5: Colligative Properties

Correct answer: A. Relative lowering of vapour pressure

Explanation: Relative lowering of vapour pressure is one of the colligative properties. For a non-volatile solute it depends on the solute mole fraction, which represents the relative number of solute particles, rather than on the identity of those particles. Colour and odour are characteristic properties, and viscosity can depend on molecular interactions and shape, so they are not defined as colligative properties.

Correct answer: B. In the higher temperature solution

Explanation: For dilute solutions, osmotic pressure is given by π = iCRT. If the effective concentration and particle factor are the same, π is directly proportional to absolute temperature T. Therefore the solution at the higher temperature has the higher osmotic pressure. Temperature must be compared in kelvin, although a higher Celsius temperature also corresponds to a higher kelvin temperature.

Correct answer: A. They depend on the number of solute particles, not on their chemical nature.

Explanation: Colligative properties depend primarily on the number of dissolved solute particles relative to the solvent, not on the chemical identity of each particle. The four standard examples are relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure. Dissociation increases particle number and association decreases it, so ionisation can change the observed value through i.

Correct answer: C. Elevation in boiling point and osmotic pressure

Explanation: The standard colligative properties are relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure. Both elevation of boiling point and osmotic pressure therefore form a correct pair. Colour, smell, shine, hardness, density and viscosity are not the defining particle-number properties listed in this classification and can depend on composition or molecular interactions.

Correct answer: B. Which property and which formula apply

Explanation: A reliable first step is to identify the physical quantity being asked: vapour-pressure lowering, boiling-point elevation, freezing-point depression or osmotic pressure. Then choose the corresponding relation and decide whether the van’t Hoff factor i is required. Convert units, substitute values and check the result’s sign and units. Colour, container material and question length have no role in selecting the formula.

Correct answer: A. 0.08

Explanation: For an ideal dilute solution containing a non-volatile solute, Raoult’s law gives (P° − P)/P° = Xsolute, where P° is the vapour pressure of the pure solvent and P is the vapour pressure of the solution. Since the solute mole fraction is 0.08, the relative lowering of vapour pressure is 0.08. It is dimensionless and may also be expressed as 8%.

Correct answer: A. 1.5 K

Explanation: The depression in freezing point is the decrease in the freezing point caused by dissolving a solute. It is calculated as ΔT_f = T_f° − T_f, where T_f° is the freezing point of the pure solvent and T_f is the freezing point of the solution. Therefore, ΔT_f = 273 K − 271.5 K = 1.5 K. Hence, option A is correct.

Correct answer: A. The solute is neither ionising nor associating.

Explanation: The van’t Hoff factor i is the ratio of the actual number of solute particles in solution to the number expected from the formula. When i = 1, there is no change in particle number: the solute neither dissociates into ions nor associates into larger particles. Therefore, it shows normal non-electrolyte-like behaviour, and option A is correct.

Correct answer: C. \(2.0\ \mathrm{mol\,kg^{-1}}\)

Explanation: For a non-electrolyte, the van’t Hoff factor is \(i=1\), so the elevation in boiling point is calculated using \(\Delta T_b=iK_bm=K_bm\). Rearranging gives \(m=\frac{\Delta T_b}{K_b}\). Substituting the data, \(m=\frac{1.20\ \mathrm{K}}{0.60\ \mathrm{K\,kg\,mol^{-1}}}=2.0\ \mathrm{mol\,kg^{-1}}\). Therefore, option C is correct.

Correct answer: A. \(0.26^{\circ}C\)

Explanation: The elevation in boiling point is the increase in the boiling temperature caused by dissolving a non-volatile solute. It is calculated as \(\Delta T_b = T_b(\text{solution}) - T_b(\text{pure solvent})\). Therefore, \(\Delta T_b = 100.26 - 100.00 = 0.26^{\circ}C\). The answer is positive because the solution boils at a higher temperature than the pure solvent; \(100.26^{\circ}C\) is the final boiling point, not the elevation.

Correct answer: A. The solute is dissociating into ions

Explanation: The van’t Hoff factor is the ratio of the actual number of solute particles in solution to the number expected from the formula units added. When i is greater than 1, more independent particles are present than expected. This occurs when an electrolyte dissociates into ions; for example, NaCl can produce Na+ and Cl− ions. Association instead gives i less than 1.

Correct answer: B. 0.050

Explanation: The relative lowering of vapour pressure is defined as (p° − p)/p°, where p° is the vapour pressure of the pure solvent and p is the vapour pressure of the solution. Substituting the given values gives (80 − 76)/80 = 4/80 = 0.050. The pure-solvent pressure must be used as the denominator. Thus, the correct answer is option B.

