5: Colligative Properties

Correct answer: A. The one whose solvent has the larger Kf value

Explanation: For a dilute non-electrolyte solution, the depression in freezing point is given by ΔTf = Kf m. In the comparison, the molality m is the same for both solutions, so the magnitude of ΔTf depends on the cryoscopic constant Kf of the solvent. The solvent with the larger Kf therefore produces the greater freezing-point depression; colour, container size, and name are irrelevant.

Correct answer: A. Osmotic pressure increases

Explanation: For a dilute solution, osmotic pressure is given by π = iCRT, or equivalently π = cRT when c is the actual particle concentration. At fixed temperature and volume, increasing the number of dissolved particles increases c and therefore increases π. Dissociation is one way to produce more particles. Osmotic pressure does not become zero or depend on colour, so A is correct.

Correct answer: A. Colligative property

Explanation: A colligative property depends on how many solute particles are present, rather than on the chemical identity of those particles. Vapour-pressure lowering, boiling-point elevation, freezing-point depression, and osmotic pressure are examples used for molecular-mass determination. Association and dissociation alter the effective particle number and therefore cause abnormal results. Colour, flame colour, and lustre do not provide this particle-count relation.

Correct answer: A. As a correction factor for particle number

Explanation: Colligative properties depend on the number of dissolved particles, not merely on the chemical formula of the solute. Dissociation increases particle number, while association decreases it. The van’t Hoff factor i accounts for this change, so it is used as a particle-number correction factor in equations for osmotic pressure, boiling-point elevation and freezing-point depression.

Correct answer: A. The number of solute particles

Explanation: Colligative properties depend on how many dissolved particles are present, not on the chemical identity of those particles in the ideal dilute limit. Vapour-pressure lowering, boiling-point elevation, freezing-point depression and osmotic pressure all respond to particle number. Association and dissociation matter because they change this number.

Correct answer: A. Colligative properties

Explanation: Abnormal molar mass is usually inferred from measurements such as osmotic pressure, boiling-point elevation or freezing-point depression. These are colligative properties and depend on the number of dissolved particles. Association and dissociation change that number, so the inferred molar mass becomes abnormal.

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

Explanation: Colligative properties depend on the number of dissolved particles, not on their individual colour, smell, or chemical identity in the ideal approximation. Association and dissociation change that number, so they alter the measured property and can produce an abnormal calculated molar mass. Container shape also does not determine a colligative property when concentration and temperature are fixed.

Correct answer: A. It will increase

Explanation: Depression in freezing point is a colligative property, so it depends on the total number of dissolved particles. Dissociation changes one solute unit into two or more ions and therefore increases the particle count. Since ΔTf = iKf m, a value of i greater than one makes the observed depression larger than the normal calculated value. It does not make the depression zero.

Correct answer: A. It will decrease

Explanation: Elevation in boiling point is a colligative effect and is proportional to the number of dissolved particles. Association combines separate solute molecules into larger species, thereby reducing the effective particle count. In ΔTb = iKb m, association gives i less than one, so the observed elevation is smaller than the value calculated without association. It cannot become infinite or equal a mass.

Correct answer: A. 2

Explanation: The wording of the question states that the measured freezing-point depression is twice the value expected for unchanged particles. Since ΔTf(observed) = i × ΔTf(normal), the ratio observed/normal is directly i. Therefore i = 2. Such a value indicates an increased number of particles, usually due to dissociation rather than association.

Correct answer: A. 0.75

Explanation: The abnormal boiling-point elevation is related to the normal value by ΔTb(observed) = i × ΔTb(normal). The question gives the observed-to-expected ratio as 0.75, so i = 0.75. Since this is less than one, the effective particle number is reduced, which is consistent with association. It is not the complement 0.25.

Correct answer: A. van’t Hoff factor

Explanation: The van’t Hoff factor i corrects the ideal colligative-property equation for the actual number of solute particles. It is multiplied into expressions such as ΔTb = iKbm, ΔTf = iKfm, and π = iCRT. Association gives i below one, while dissociation gives i above one. Acidity, colour, and evaporation factors do not provide this particle-number correction.

