5: Colligative Properties

Correct answer: C. 0.4 m

Explanation: Colligative effects depend on the product i × m rather than on the uncorrected molality alone. Here, the true molality is 0.2 m and i = 2, so the observed effect corresponds to an effective molality of 2 × 0.2 = 0.4 m. If the i factor is ignored, this effective value would incorrectly be reported as the molality. Thus option C is correct.

Correct answer: A. 0.1 K

Explanation: For a given solute and solvent, boiling-point elevation is directly proportional to molality: ΔT_b = K_b m. Molality is the number of moles of solute per kilogram of solvent. Here, both the solute mass and solvent mass are doubled, so the mole-to-solvent-mass ratio remains unchanged. Therefore, the molality and boiling-point elevation remain the same, equal to 0.1 K. Hence, option A is correct.

Correct answer: A. 0.246 atm

Explanation: For a dilute non-electrolyte solution, osmotic pressure is calculated from π = nRT/V. Substituting n = 0.01 mol, R = 0.082 L atm K⁻¹ mol⁻¹, T = 300 K, and V = 1 L gives π = (0.01 × 0.082 × 300)/1 = 0.246 atm. The units are consistent because volume is in litres and R is expressed in L atm units.

Correct answer: A. Greater

Explanation: Depression in freezing point is a colligative property, so it depends mainly on the number of dissolved particles. For a given amount of solute, dissociation produces more particles than the undissociated solute would provide. In the equation ΔTf = iKf m, an increase in i increases ΔTf, provided concentration and solvent conditions are otherwise comparable. Therefore, dissociation may make the freezing-point depression greater than the normal value; it is not unrelated to particle number.

Correct answer: A. Lower

Explanation: Osmotic pressure is a colligative property and follows π = iCRT for a dilute solution. Association combines separate solute particles, so i becomes less than one and the effective particle concentration decreases. Consequently, the observed osmotic pressure is lower than the normal value calculated by assuming no association.

Correct answer: A. Molar mass calculated from colligative properties

Explanation: Molar mass obtained from boiling-point elevation, freezing-point depression or osmotic pressure assumes a certain number of solute particles. Association or dissociation changes that number. Including i corrects the calculation for the actual particle count; it does not alter atomic number, nuclear charge or chemical notation.

Correct answer: A. ΔTb = iKb m

Explanation: For a dilute solution without abnormal behavior, the elevation in boiling point is ΔTb = Kb m, where Kb is the ebullioscopic constant and m is molality. If dissociation or association changes the particle number, the correction is made by multiplying by i: ΔTb = iKb m. Thus i greater than one increases the elevation, while i less than one decreases it.

Correct answer: A. ΔTf = iKf m

Explanation: The normal dilute-solution equation is ΔTf = Kf m. Since the measured effect is proportional to the actual number of particles, abnormal behavior is included by multiplying by i. Hence the correct equation is ΔTf = iKf m. A factor in the denominator would predict a smaller effect for dissociation, contrary to the increased particle number.

Correct answer: A. π = iCRT

Explanation: For a dilute ideal solution, osmotic pressure follows the van’t Hoff equation π = CRT. If the solute dissociates or associates, the effective concentration of particles is changed, so the van’t Hoff factor is included as a multiplier: π = iCRT. Dissociation gives i greater than one and raises the pressure; association gives i below one and lowers it relative to the ideal calculation.

Correct answer: A. 0.2 molal

Explanation: Complete dissociation of NaCl produces two ions per formula unit, so the ideal van't Hoff factor is i = 2. The effective particle molality is i times the formal molality: 2 × 0.1 = 0.2 molal. This is an ideal approximation; real ionic interactions can cause deviations, but they are excluded by the stated complete-dissociation assumption.

Correct answer: A. Osmotic pressure will increase

Explanation: Osmotic pressure is a colligative property and depends on the concentration of solute particles, according to π = iCRT. Dissociation converts one solute unit into two or more particles, so i becomes greater than one. At the same analytical concentration, the effective particle concentration increases and the osmotic pressure rises. It does not become zero.

Correct answer: A. It decreases the depression

Explanation: The depression in freezing point is ΔTf = iKf m. Association combines solute molecules and reduces the effective number of particles, making i less than one. Therefore, the observed depression is smaller than the value calculated for non-associating particles. It does not necessarily halve or double the depression; the exact change depends on the extent of association.

