5: Colligative Properties

Correct answer: B. Water leaves the cell and it shrinks

Explanation: A highly concentrated salt solution is hypertonic relative to the fluid inside a blood cell. Across the selectively permeable cell membrane, water moves by osmosis from the region of lower solute concentration inside the cell toward the region of higher solute concentration outside. As water leaves, the cell loses volume and shrinks. This process is called crenation in red blood cells.

Correct answer: A. The vapour pressure of the solution decreases.

Explanation: A liquid boils when its vapour pressure becomes equal to the external pressure. When a non-volatile solute is dissolved in a solvent, the mole fraction and escaping tendency of the solvent decrease, so the solution has a lower vapour pressure than the pure solvent at the same temperature. The solution must therefore be heated to a higher temperature to reach external pressure. Thus its boiling point rises, making option A correct.

Correct answer: C. MgCl₂

Explanation: For a dilute solution, osmotic pressure is π = iCRT. Because concentration C and temperature T are the same for every option, osmotic pressure depends on the van’t Hoff factor i. Urea and glucose are non-electrolytes, so i = 1; NaCl gives two ions, so i = 2; MgCl₂ gives Mg²⁺ and two Cl⁻ ions, so i = 3 under complete dissociation. Therefore MgCl₂ has the highest osmotic pressure, option C.

Correct answer: B. 0.50 m

Explanation: The freezing-point depression equation is ΔTf = iKf m. Since the solute is non-dissociated, i = 1. Rearranging gives m = ΔTf/(iKf) = 0.93/(1 × 1.86) mol kg⁻¹ = 0.50 mol kg⁻¹. Molality is normally written as 0.50 m. The negative sign in the actual freezing temperature change only indicates lowering; the given depression is already its positive magnitude. Thus option B is correct.

Correct answer: B. Equal to 75% of it

Explanation: For a given solvent, concentration, and temperature, each colligative effect is proportional to the van’t Hoff factor i. A normal non-dissociated solute has i = 1, whereas the associated solute has i = 0.75 because some solute particles combine and the total number of particles decreases. Therefore its colligative effect is 0.75 times, or 75% of, the normal effect. Hence option B is correct.

Correct answer: C. Osmotic pressure

Explanation: A colligative property depends primarily on the number of dissolved particles, not on the chemical identity of those particles. Osmotic pressure is one of the four principal colligative properties and is represented for a dilute solution by π = CRT, where C indicates the concentration of solute particles. Viscosity, surface tension, and refractive index also depend strongly on molecular nature.

Correct answer: B. Because it depends on the number of solute particles, not their nature

Explanation: A colligative property is controlled by the number of dissolved particles in a specified amount of solvent, rather than by the chemical identity of the solute. Osmotic pressure follows π = CRT for a dilute solution, so it depends on particle concentration. If a solute dissociates, the increased number of particles increases osmotic pressure; if particles associate, the effect decreases.

Correct answer: A. Vapour pressure decreases, so a higher temperature is needed for boiling

Explanation: Adding a non-volatile solute lowers the vapour pressure of the solution at every given temperature because fewer solvent molecules can escape into the vapour phase. Boiling occurs when vapour pressure equals external pressure. Therefore, the solution must be heated to a higher temperature to reach that pressure, producing elevation of the boiling point. Thus the two effects are directly connected.

Correct answer: A. The mole fraction of solute particles

Explanation: For a solution containing a non-volatile solute, Raoult's law gives the relative lowering of vapour pressure as (p° − p)/p° = x_solute, for an ideal dilute solution. Therefore, the effect depends on the mole fraction, or effective number, of solute particles. It does not depend on colour, smell, or container shape. If the solute dissociates or associates, the effective particle number must be considered.

Correct answer: A. The solute produces more particles in the solution

Explanation: Freezing-point depression is a colligative property and is determined by the number of dissolved particles. The relation ΔT_f = iK_fm shows that, at a fixed solvent and molality, a larger van't Hoff factor i produces a larger depression. Ionic compounds can dissociate into several ions, increasing the effective particle count. Colour, constant volume, or low solubility does not by itself explain a larger depression.

Correct answer: A. They have the same osmotic pressure

Explanation: Two solutions are isotonic when, at the same temperature, they exert equal osmotic pressure across a semipermeable membrane. Their solutes need not be chemically identical; different solutes can produce the same effective concentration of particles. Equal osmotic pressure does not mean that colour, composition, or every other physical property is identical. Thus, option A gives the defining property of isotonic solutions.

Correct answer: A. Osmosis

Explanation: Osmosis is the net movement of solvent molecules through a semipermeable membrane from the side with lower solute concentration or higher solvent potential to the side with higher solute concentration or lower solvent potential. The membrane permits solvent molecules to pass while restricting solute particles. Diffusion does not specifically require a semipermeable membrane, whereas evaporation and boiling are phase changes rather than this membrane-mediated transport.

