5: Colligative Properties

Correct answer: A. From pure solvent to solution

Explanation: Pure solvent represents the side with zero solute concentration, while the solution is the more concentrated side. Across a membrane permeable to solvent but not solute, solvent moves toward the solution in order to reduce the concentration difference. This net movement is osmosis. It would stop only when balanced by an opposing pressure or when the sides become isotonic.

Correct answer: A. Which property and which formula apply

Explanation: A reliable first step is to identify whether the question concerns vapour pressure, boiling-point elevation, freezing-point depression or osmotic pressure. Then choose the matching relation, check whether molarity or molality is required, and include the van’t Hoff factor when appropriate. Colour, container shape and question length do not select a scientific formula.

Correct answer: C. Colligative property

Explanation: A colligative property depends on the number of dissolved particles present, not on the chemical identity of those particles. Vapour-pressure lowering, boiling-point elevation, freezing-point depression, and osmotic pressure are examples. Colour, smell, and chemical reactivity usually depend on the nature of the substance, so they are not the correct choice.

Correct answer: A. Depression in freezing point

Explanation: Depression in freezing point is colligative because it depends on the number of dissolved particles. Colour, taste, and flammability depend mainly on the chemical nature and interactions of substances. The other standard colligative properties are boiling-point elevation, vapour-pressure lowering, and osmotic pressure.

Correct answer: A. Freezing point decreases

Explanation: Sugar is a non-volatile, non-electrolyte solute, so it remains as dissolved molecules in water. These particles lower the escaping tendency and chemical potential of liquid water, making formation of the solid structure more difficult. Therefore the solution must be cooled below 0 °C before freezing; its freezing point is depressed.

Correct answer: A. Boiling point increases

Explanation: Dissolved salt particles reduce the vapour pressure of water at a given temperature. Boiling occurs only when vapour pressure equals the external pressure, so the salt solution must be heated to a higher temperature to reach that condition. Thus its boiling point rises. The effect is a colligative property and depends on effective particle concentration.

Correct answer: A. Very dilute solution

Explanation: Osmotic pressure is particularly useful for very dilute solutions because it can be measured without requiring a large concentration of solute. It is also suitable for finding molar masses of proteins and other large molecules, for which boiling or freezing measurements may be inconvenient.

Correct answer: A. Only solvent molecules

Explanation: A semipermeable membrane is selectively permeable: in the ideal model used for osmosis, it allows solvent molecules to pass while preventing solute particles from crossing. This unequal passage produces a net movement of solvent toward the solution with higher effective solute concentration. Therefore only solvent molecules are represented by the correct option.

Correct answer: A. From dilute to concentrated solution

Explanation: Through a semipermeable membrane, solvent moves spontaneously from the dilute side, where solvent chemical potential is higher, toward the concentrated side. The movement continues until equilibrium is reached or an opposing pressure prevents further flow. The direction is not determined by physical height.

Correct answer: A. Isotonic

Explanation: Solutions having equal osmotic pressure at the same temperature are called isotonic solutions. If they are separated by a suitable semipermeable membrane, there is no net movement of solvent from one side to the other. Hypertonic and hypotonic describe relative osmotic concentrations, while non-volatile describes vapour behaviour, so those choices do not name equal osmotic pressure.

Correct answer: A. π

Explanation: Osmotic pressure is conventionally represented by the Greek letter π. In the dilute-solution relation πV = nRT, π denotes the pressure that would oppose osmosis. Kf is the molal depression constant for freezing point, Kb is the molal elevation constant for boiling point, and x commonly denotes a mole fraction, so only π fits the question.

Correct answer: A. Depression increases

Explanation: For a dilute solution, freezing-point depression is ΔTf = iKf m. If the solvent and the nature of the solute remain unchanged, i and Kf are fixed, so ΔTf increases directly with molality m. Increasing molality means more effective solute particles per kilogram of solvent. Hence the freezing point moves farther below that of the pure solvent.

