5: Colligative Properties

Correct answer: C. Nearly twice that of urea

Explanation: For a dilute solution, boiling-point elevation is given by ΔTb = iKb m. Urea is a non-electrolyte and does not dissociate, so its van’t Hoff factor is approximately 1. Completely dissociated NaCl produces Na+ and Cl−, giving i approximately 2. Since the molality and solvent are the same, the NaCl solution has nearly twice the boiling-point elevation of the urea solution.

Correct answer: B. When the solution is very dilute and contains biomolecules

Explanation: Osmotic pressure is especially useful for determining the molar mass of proteins, polymers, and other biomolecules because it can be measured in very dilute solutions at ordinary temperatures. These substances may decompose, denature, or undergo other changes when heating is required for boiling-point or freezing-point methods. Since π = CRT for an ideal dilute solution, even a small concentration gives a measurable pressure.

Correct answer: A. Association of solute molecules

Explanation: Colligative effects are proportional to the number of independent solute particles. If solute molecules associate, two or more particles combine to form one larger species, reducing the number of particles in solution. The measured effect then becomes smaller than the ideal prediction, and the van’t Hoff factor is less than one. Complete dissociation would instead increase the effect and give i greater than one.

Correct answer: B. It becomes nearly double

Explanation: In benzene, acetic acid molecules can associate through hydrogen bonding to form dimers. Thus, two solute molecules behave as one particle, reducing the number of independent particles in the solution. The colligative effect becomes smaller than expected for the unassociated solute. When molar mass is calculated from that reduced effect, the observed molar mass is higher, approximately twice the normal molar mass for complete dimerisation.

Correct answer: A. Benzoic acid in benzene

Explanation: Benzoic acid associates through hydrogen bonding in non-polar benzene and commonly forms dimers. This decreases the effective number of solute particles, so the van’t Hoff factor becomes less than one. Since the observed molar mass is related by Mobserved = Mactual/i, a value of i below one makes the observed molar mass greater than the actual value. KCl, in contrast, dissociates.

Correct answer: B. The solute is associating

Explanation: For a fixed solvent and molality, the colligative effect is proportional to the van’t Hoff factor. The observed depression is 40% of the non-dissociated value, so the effective factor is i = 0.40, which is less than 1. A value below 1 indicates that solute particles have associated, reducing the number of independent particles. Complete dissociation would instead produce i greater than 1. Hence, option B is correct.

Correct answer: C. 1.116 K

Explanation: For the first non-dissociated solution, i = 1 and ΔT_f = K_f m. Therefore, K_f = 0.744/0.4 = 1.86 K kg mol⁻¹. Complete dissociation of CaCl₂ produces three ions, so i = 3. For the second solution, ΔT_f = iK_fm = 3 × 1.86 × 0.2 = 1.116 K. Hence, option C is correct.

Correct answer: A. Association of solute molecules

Explanation: Boiling-point elevation is expressed as ΔT_b = iK_bm. For a non-dissociated solute, i = 1. If molecules associate, several solute molecules combine to form fewer independent particles, so i becomes less than 1 and the observed elevation decreases. Dissociation or ionisation would increase the number of particles and therefore increase ΔT_b, not lower it.

Correct answer: D. 0.25 M

Explanation: The salt K₄[Fe(CN)₆] completely dissociates as K₄[Fe(CN)₆] → 4K⁺ + [Fe(CN)₆]⁴⁻. Thus, one formula unit produces five independent solute particles. The effective molarity is therefore iC = 5 × 0.05 M = 0.25 M. The complex ion [Fe(CN)₆]⁴⁻ remains one particle and is not counted as seven separate particles.

Correct answer: C. 1.18 times

Explanation: For dimerisation, two original solute particles combine to form one particle, so the number of particles decreases. If the degree of association is α = 0.30, the van’t Hoff factor is i = 1 − α/2 = 1 − 0.15 = 0.85. Since the observed molar mass is related by Mobserved = Mtrue/i, the required ratio is 1/0.85 = 1.176, approximately 1.18. Thus association makes the observed molar mass larger than the true value.

Correct answer: B. 0.60

Explanation: For association in which two molecules form one dimer, the van’t Hoff factor is i = 1 − α/2, where α is the fraction associated. With α = 80% = 0.80, i = 1 − 0.80/2 = 1 − 0.40 = 0.60. Association decreases the total number of particles, so the factor must be less than one. Therefore, option B is correct.

Correct answer: B. The solute has associated

Explanation: For a non-dissociated solute, the expected van’t Hoff factor is i = 1, so ΔTf = Kf m. If the observed depression is smaller while the solvent and molality are otherwise comparable, the effective particle factor is less than one. This occurs when solute molecules associate to form dimers or larger aggregates. Association reduces the number of independent particles and therefore lowers the colligative effect.

Correct answer: A. Because osmotic pressure is easier to measure than the extremely small boiling-point change

Explanation: For a dilute solution, osmotic pressure is π = CRT, and it can remain measurable even when the concentration is very small. The boiling-point elevation is ΔTb = iKb m; in a dilute protein solution, this temperature change is often extremely small and difficult to determine accurately. Proteins are also non-volatile and may decompose on heating, so avoiding boiling is advantageous. Osmotic measurements are therefore preferred for biomolecules.

Correct answer: A. It will be equal

Explanation: For a dilute solution, osmotic pressure is given by π = iCRT, where i is the van’t Hoff factor, C is molar concentration, R is the gas constant, and T is absolute temperature. For the first solution, iC = 4 × 0.1 = 0.4 M. For the non-dissociating solute, i = 1, so iC = 1 × 0.4 = 0.4 M. Since temperature is the same, both solutions have equal osmotic pressure and are isotonic.

Correct answer: A. The number of particles will be underestimated, giving an incorrect molar mass

Explanation: Calcium chloride ideally dissociates according to CaCl2 → Ca2+ + 2Cl−, producing three particles from one formula unit. Therefore its ideal van’t Hoff factor is approximately 3. If the student assumes no dissociation, i = 1 is used, so the effective number of solute particles is underestimated. Because ΔTf = iKf m, the calculation ignores the larger particle effect and produces an erroneous molar mass.

Correct answer: B. Three-fourths of the ideal value

Explanation: Let the original number of solute particles be N. If half, N/2, remain single and the other half, N/2, associate in pairs, those associated particles form N/4 dimers. The total effective particles are therefore N/2 + N/4 = 3N/4. Since freezing-point depression is proportional to the effective particle concentration, ΔTf observed is three-fourths of the ideal value. Thus association gives i = 0.75.

Correct answer: A. Solution A

Explanation: Osmotic pressure is π = iCRT. For completely dissociated NaCl, i is approximately 2, whereas for a nonelectrolyte i = 1. Equal osmotic pressures at the same temperature require equal values of iC. Therefore, the NaCl concentration must be about half the nonelectrolyte concentration: 2C_A = C_B. Since the volumes are equal, solution A contains fewer actual moles of solute, even though both solutions have the same effective particle concentration.

Correct answer: B. 0.15 m

Explanation: The original amount is 0.2 mol in 1 kg of solvent. Half, or 0.1 mol, remains as individual molecules. The other 0.1 mol of molecules associates in pairs, producing only 0.05 mol of dimers. Therefore, effective solute particles are 0.10 + 0.05 = 0.15 mol, and effective molality is 0.15/1 = 0.15 m. Association decreases the number of particles.