7: Abnormal Molecular Mass

Correct answer: A. Colligative effect lower than expected

Explanation: Association combines solute molecules into larger units, reducing the number of independent particles. Since colligative properties depend on particle number, the observed effect becomes lower than the value expected for the same amount of unassociated solute. Using that smaller effect in calculations gives an apparently higher molar mass.

Correct answer: A. 2.70

Explanation: MgCl2 dissociates into one Mg2+ ion and two Cl− ions, giving three particles on complete dissociation. For fractional dissociation, i = 1 + alpha(n − 1). Thus i = 1 + 0.85(3 − 1) = 1 + 1.70 = 2.70, so option A is correct. The limiting value 3.00 applies only to complete dissociation; 2.85 incorrectly adds the percentage directly.

Correct answer: A. 1.67 times

Explanation: At equal concentration, the ratio of the colligative effect is i, so i = 0.6. The observed molar mass obeys Mobserved = Mtrue/i. Therefore Mobserved/Mtrue = 1/0.6 = 1.6667, or approximately 1.67. A reduced particle effect gives a larger apparent molar mass because of association.

Correct answer: D. 0.16 mol

Explanation: The particle-count principle states that one AB3 formula unit produces four ions on complete dissociation: one A-containing ion and three B-containing ions. Therefore the amount of effective particles is 0.04 mol × 4 = 0.16 mol. Option D is correct. The values 0.04 and 0.08 fail to count all products, while 0.12 corresponds to only three particles per formula unit.

Correct answer: A. Strong dissociation of a multi-ion salt

Explanation: i exceeds one when the number of solution particles is greater than the number of original formula units. A multi-ion salt that dissociates strongly can produce several ions per formula unit, giving a large i. Dimerisation and other association lower i, while glucose dissolves as intact molecules and has i close to one.

Correct answer: C. 72%

Explanation: A 1:1 electrolyte produces two ions when completely dissociated. Therefore, for degree of dissociation alpha, i = 1 + alpha(2 − 1) = 1 + alpha. Given i = 1.72, alpha = 1.72 − 1 = 0.72, or 72%. Hence option C is correct. The value 86% would incorrectly divide the factor by two rather than using the increase above unity.

Correct answer: C. 2.8

Explanation: One formula unit of Na2SO4 gives two Na+ ions and one SO4^2− ion, so complete dissociation would produce three particles. For degree of dissociation alpha, i = 1 + alpha(3 − 1). With alpha = 0.90, i = 1 + 0.90 × 2 = 2.80. Therefore option C is correct. The value 3.0 applies only to complete 100% dissociation, while 1.9 is too low.

Correct answer: A. i = 1.5, dissociation

Explanation: The van’t Hoff factor is the ratio of the actual effective particle amount to the original solute amount. Thus i = 0.045/0.03 = 1.5. Since i exceeds one, more particles are present than in the undissociated state, indicating dissociation into smaller ions or molecules. Association would reduce the particle count and give i below one. Therefore option A is correct.

Correct answer: A. Change in the number of solute particles

Explanation: Colligative properties depend on the number of dissolved particles rather than their chemical identity alone. Dissociation increases the number of particles, whereas association decreases it. If this change is ignored while calculating molar mass from a colligative property, the result differs from the true molar mass; this is called abnormal molar mass.

Correct answer: A. Dissociation is very small

Explanation: AB2 can produce three particles on complete dissociation, for which i would approach 3. The measured value 1.1 is only slightly above one, so the increase in particle number is small. Under the stated salt model, this corresponds to slight dissociation, not complete dissociation or association. No exact percentage is requested.

Correct answer: A. 0.75

Explanation: The van’t Hoff factor is defined as the number of effective particles divided by the initial number of solute particles. Therefore i = 0.015/0.02 = 0.75. Because i is less than one, association has reduced the number of independently moving particles. Dimerisation or another association process can cause this result; dissociation would produce i greater than one. Hence option A is correct.

Correct answer: A. i = 0.5

Explanation: In complete dimerisation, every two original solute molecules combine to form one dimer particle. Thus the final number of particles is one-half of the original number. Since i is the final particle count divided by the initial count, its limiting value is 1/2 = 0.5. The value 1 would mean no association.

Correct answer: A. 30 g mol−1

Explanation: The observed molar mass is related to the true molar mass by Mobserved = Mtrue/i. Substitution gives Mobserved = 72/2.4 = 30 g mol−1. Since i is greater than one, dissociation has increased the number of particles, so the colligative measurement makes the molar mass appear lower than its true value.

Correct answer: A. The solute underwent dissociation

Explanation: Because Mobserved = Mtrue/i, an observed mass lower than the true mass means i > 1. A factor greater than one indicates that the number of effective particles has increased relative to the original solute units. Dissociation is the usual explanation. Dimerisation and particle decrease would give i < 1 and a higher observed molar mass.

Correct answer: A. Association of solute particles in solution

Explanation: A higher observed molar mass follows from Mobserved = Mtrue/i, so it requires i < 1. Association combines two or more solute molecules into fewer independent particles, lowering osmotic pressure and making the calculated mass too high. Ionisation would increase particles and produce i > 1 instead.

Correct answer: A. Approximately 1

Explanation: For a normal solute, each dissolved formula unit remains one particle: there is neither dissociation into ions nor association into larger molecules. Since colligative properties depend on the number of solute particles, the actual particle concentration equals the calculated concentration. Therefore, the van’t Hoff factor i is approximately 1 in all four methods. Values below or above 1 indicate association or dissociation, respectively.

Correct answer: A. i = 1 − α + α/m

Explanation: Let α be the fraction of solute molecules that associates to form groups of m molecules. The unassociated fraction, 1 − α, contributes the same number of particles, while the associated fraction α becomes α/m groups. Therefore, the total particle factor is i = (1 − α) + α/m = 1 − α + α/m. Option B describes the corresponding dissociation relation, and option D is the inverse relation for molar masses.

Correct answer: A. i = 0.8, association

Explanation: For the same concentration and temperature, osmotic pressure is proportional to the effective number of solute particles: π = iCRT. A value 20% below normal means π is 80% of the normal value, so i = 0.80. Since i is less than 1, the number of particles has decreased, which is characteristic of association such as dimerisation. Therefore A is correct.

Correct answer: A. i < 1 indicates association

Explanation: The van’t Hoff factor compares the actual number of dissolved particles with the number expected from the formula. Dissociation creates more particles and therefore gives i greater than 1. Association combines particles into larger units, reducing their number and giving i less than 1. It also makes the observed molar mass higher, not lower. Hence A is the correction.