7: Abnormal Molecular Mass

Correct answer: A. 200

Explanation: Use i = normal molecular mass / observed molecular mass. On rearrangement, observed molecular mass = normal molecular mass / i. Therefore the observed value is 100 / 0.5 = 200. The factor is below one because association decreases the number of particles; accordingly, the observed molecular mass becomes larger than the normal molecular mass rather than smaller.

Correct answer: A. The number of solute particles changes

Explanation: Colligative properties such as osmotic pressure and freezing-point depression depend on the number of dissolved particles. If solute molecules associate, the number decreases; if they dissociate, it increases. The measured property then differs from the value expected for unchanged molecules, producing an abnormal calculated molecular mass. Colour and container shape are irrelevant to this effect.

Correct answer: A. 3

Explanation: On complete dissociation, sodium sulfate gives Na₂SO₄ → 2Na⁺ + SO₄²⁻. Thus, one formula unit produces two sodium ions and one sulfate ion, making three particles in solution. For an ideal completely dissociated electrolyte, the van’t Hoff factor equals this particle count, so i = 3. The sulfate ion remains one particle despite containing several atoms.

Correct answer: A. 1.30

Explanation: For the dissociation AB → A + B, the van’t Hoff relation is i = 1 + α, because one original particle becomes two and the particle count increases by one for each dissociated fraction. A 30% dissociation means α = 30/100 = 0.30. Substitution gives i = 1 + 0.30 = 1.30. Thus option A is correct; 0.30 is the degree of dissociation itself, not the van’t Hoff factor.

Correct answer: A. 75%

Explanation: For an AB-type electrolyte, the relation between the van’t Hoff factor and degree of dissociation is i = 1 + α, because one formula unit can produce two particles. Rearranging gives α = i − 1 = 1.75 − 1 = 0.75. Multiplying by 100 converts this fraction to 75%. Thus, the solute is 75% dissociated.

Correct answer: A. 0.80

Explanation: The governing concept is the van’t Hoff factor for association. During dimerization, two solute molecules combine to form one particle, so for an association fraction α, i = 1 − α/2. Here α = 40/100 = 0.40. Therefore, i = 1 − 0.40/2 = 1 − 0.20 = 0.80. Option A is correct. The value is below one because dimer formation decreases the total number of solute particles; values above one would indicate dissociation.

Correct answer: A. Greater than 1

Explanation: The factor is i = M normal / M observed. If the observed molecular mass is lower than the normal mass, the denominator is smaller than the numerator, so the ratio is greater than one. This indicates an increased number of particles, usually because of dissociation. The exact percentage is not needed to decide the direction of i.

Correct answer: A. i < 1 and association is possible

Explanation: Since i = M normal / M observed, an observed mass greater than the normal mass makes i less than one. A value below one means fewer particles are present than expected, which is consistent with association. The observation alone does not prove a particular mechanism with absolute certainty, so “association is possible” is the appropriately cautious statement.

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

Explanation: Colligative properties depend on the number of dissolved particles rather than directly on their chemical identity. Association and dissociation change that particle number, so measurements such as osmotic pressure can differ from the value expected for unchanged molecules. This difference leads to an abnormal calculated molecular mass. Colour, smell, and instrument size do not govern the colligative relation.

Correct answer: A. When i = 1

Explanation: The van’t Hoff factor compares the actual number of solute particles with the number expected if each formula unit remained unchanged. Normal behaviour means there is neither dissociation nor association, so the actual and expected particle numbers are equal. Their ratio is therefore i = 1. Values below one indicate association, while values above one usually indicate dissociation.

Correct answer: A. One-fourth

Explanation: The relation is i = normal molecular mass / observed molecular mass. Rearranging gives observed molecular mass = normal molecular mass / i. For i = 4, the observed mass is normal mass divided by four, or one-fourth of it. A larger i means more effective particles and therefore a smaller observed molecular mass in this relation.

Correct answer: A. It will be greater than normal

Explanation: Since i = normal molecular mass / observed molecular mass, a value of i less than one means the numerator is smaller than the denominator. Therefore the observed molecular mass must be greater than the normal molecular mass. This commonly results from association, where several solute units combine and the number of particles decreases. It cannot make the mass zero.

Correct answer: C. Tetramer

Explanation: In complete association, n original particles combine to form one group, so the van’t Hoff factor is i = 1/n. Since 0.25 = 1/4, n equals 4. Thus four solute particles form one associated group, called a tetramer. A dimer, trimer, or pentamer would give factors 0.50, 0.333, and 0.20 respectively.