Correct answer: B. 0.2 m

Explanation: For freezing-point depression, the equation is ΔTᶠ = iKᶠm, where i is the van’t Hoff factor. Because the solute is non-dissociated, i = 1. Therefore, m = ΔTᶠ/Kᶠ = 0.372/1.86 = 0.2 mol kg⁻¹, or 0.2 m. Hence, option B is correct. The calculation also shows that a 0.1 m solution would produce only 0.186 K depression under the same conditions.

Correct answer: B. 0.10 mol

Explanation: Molality is defined as m = moles of solute divided by kilograms of solvent. Rearranging gives moles of solute = molality × mass of solvent in kilograms. Thus, n = 0.2 mol kg⁻¹ × 0.5 kg = 0.10 mol. Therefore, option B is correct. The mass used in this formula is the mass of solvent, not the total mass of the solution, and it must be expressed in kilograms.

Correct answer: B. 0.20 m

Explanation: For boiling-point elevation, the relation is ΔTb = iKb m. Because the solute is non-dissociated, its van’t Hoff factor is i = 1. Thus, m = ΔTb ÷ Kb = 0.104 ÷ 0.52 = 0.20 mol kg⁻¹, or 0.20 m. Hence, option B is correct. The absence of dissociation is important because it prevents any increase in the number of solute particles.

Correct answer: A. Greater

Explanation: Colligative properties depend on the number of solute particles present in a solution, not on their chemical identity. If the same mass of solute produces more particles because of dissociation or ionisation, the particle concentration increases. Consequently, effects such as boiling-point elevation, freezing-point depression, vapour-pressure lowering, or osmotic pressure become greater. Therefore, option A is correct.

Correct answer: A. They have equal osmotic pressure

Explanation: Two solutions are called isotonic when they exert the same osmotic pressure at the same temperature. If they are separated by a suitable semipermeable membrane, there is no net movement of solvent from one solution to the other. Isotonicity does not require equal colour, equal total mass, or a zero boiling point. Therefore, option A states the defining property.

Correct answer: A. Osmotic pressure

Explanation: Osmotic pressure is especially useful because it can be measured at room temperature using a dilute solution. This avoids heating or cooling the sample and is particularly valuable for proteins, polymers, and other macromolecules that may decompose on heating. Freezing-point depression and boiling-point elevation require temperature changes, while heat of combustion is not a colligative property. Hence, option A is correct.

Correct answer: A. High-molar-mass proteins or polymers

Explanation: Osmotic pressure is measured using a very dilute solution and can be determined at room temperature. Its value remains appreciable even when the solute concentration is extremely small, which makes it particularly useful for proteins, polymers, and other macromolecules with high molar masses. Heating methods may decompose such substances, and osmotic pressure does not require the solute to be volatile. Thus, option A is correct.

Correct answer: A. Relative lowering of vapour pressure

Explanation: Relative lowering of vapour pressure is a colligative property because, for a dilute solution with a non-volatile solute, it depends on the number of solute particles rather than their chemical identity. The measured lowering gives the solute mole fraction; with known solute and solvent masses, the number of moles and hence molar mass can be calculated. The other choices are not colligative properties. Therefore, A is correct.

Correct answer: A. π = CRT

Explanation: Osmotic pressure is analogous to the pressure equation for an ideal dilute gas and is given by π = iCRT. For a non-dissociating solute, the van’t Hoff factor i equals 1, so the equation simplifies to π = CRT. Here C is the molar concentration, R is the gas constant, and T is the absolute temperature in kelvin. The other formulas describe freezing-point depression or boiling-point elevation, not osmotic pressure.

Correct answer: B. 0.372 K

Explanation: For a non-dissociating solute, the van’t Hoff factor is i=1, so the freezing-point depression is calculated using ΔT_f = iK_fm = K_fm. Substituting the given values gives ΔT_f = 1.86 K kg mol⁻¹ × 0.2 mol kg⁻¹ = 0.372 K. The kilogram and mole units cancel, leaving kelvin. Therefore, option B is correct.

Correct answer: B. 0.372 K

Explanation: For freezing-point depression, the relation is ΔTf = iKf m. A nonelectrolyte does not ionize in solution, so its van’t Hoff factor is i = 1. Substituting the given values gives ΔTf = 1 × 1.86 × 0.2 = 0.372 K. Thus the freezing point is lowered by 0.372 K, making option B the correct answer.

Correct answer: A. \(0.615\,\mathrm{atm}\)

Explanation: For a dilute solution at constant temperature, osmotic pressure follows \(\pi=iCRT\). If the solute concentration and van’t Hoff factor remain unchanged except for the concentration being halved, \(\pi\) is also halved. Thus, the new pressure is \(1.23/2=0.615\,\mathrm{atm}\). Therefore, option A is correct.

Correct answer: A. The number of solute particles

Explanation: Colligative properties, such as relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure, depend primarily on the number of dissolved solute particles relative to the solvent. They do not primarily depend on colour or container thickness. This dependence allows an unknown molar mass to be found from a measured colligative effect, provided the solution is sufficiently dilute and the particle behaviour is understood.