: A. Less than one

Explanation: Association joins two or more solute particles to form a larger species. Consequently, the actual number of particles becomes lower than the number predicted if no association occurred. Since i is the ratio of actual to ideal particle number, its value is generally less than one. The exact value depends on the extent of association and is not always zero.

Correct answer: A. It will be higher

Explanation: For a dilute ideal solution, π = CRT. When temperature and the gas constant are fixed, osmotic pressure is directly proportional to molar concentration. Therefore the solution with the larger osmotic pressure has the greater effective solute concentration, assuming comparable solute behaviour. Colour does not determine osmotic pressure, and zero concentration would give zero pressure in this model.

Correct answer: A. Osmotic pressure

Explanation: Two solutions are called isotonic when they have equal osmotic pressure at the same temperature. If separated by a suitable semipermeable membrane, there is no net movement of solvent between them. Their colours, densities and boiling points need not be equal, because those properties depend on additional molecular and physical factors.

Correct answer: A. From lower concentration to higher concentration

Explanation: In osmosis, solvent moves through the membrane from the side with lower effective solute concentration to the side with higher effective concentration. The higher-concentration side generally has the higher osmotic pressure, and incoming solvent tends to dilute it. The direction is not determined by upward or downward position; equal osmotic pressures would produce no net flow.

Correct answer: A. K_b

Explanation: K_b is called the ebullioscopic constant because it is used in calculating elevation of the boiling point: ΔT_b = K_b m for a non-electrolyte, or ΔT_b = iK_b m when particle association or dissociation is considered. K_f is the cryoscopic constant for freezing-point depression, while R and π have different meanings.

Correct answer: A. \(K_f\)

Explanation: The cryoscopic constant is represented by \(K_f\), where the subscript \(f\) refers to freezing. It appears in the relation \(\Delta T_f = K_fm\) for a non-electrolyte dilute solution, or \(\Delta T_f = iK_fm\) when the van’t Hoff factor is included. \(K_b\) is the ebullioscopic constant for boiling-point elevation; \(R\) is the gas constant and \(C\) denotes concentration.

Correct answer: A. Elevation in boiling point

Explanation: The boiling-point relation is ΔTb = iKb m. If the solute is a non-electrolyte, i is approximately one; if it is an electrolyte, i must also be known. Thus Kb and molality, together with the appropriate i, determine the elevation in boiling point. Kf is needed for freezing depression, while osmotic pressure requires concentration and temperature.

Correct answer: A. Depression in freezing point

Explanation: Freezing-point depression is calculated from ΔTf = iKf m. For a non-electrolyte, i is generally taken as one; for an electrolyte or associating solute, the appropriate factor must be included. Kb is used instead for boiling-point elevation. The given quantities do not by themselves determine colour or necessarily the mole fraction.

Correct answer: A. It increases

Explanation: Adding a non-volatile solute lowers the vapour pressure of the solvent at a given temperature. A liquid boils when its vapour pressure equals the external pressure, so the solution must be heated to a higher temperature to reach that condition. Therefore, its boiling point increases. Option B describes the opposite trend, while C ignores the vapour-pressure lowering and D has no relation to this colligative effect.

Correct answer: A. It will decrease

Explanation: A non-volatile solute lowers the chemical potential of liquid water and makes formation of the pure solid phase less favourable at a given temperature. Consequently, solid–liquid equilibrium occurs only at a lower temperature. The change is called freezing-point depression and is represented by ΔTf = iKf m; it does not make the freezing and boiling points equal.

Correct answer: A. It will be lower

Explanation: The number of moles is n = mass/molar mass. If the sample mass is fixed, increasing molar mass makes n smaller, so there are fewer dissolved molecules or formula units. Because colligative properties depend on effective particle number, the corresponding effect is generally smaller, provided the solutes have comparable ionisation or association behaviour.