Correct answer: A. The actual number of solute particles changes by association or dissociation

Explanation: Colligative properties depend on the number of dissolved particles, not directly on their chemical identity. Association joins particles and lowers their number, whereas dissociation separates them and increases their number. If this change is ignored, the calculated molar mass differs from the true value; this is called abnormal molar mass.

Correct answer: A. i = 3, dissociation

Explanation: The governing relation for abnormal molar mass is M_app = M/i, where M is the normal molar mass. Given M_app = M/3, equating M/i with M/3 gives i = 3. A van’t Hoff factor greater than one means that the number of solute particles has increased, which occurs through dissociation. Thus option A is correct. Association would produce i below one, normal behaviour gives i = 1, and i = 0 is physically inappropriate for a dissolved solute.

Correct answer: C. 300 g mol−1

Explanation: Use the relation Mapparent = Mtrue/i. Substitution gives Mapparent = 180/0.6 = 300 g mol−1. Since i is less than one, the effective number of particles has decreased because of association, so the apparent molar mass must be greater than the true molar mass. This also rules out 108 and 60.

Correct answer: B. 5:1

Explanation: Complete pentamerisation means five original solute molecules combine into one effective particle. Therefore the van’t Hoff factor is i = 1/5. Apparent molar mass varies inversely with i, so M_app = M/i = M/(1/5) = 5M. Hence M_app : M = 5 : 1, making option B correct. The ratio 1:5 reverses the inverse relation and would incorrectly treat the apparent mass as smaller during association.

Correct answer: A. Complete dimer association

Explanation: The van’t Hoff factor is i = M/M_app = 120/240 = 0.50. Thus the effective number of particles has become half the original number. Complete dimer association converts every pair of particles into one particle and gives i = 1/2. Dissociation would produce i greater than one, while no change would give i = 1.

Correct answer: A. 0.42M

Explanation: The apparent molar mass is related to the true mass by M_app = M/i. Substituting i = 2.4 gives M_app = M/2.4 = 0.4167M, which rounds to 0.42M. The statement about three ions confirms that i can arise from partial dissociation, but once i is given, the mass calculation uses only the inverse relation.

Correct answer: A. 0.80

Explanation: A 25% increase means M_app = 1.25M. Since M_app = M/i, we have 1.25M = M/i. Cancelling M gives i = 1/1.25 = 0.80. A value below one indicates association or some other process that reduces the effective particle count; dissociation would instead produce an apparent mass below M.

Correct answer: C. 0.135 M

Explanation: The van’t Hoff factor converts formal concentration into the effective concentration of solute particles: Ceffective = iC. Substitution gives Ceffective = 1.35 × 0.10 = 0.135 M. Thus option C is correct. The value exceeds 0.10 M because partial dissociation of XY creates additional particles; 0.035 M would represent only the increase, not the total effective concentration.

Correct answer: D. 200 g mol⁻¹

Explanation: Use M_app = M/i. Substitution gives M_app = 110/0.55 = 200 g mol⁻¹. Because i is less than one, the effective number of particles is reduced, so the apparent molar mass is greater than the true mass. Multiplying 110 by 0.55 would give the wrong direction and does not satisfy the defining relation.

Correct answer: C. 1.33M

Explanation: A 25% lower colligative property means i = 0.75. Since apparent molar mass is inversely related to i, M_app = M/i = M/0.75 = 1.333M, approximately 1.33M. The reduction in boiling-point elevation reflects fewer effective particles due to association. It is therefore incorrect to multiply M by 0.75.

Correct answer: A. i = 2, particles doubled

Explanation: The van’t Hoff factor is the ratio of observed to normal colligative property: i = 0.372/0.186 = 2. Thus the effective particle concentration is twice the normal value. This is consistent with dissociation or another particle-increasing process. A factor of 0.5 would correspond to a depression smaller than normal, while i = 1 would indicate no abnormality.

Correct answer: B. 33.3%

Explanation: The governing relation for dissociation into two ions is i = 1 + α, because each dissociated solute unit changes from one particle into two, producing one additional particle. Substituting i = 4/3 gives α = 4/3 − 1 = 1/3. Converting this fraction to a percentage gives 33.3%. Thus option B is correct. Option A is half of the required value, while option D is the undissociated fraction, not the degree of dissociation.