Correct answer: A. The nature of the solvent

Explanation: Kb is the molal elevation constant, also called the ebullioscopic constant. It appears in the relation ΔTb = iKb m. Its value is a characteristic property of the solvent because it depends on solvent properties such as its boiling point, enthalpy of vaporisation, and molar mass. For a specified solvent, Kb does not depend on the chemical identity or colour of the dissolved solute. The solute affects the factor i and molality m instead.

Correct answer: A. Freezing-point depression for a one-molal solution

Explanation: Kf is the molal depression constant, also called the cryoscopic constant. It is defined as the decrease in the freezing point produced when one mole of a non-volatile, non-electrolyte solute is dissolved in one kilogram of solvent, under ideal dilute-solution conditions. The equation is ΔTf = iKf m. Its value depends on the solvent, and molality—not molarity—is used because molality is independent of temperature-related volume changes.

Correct answer: A. 300 K

Explanation: Temperature in the osmotic-pressure equation π = iCRT must be expressed on the absolute Kelvin scale. To convert Celsius to Kelvin, add 273.15, or approximately 273 for school-level calculations. Thus, 27 °C = 27 + 273 = 300 K, approximately 300.15 K if greater precision is required. Using 27 directly would be incorrect because gas-law and osmotic-pressure relations require absolute temperature.

Correct answer: A. The vapour pressure of solution decreases

Explanation: A non-volatile solute lowers the vapour pressure of the solvent in the solution. Boiling occurs when vapour pressure becomes equal to external pressure. Because the solution has a lower vapour pressure at the same temperature, it must be heated to a higher temperature to boil. This increase is called elevation of boiling point, a colligative property.

Correct answer: A. Solute particles hinder crystal formation of solvent

Explanation: Freezing requires solvent molecules to arrange themselves into an ordered solid crystal. Dissolved solute particles disturb this arrangement and reduce the escaping tendency of solvent molecules into the solid phase. Therefore, the solution must be cooled to a lower temperature before equilibrium between liquid and solid is reached. This is called depression of freezing point.

Correct answer: A. 3

Explanation: Complete dissociation of one formula unit of AB₂ produces one A ion and two B ions: AB₂ → A + 2B. Thus, one original particle becomes three particles in solution. The ideal van't Hoff factor is the ratio of the total particles after dissociation to the original number of formula units, so i = 3/1 = 3, assuming no association or other non-ideal effects.

Correct answer: A. 2

Explanation: The van't Hoff factor measures the change in the number of dissolved particles: i = actual number of particles divided by the number expected if no dissociation or association occurred. Since the actual number is twice the expected number, i = 2N/N = 2. A value greater than one generally indicates particle formation through dissociation, although the numerical definition itself is based on the particle ratio.

Correct answer: A. From concentrated solution to dilute solution

Explanation: In ordinary osmosis, solvent passes through a semipermeable membrane from the dilute side, which has lower solute concentration, toward the concentrated side. In reverse osmosis, an external pressure greater than the osmotic pressure is applied on the concentrated solution. This forces solvent to move in the opposite direction, from the concentrated solution toward the dilute or purified side.

Correct answer: A. Osmotic pressure

Explanation: Osmotic pressure is the minimum pressure that must be applied to the solution side to stop the natural flow of solvent through a semipermeable membrane. To make the solvent flow in the reverse direction, the applied pressure must exceed this osmotic pressure. Vapour pressure, temperature, and freezing point do not provide the relevant pressure comparison for reverse osmosis.

Correct answer: A. 2

Explanation: The van't Hoff factor is defined as i = effective number of solute particles actually present divided by the number expected for the undissociated solute. In this case, the effective particle amount is 2 mol and the expected amount is 1 mol. Therefore, i = 2/1 = 2. The value above one indicates that the number of particles has increased, as in dissociation.

Correct answer: A. Chemical nature of the particles

Explanation: Colligative properties depend primarily on the number of independently moving solute particles present in a solution, not on the chemical identity of those particles. The number of moles and the van’t Hoff factor help determine the effective particle concentration. Thus, under the ideal dilute-solution assumption, chemical nature is not the determining factor.

Correct answer: A. Depression of freezing point and osmotic pressure

Explanation: The four standard colligative properties are relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure. Freezing-point depression and osmotic pressure therefore form a pair consisting entirely of colligative properties. Viscosity, colour, density, and refractive index depend on factors other than only the number of solute particles.

Correct answer: A. Because of dissolved salt particles

Explanation: Seawater contains dissolved sodium chloride and other salts. These solutes increase the number of particles in the liquid phase and lower its freezing point, according to the colligative property called depression of freezing point. Therefore seawater must be cooled below the freezing point of pure water before it solidifies. The colour and depth are not the primary explanation.