Correct answer: A. Because they dissociate into ions and increase particle number

Explanation: An electrolyte can dissociate into two or more ions in solution. Thus one formula unit may produce several effective particles, increasing i and magnifying a colligative effect at the same formal concentration. The increase is not automatic in exactly the ideal integer ratio because incomplete dissociation and ion interactions may occur, but particle multiplication is the governing idea.

Correct answer: A. Association of solute particles

Explanation: The van’t Hoff factor compares actual effective particles with the number expected from undissociated solute units. If particles associate, several original particles combine into fewer larger species, so the actual count falls and i becomes less than one. Complete ionisation and a large particle increase would instead tend to produce i greater than one.

Correct answer: A. Ionisation of solute

Explanation: An i value greater than one means that the solution contains more effective particles than expected from intact solute units. Dissociation or ionisation produces several ions from one formula unit, causing this increase. Association would reduce the particle number and generally give i less than one. The value does not mean that the solute disappears or that freezing becomes impossible.

Correct answer: A. By adding ΔT_b to the boiling point of pure solvent

Explanation: The elevation in boiling point is defined as ΔT_b = T_b(solution) − T_b(pure solvent), so rearranging gives T_b(solution) = T_b(pure solvent) + ΔT_b. Thus the increase must be added, not subtracted. ΔT_f concerns freezing point, and colour cannot determine a solution’s boiling temperature. This reasoning assumes the usual dilute-solution treatment.

Correct answer: A. By subtracting ΔT_f from the freezing point of pure solvent

Explanation: Depression in freezing point is defined as ΔT_f = T_f(pure solvent) − T_f(solution). Therefore, the solution’s actual freezing point is T_f(solution) = T_f(pure solvent) − ΔT_f. Subtraction is required because the solution freezes at a lower temperature. Adding ΔT_f would give the opposite trend, while ΔT_b and osmotic pressure do not directly answer this calculation.

Correct answer: A. \(-2^\circ C\)

Explanation: Depression in freezing point means that the solution freezes at a temperature lower than the pure solvent. The relation is \(T_f(\text{solution}) = T_f^0 - \Delta T_f\). Substituting the given values gives \(0^\circ C - 2^\circ C = -2^\circ C\). Thus option A is correct. Adding the values would incorrectly treat depression as an elevation, while 0°C would mean no change.

Correct answer: A. 100.5 °C

Explanation: Elevation in boiling point means that the solution boils at a temperature higher than that of the pure solvent. Apply T_b(solution) = T_b(pure solvent) + ΔT_b. Hence, 100 °C + 0.5 °C = 100.5 °C. Subtracting would describe a decrease, while 50 °C and 0.5 °C ignore the given pure-solvent boiling point.

Correct answer: A. 1.0

Explanation: For a non-electrolyte, or when i is already included as one, use ΔTb = Kb m. Substituting the given values gives ΔTb = 0.5 × 2 = 1.0. The result has the temperature unit associated with Kb and m. The other options arise from adding, squaring or dividing the values rather than applying the required multiplication.

Correct answer: A. 1.8

Explanation: For a non-electrolyte solution, the freezing-point depression is calculated with ΔT_f = K_f m. Substituting K_f = 1.8 and m = 1 gives ΔT_f = 1.8 × 1 = 1.8. The value is a temperature difference, normally expressed in °C or K. The van’t Hoff factor is implicitly one here because no ionisation is specified.

Correct answer: A. Depression in freezing point

Explanation: When salt dissolves in the thin liquid layer on ice, it produces a solution whose freezing point is lower than that of pure water. At temperatures near 0 °C, the ice–solution equilibrium shifts so that ice can melt. This everyday application is freezing-point depression, not boiling-point elevation, osmotic pressure or a colour effect.

Correct answer: A. To lower the freezing point of coolant

Explanation: An antifreeze dissolves in the vehicle coolant and lowers its freezing point through a colligative effect. The coolant can therefore remain liquid at temperatures below the freezing point of pure water, helping prevent expansion-related damage in the cooling system. Its purpose is not to alter fuel colour, transform metal or create infinite vapour pressure.