Correct answer: A. Colligative effect

Explanation: Colligative effects become larger when the number of effective solute particles increases. Thus a solution with more particles generally shows greater vapour-pressure lowering, boiling-point elevation, freezing-point depression, or osmotic pressure, with the exact comparison requiring the same relevant conditions. Colourlessness, metallic lustre, and electromagnetic waves are not general consequences of particle count.

Correct answer: A. Depression in freezing point

Explanation: When a suitable solute is dissolved, the solution generally freezes at a lower temperature than the pure solvent. The magnitude of this lowering is called depression in freezing point and is written as ΔTf = Tf° − Tf. Boiling-point elevation concerns boiling temperatures, osmotic pressure concerns solvent flow through a membrane, and heat of vaporisation is an energy change.

Correct answer: A. Elevation in boiling point

Explanation: A non-volatile solute lowers the vapour pressure of the solvent, so the solution must be heated to a higher temperature before its vapour pressure reaches the external pressure. The increase is called elevation in boiling point and is represented by ΔTb = Tb − Tb°. Freezing-point depression, vapour pressure, and mole fraction are different quantities, not this temperature difference.

Correct answer: A. Its size remains nearly unchanged

Explanation: An isotonic external solution has approximately the same effective osmotic pressure as the cell fluid. Water may move in both directions, but there is no net solvent movement, so the cell neither gains enough water to burst nor loses enough to shrink. Its size therefore remains nearly unchanged.

Correct answer: A. Out of the cell

Explanation: A more concentrated external solution is hypertonic relative to the cell. Through the cell membrane, water moves from the side with lower effective solute concentration toward the side with higher effective solute concentration. Thus water leaves the cell, causing it to lose volume and shrink if the condition persists.

Correct answer: A. Into the cell

Explanation: A less concentrated external solution is hypotonic relative to the cell interior. Water crosses the selectively permeable membrane from the dilute outside toward the more concentrated cell fluid. Consequently the cell gains water and swells; in cells lacking a strong wall, excessive inflow may cause rupture.

Correct answer: A. A lower temperature is needed to freeze

Explanation: Adding a non-volatile solute lowers the vapour pressure and chemical potential of the liquid solvent. For the liquid and solid phases to reach equilibrium, the temperature must be reduced further than for the pure solvent. Thus the solution freezes at a lower temperature. The change is called freezing-point depression; it does not mean that the solution can never freeze or freezes instantly.

Correct answer: A. Because vapour pressure decreases

Explanation: For a non-volatile solute, the escaping tendency of solvent molecules is reduced, so the solution has lower vapour pressure at a given temperature. Boiling requires vapour pressure to equal external pressure; reaching that pressure therefore needs a higher temperature. This produces boiling-point elevation, not a colour or mass effect.

Correct answer: A. Refractive index

Explanation: The standard colligative properties are relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure. Refractive index depends on the nature and optical response of the material, not simply on the number of solute particles. Thus it is the only non-colligative option.

Correct answer: A. Kf

Explanation: The molal freezing-point depression constant is denoted by Kf and appears in ΔTf = iKf m. Its value depends on the solvent, not on the identity of the solute under the ideal dilute treatment. Kb belongs to boiling-point elevation, π denotes osmotic pressure, and R is the universal gas constant. Therefore Kf is the only suitable symbol.

Correct answer: A. Kb

Explanation: The molal boiling-point elevation constant is represented by Kb and occurs in the relation ΔTb = iKb m. For a specified solvent it has a characteristic value. Kf is used for freezing-point depression, π for osmotic pressure, and x₂ usually represents the solute mole fraction. The subscript b is a useful reminder of boiling point.

Correct answer: A. They depend on the number of solute particles

Explanation: The defining feature of a colligative property is dependence on the number of effective solute particles in a given amount of solvent. Particle identity matters indirectly only when it changes particle number through dissociation or association. Colour, smell, and container shape do not determine these properties.

Correct answer: A. Salt lowers the freezing point

Explanation: Salt dissolves in the thin liquid-water film around ice and forms a solution whose freezing point is below 0°C. At ordinary temperatures, this ice-salt mixture is therefore not at its freezing equilibrium, so more ice melts to produce liquid brine. The effect is freezing-point depression, not a colour or gas change.