Correct answer: A. ΔTb = iKbm

Explanation: For a solution without abnormal behavior, the elevation is ΔTb = Kb m. If the solute associates or dissociates, the effective particle number must be corrected by the van’t Hoff factor. Therefore, the modified relation is ΔTb = iKb m. The factor multiplies the ordinary expression; it is not subtracted, added, or placed in the denominator.

Correct answer: A. ΔTf = iKf m

Explanation: For a nonelectrolyte, the freezing-point depression is ΔTf = Kf m. If the solute dissociates or associates, the actual number of particles differs from the expected number, and the van’t Hoff factor corrects this effect. Therefore the modified expression is ΔTf = iKf m. Division by i would reverse the particle-number correction.

Correct answer: A. π = iCRT

Explanation: For a dilute solution, the ideal osmotic-pressure equation is π = CRT, where C is molar concentration, R is the gas constant, and T is absolute temperature. If association or dissociation changes the number of particles, the van’t Hoff factor i corrects the equation: π = iCRT. Thus, i multiplies the normal expression; it is not placed in the denominator or subtracted.

Correct answer: A. 0.4 molal

Explanation: The governing idea is that colligative effects depend on the total number of dissolved particles. Complete dissociation follows KCl → K⁺ + Cl⁻, producing two particles per formula unit, so the ideal van’t Hoff factor is i = 2. Effective particle molality = i × m = 2 × 0.2 = 0.4 molal. Thus option A is correct. The value 0.2 molal refers only to KCl formula units, while 0.1 is a division error and 0.6 has no basis in the dissociation equation.

Correct answer: A. i = 3 and dissociation has occurred

Explanation: The elevation in boiling point follows ΔTb = iKb m, so it is directly proportional to the van’t Hoff factor when solvent and concentration are fixed. A value three times the expected non-associated value means i = 3. Since i is greater than one, the number of particles has increased, which is consistent with dissociation. Association would instead give i less than one.

Correct answer: B. 2.2 times

Explanation: The governing concept is the van’t Hoff factor, which accounts for the number of solute particles. AB₃ gives four ions on complete dissociation. For 40% dissociation, i = 1 + α(n − 1) = 1 + 0.40(4 − 1) = 2.2. Since ΔTf = iKf m and glucose has i = 1, at equal molality and solvent the electrolyte produces 2.2 times the freezing-point depression. Thus option B is correct.

Correct answer: B. 50%

Explanation: For fixed solvent and molality, boiling-point elevation is proportional to the van’t Hoff factor: ΔTb = iKb m. The observed elevation is 1.5 times the value for undissociated particles, so i = 1.5. An AB electrolyte forms two ions when fully dissociated, giving i = 1 + α(2 − 1) = 1 + α. Therefore α = 1.5 − 1 = 0.50, or 50%. Hence option B is correct.

Correct answer: C. 80%

Explanation: When one formula unit of a solute can produce four ions on complete dissociation, the van’t Hoff relation is i = 1 + α(n − 1) = 1 + 3α. Substituting i = 3.4 gives 3.4 = 1 + 3α, so 3α = 2.4 and α = 0.80. The degree of dissociation is therefore 0.80 × 100 = 80%. Since complete dissociation would give i = 4, the result is physically consistent. Option C is correct.

Correct answer: D. Al2(SO4)3

Explanation: At the same concentration, a colligative effect is proportional to the number of particles produced, represented by i. Sucrose gives one particle, Na2SO4 gives three, MgCl2 gives three, and Al2(SO4)3 gives five ions on complete dissociation: two Al3+ and three sulfate ions. Therefore aluminium sulfate gives the greatest effect.

Correct answer: B. 1.8 times

Explanation: The governing concept is the van’t Hoff factor, which measures the effective number of solute particles. One A2B unit forms three particles on complete dissociation, so i = 1 + alpha(n − 1) = 1 + 0.40(3 − 1) = 1.80. Since osmotic pressure is proportional to i at fixed molality and temperature, its value is 1.8 times that of a non-electrolyte, for which i = 1. The other options use an incorrect particle count or ignore partial dissociation.

Correct answer: D. AlCl3

Explanation: Depression in freezing point follows ΔTf = iKf m. Since molality, solvent, and temperature conditions are the same, the solution with the greatest van’t Hoff factor has the greatest depression and therefore the lowest freezing point. Glucose gives i = 1, NaCl gives 2, CaCl2 gives 3, and AlCl3 gives 4 on complete dissociation. Thus AlCl3 is correct.