Correct answer: A. 1.5

Explanation: For a fixed solvent, solute concentration, and temperature conditions, depression in freezing point is proportional to the number of dissolved particles: ΔTf = iKf m. The expected value corresponds to i = 1. If the measured depression is 1.5 times that value, then the multiplying factor is i = 1.5. This indicates more particles than predicted for a non-associated, non-dissociated solute.

Correct answer: A. 0.5

Explanation: Osmotic pressure is given by π = iCRT. When concentration, gas constant, and temperature are unchanged, the ratio of actual pressure to the normal expected pressure equals i. A pressure equal to one-half of the expected value therefore gives i = 0.5. Such a value indicates fewer effective particles, commonly because of association rather than dissociation.

Correct answer: D. AlCl₃

Explanation: For equal molality, ΔTf = iKf m, so the solution with the largest van’t Hoff factor gives the greatest depression. Glucose is a non-electrolyte and gives i = 1. NaCl gives two ions, CaCl₂ gives three, and AlCl₃ gives four ions on complete dissociation. Therefore AlCl₃ has the largest i and produces the maximum freezing-point depression, assuming ideal complete dissociation.

Correct answer: A. The effective number of particles doubled

Explanation: Colligative properties depend on the number of dissolved particles, not on their colour or chemical name alone. With the same masses and concentration conditions, a doubled effect corresponds to approximately i = 2, meaning twice as many effective particles as in the normal molecular case. Dissociation of an electrolyte is a common way to produce this increase.

Correct answer: A. Completely dimerised solute

Explanation: At fixed temperature and molality, osmotic pressure is proportional to i. Complete dimerisation changes two solute molecules into one particle, so i = 1/2. Glucose has i about 1, NaCl gives i about 2 when fully dissociated, and AlCl3 gives i about 4. Therefore the completely dimerised solute has the fewest particles and the lowest osmotic pressure.

Correct answer: A. i = actual effective particle number / normal expected particle number

Explanation: The van’t Hoff factor compares the number of particles actually present in solution with the number expected if the solute neither dissociated nor associated. Thus i is the ratio of actual effective particle number to normal expected particle number. Dissociation gives i greater than one, association gives i less than one, and no abnormality gives i equal to one.

Correct answer: C. 0.06 mol

Explanation: The governing idea is that complete dissociation changes one formula unit into all of its constituent ions. Each AB₂ unit produces one A ion and two B ions, so the total number of particles is three per formula unit. Therefore, effective particle moles = 0.02 × 3 = 0.06 mol. Option A ignores dissociation, while options B and D use incorrect particle multipliers. Thus option C is correct.

Correct answer: A. 1.6 times

Explanation: For dilute solutions at the same temperature and molality, osmotic pressure is proportional to the van’t Hoff factor i. A normal nonelectrolyte has i = 1, while the given solute has i = 1.6. Therefore its osmotic pressure is 1.6 times that of the normal solution. The increase indicates more effective particles, usually because of dissociation.

Correct answer: A. 0.8 times

Explanation: For dilute solutions, boiling-point elevation is ΔTb = iKb m. At the same molality and with the same solvent, Kb and m are unchanged, so the ratio of elevations is simply the ratio of i values. A normal nonelectrolyte has i = 1, whereas this solute has i = 0.8; its elevation is therefore 0.8 times as large.

Correct answer: A. 0.9 times

Explanation: The boiling-point elevation is Delta Tb = i Kb m. For a normal non-electrolyte, i = 1. At the same molality and in the same solvent, the ratio of elevations is therefore 0.9/1 = 0.9. Thus the solution shows 0.9 times the normal elevation, reflecting a slight reduction in particles due to association.

Correct answer: C. 1.75 times

Explanation: KCl dissociates into two ions, so for degree of dissociation alpha its van’t Hoff factor is i = 1 + alpha. With alpha = 0.75, iKCl = 1.75. Glucose is a non-electrolyte and does not dissociate, so iGlucose = 1. At the same concentration and temperature, osmotic pressure is proportional to i; therefore the ratio is 1.75. Option C is correct.

Correct answer: A. 25% higher

Explanation: For fixed concentration, solvent, and temperature, a colligative effect is proportional to the van’t Hoff factor i. A normal non-electrolyte has i = 1, whereas this solute has i = 1.25. Its effect is therefore 1.25 times the normal value. The increase is (1.25 − 1) × 100 = 25%. Hence option A is correct; 125% describes the total relative value, not the increase.