5: Colligative Properties

Correct answer: B. Because it is a ratio of two pressures

Explanation: Relative lowering is defined as (p° − p)/p°, where both p° and p are pressures. The numerator and denominator therefore carry the same pressure unit, such as pascal or kilopascal, and the units cancel during division. The resulting number has no unit. It should not be confused with vapour pressure itself, which does have pressure units.

Correct answer: B. The mole fraction of the solute

Explanation: For a non-volatile solute, Raoult’s law gives Δp/p° = x_solute, or Δp = p°x_solute. At a fixed temperature and fixed pure-solvent pressure, the lowering is therefore directly proportional to the solute mole fraction. The relation depends on the number of solute particles, not on colour, smell, or container mass. This particle-based dependence is why the property is colligative.

Correct answer: A. It becomes approximately double

Explanation: For a non-volatile solute, the relative lowering of vapour pressure is Δp/p° = x_solute. This is a direct proportionality in the dilute-solution treatment. Therefore, if the solute mole fraction changes from x to 2x while temperature and the solvent remain otherwise comparable, the relative lowering changes from x to approximately 2x. It does not become half or stay unchanged; the statement that the solution remains dilute supports this approximation.

Correct answer: A. Number of non-volatile solute particles has increased

Explanation: At fixed temperature and with the same solvent, vapour-pressure lowering is a colligative effect. Increasing the number of effective non-volatile solute particles lowers the solvent mole fraction and its escaping tendency, so the lowering becomes larger. Container colour, solvent identity disappearing, or external pressure being zero does not explain the stated colligative change.

Correct answer: A. Mole fraction of the solute

Explanation: Let p⁰ be the vapour pressure of the pure solvent and p its pressure in the solution. For a non-volatile solute, Raoult’s law gives p = x_solvent p⁰. Therefore, the relative lowering is (p⁰ − p)/p⁰ = 1 − x_solvent. Since x_solvent + x_solute = 1, this becomes x_solute. It depends on the number of solute particles, not on colour, container volume, or solvent mass alone.

Correct answer: A. Boiling point increases

Explanation: A liquid boils when its vapour pressure reaches the external pressure. If adding a non-volatile solute lowers the vapour pressure at every given temperature, the solution must be heated to a higher temperature before its pressure can reach that same external pressure. Consequently, the boiling point rises; this is called elevation of boiling point. It does not fall to zero, and boiling point is directly related to vapour pressure.

Correct answer: A. Non-volatile solute

Explanation: A non-volatile solute contributes negligibly to the vapour phase, while its presence lowers the mole fraction of the solvent in the liquid. Raoult’s law therefore gives a lower solvent partial pressure than for the pure solvent. A volatile solute contributes its own vapour and requires a different total-pressure treatment. Colour and the physical state of the solvent are irrelevant to this principle.

Correct answer: A. Colligative property

Explanation: Lowering of vapour pressure is a colligative property because, for dilute solutions, its magnitude depends primarily on the number of dissolved solute particles relative to solvent particles, not on the chemical identity of those particles. Association and dissociation can change the effective particle number, which explains abnormal colligative behaviour. Colour, shape, and nuclear properties are unrelated.

Correct answer: A. In the solution having more solute particles

Explanation: For comparable solutions with a non-volatile solute, more solute particles produce a larger solute mole fraction and a smaller solvent mole fraction. From Δp = p°xsolute, the lowering therefore increases with the effective number of particles. A pure solvent has no solute and hence no solute-caused lowering. Container size alone does not determine the equilibrium vapour pressure.

Correct answer: A. Difference between vapour pressure of pure solvent and solution divided by vapour pressure of pure solvent

Explanation: If p° is the vapour pressure of the pure solvent and p is the pressure over the solution, the lowering is p° − p. Relative lowering compares this decrease with the original reference value p°, so it is (p° − p)/p°. The denominator must be the pure-solvent pressure; multiplying by mass or using colour, volume, or temperature is not the definition.

Correct answer: A. 0.20

Explanation: For an ideal solution with a non-volatile solute, Raoult’s law gives the relative lowering of vapour pressure as (p° − p)/p° = x_solute, where p° is the vapour pressure of the pure solvent and p is its vapour pressure in the solution. Since x_solute = 0.20, the relative lowering is 0.20, or 20%. The value 0.80 is the solvent’s mole fraction, not the relative lowering.

Correct answer: A. Because it is a ratio of two pressures of the same kind

Explanation: Relative lowering of vapour pressure is defined as (p° − p)/p°, where p° is the vapour pressure of the pure solvent and p is that of the solution. Both the numerator and denominator are pressures, measured in the same unit. Their units cancel during division, leaving a pure number. It may be written as a decimal or percentage, but it is not measured in pascals, grams, or any other physical unit. Hence option A is correct.

Correct answer: A. Lowering increases

Explanation: For a non-volatile solute, Raoult’s law gives relative lowering Δp/p° = x_solute in the dilute ideal limit. Increasing x_solute therefore increases the fractional lowering and lowers the solution pressure further. The statement assumes the same solvent and temperature and no unusual association or dissociation; in such cases the effective particle concentration must be considered.

Correct answer: A. 14 kPa

Explanation: The solute mole fraction is x_solute = 1 − x_solvent = 1 − 0.65 = 0.35. For a non-volatile solute, the lowering in vapour pressure is Δp = x_solute p°_solvent. Therefore, Δp = 0.35 × 40 kPa = 14 kPa. Equivalently, the solution pressure is 0.65 × 40 = 26 kPa, and 40 − 26 = 14 kPa. Thus 26 kPa is the solution pressure, not the lowering, so option A is correct.

Correct answer: B. It lowers vapour pressure by decreasing solvent mole fraction

Explanation: A non-volatile solute has negligible vapour pressure under the stated conditions, so it does not add a significant partial pressure to the vapour. Its presence reduces the mole fraction of the solvent, and Raoult’s law then gives a lower solvent partial pressure and lower total pressure. This is a physical solution effect, not necessarily a chemical reaction with the solvent.

Correct answer: A. Colligative property

Explanation: A colligative property depends on the number of dissolved particles relative to the solvent, provided the solution is sufficiently dilute and particle interactions are handled appropriately. Important examples are relative lowering of vapour pressure, boiling-point elevation, freezing-point depression, and osmotic pressure. Dissociation and association matter because they change the effective number of particles.

Correct answer: A. Colligative property

Explanation: Elevation in boiling point is a colligative property because, for a dilute solution, its magnitude depends mainly on the number of dissolved particles relative to the solvent, not on the chemical identity of those particles. The other choices describe unrelated categories and do not represent the standard solution properties used in this chapter.

Correct answer: A. Freezing point of solution is lower than pure solvent

Explanation: Depression in freezing point means that the solution freezes at a temperature lower than the freezing point of the pure solvent. A non-volatile solute lowers the chemical potential or escaping tendency of the solvent, so solid–liquid equilibrium is reached only at a lower temperature. It is not a fall in boiling point or a colour change.

Correct answer: A. Flow of solvent through a semipermeable membrane

Explanation: Osmosis is the movement of solvent molecules through a semipermeable membrane from the dilute solution, or pure solvent, toward the more concentrated solution. Osmotic pressure is the external pressure required to stop this movement. It depends mainly on the number of dissolved particles, concentration, and temperature, not on colour, burning, or melting.

Correct answer: A. \(\pi = CRT\)

Explanation: For a dilute ideal solution, osmotic pressure follows the van’t Hoff equation \(\pi = CRT\). Here, \(\pi\) is osmotic pressure, \(C\) is the molar concentration of the solute, \(R\) is the gas constant, and \(T\) is absolute temperature in kelvin. Option B describes boiling-point elevation, option C describes freezing-point depression, and option D is Raoult’s law for vapour pressure.

Correct answer: B. 1

Explanation: Glucose behaves as a non-electrolyte in this idealised situation and remains as intact molecules rather than splitting into ions. Therefore one dissolved molecule gives one effective solute particle, so the ratio of actual to calculated particles is i = 1. The amount of water and glucose affects molality, but not this factor when there is no association or dissociation.

Correct answer: A. \(\Delta T_f = iK_fm\)

Explanation: The depression in freezing point is given by \(\Delta T_f = iK_fm\). In this expression, \(i\) is the van’t Hoff factor, \(K_f\) is the cryoscopic constant of the solvent, and \(m\) is the molality of the solution. The symbol \(f\) identifies freezing-point depression. Option B is the corresponding formula for elevation in boiling point, while C and D describe osmotic pressure and the ideal-gas equation.

Correct answer: A. Because it does not depend on volume change with temperature

Explanation: Molality is defined as moles of solute per kilogram of solvent. It uses mass, which remains essentially unchanged when temperature changes, whereas solution volume can expand or contract. This makes molality convenient for temperature-dependent colligative-property calculations. It does include solvent mass and is not restricted to gases.

Correct answer: A. Elevation increases

Explanation: For a dilute solution, ΔTb = iKb m. If the solvent and solute behaviour remain the same, i and Kb are constant, so ΔTb is directly proportional to molality. Increasing m therefore increases the boiling-point elevation. It does not make the elevation zero or remove the relation; the statement assumes the usual dilute-solution conditions.

Correct answer: A. Colour of solution

Explanation: The standard colligative properties are relative lowering of vapour pressure, elevation in boiling point, depression in freezing point and osmotic pressure. They depend primarily on the number of dissolved particles. Colour depends on the nature and electronic structure of the substance, so it is not classified as a colligative property.