7: Abnormal Molecular Mass

Correct answer: A. 2

Explanation: Use the abnormal-mass relation i = normal molecular mass / observed molecular mass. Substituting the given values gives i = 120/60 = 2. Since the factor is greater than one, the solution contains twice as many effective particles as expected from undissociated molecules. This is consistent with dissociation, although the exact extent depends on the chemical species involved.

Correct answer: A. 61

Explanation: The relation between normal and observed molecular mass is i = normal mass / observed mass. Rearranging gives observed mass = normal mass/i. Therefore, observed mass = 122/2 = 61. Because i is greater than one, the observed mass is lower than the normal mass, which is the characteristic direction for dissociation and increased particle number.

Correct answer: A. Because it dissociates into ions

Explanation: In water, NaCl dissociates into Na⁺ and Cl⁻ ions. One formula unit therefore produces two solute particles instead of one, increasing the colligative effect. If that enhanced effect is used to calculate molecular mass, the result is lower than the normal formula-unit mass. Dimer formation would reduce particles and produce the opposite trend.

Correct answer: A. When the solute undergoes association or dissociation

Explanation: Colligative-property calculations assume that each solute formula unit remains an independent particle. If molecules associate, the particle number decreases; if an electrolyte dissociates, it increases. Either change makes the calculated molecular mass differ from the normal value. Pure solvent, constant temperature, or non-volatility alone does not create abnormal molecular mass.

Correct answer: A. Dimerization

Explanation: An i value of 0.5 means that the effective number of particles is half the number expected from separate solute molecules. In complete dimerization, two molecules combine to form one particle, so the particle count becomes half and i = 1/2. Dissociation would increase the count and produce i greater than one, while no change would give i = 1.

Correct answer: A. 3

Explanation: One formula unit of K₂SO₄ dissociates ideally as K₂SO₄ → 2K⁺ + SO₄²⁻. Thus, one formula unit produces two potassium ions and one sulfate ion, making three particles in total. Therefore, the ideal van’t Hoff factor is i = 3. The ionic charges do not add extra particles; only the number of ions is counted.

Correct answer: A. 3

Explanation: On complete dissociation, one formula unit of CaCl₂ separates according to CaCl₂ → Ca²⁺ + 2Cl⁻. Thus, one solute unit produces three independent particles in solution. The van’t Hoff factor is the number of particles formed per original formula unit, so i = 3. The values 2 and 1 would not count all the chloride ions, while 0.5 is not possible for complete dissociation here.

Correct answer: A. 4

Explanation: AlCl₃ dissociates ideally according to AlCl₃ → Al³⁺ + 3Cl⁻. One formula unit therefore produces one aluminium ion and three chloride ions, for a total of four particles. The ideal van’t Hoff factor is consequently i = 4. The magnitude of the ionic charge is irrelevant to the count; the stoichiometric number of ions is what determines i.

Correct answer: A. 0.5

Explanation: The relation is i = normal molecular mass / observed molecular mass. If the observed mass is 2M while the normal mass is M, then i = M/(2M) = 1/2 = 0.5. Since i is below one, the effective number of particles has decreased, which is consistent with association such as dimer formation. Dissociation would produce a lower observed mass and i greater than one.

Correct answer: A. 2

Explanation: Use i = normal molecular mass / observed molecular mass. Let the normal mass be M; the observed mass is M/2. Therefore i = M/(M/2) = 2. A factor greater than one means that more particles are present than expected from intact solute molecules, which points to dissociation. Association would give i below one rather than two.

Correct answer: A. van't Hoff factor

Explanation: The abnormal value results from a difference between the expected and actual number of solute particles. The van't Hoff factor i represents this ratio and is inserted into colligative-property equations such as ΔTb = iKb m, ΔTf = iKf m, and π = iCRT. Acidity, colour, and vapour-density factors are not the general correction used here.

Correct answer: A. Acetic acid in benzene

Explanation: Acetic acid in benzene is a standard example because acetic-acid molecules form hydrogen-bonded dimers in the relatively non-polar solvent. Dimer formation lowers the number of solute particles and gives a higher observed molar mass. The aqueous salts listed are generally discussed as electrolytes that dissociate into ions, giving i greater than one under ideal assumptions.

Correct answer: A. Neither association nor dissociation

Explanation: The van't Hoff factor compares actual and expected particle numbers. If i = 1, the two numbers are equal, so the solute behaves as though each formula unit remains one independent particle. In the usual model, there is no effective association or dissociation. Complete association gives i below one, and complete dissociation gives a value greater than one.

Correct answer: A. One-third

Explanation: From i = normal molecular mass/observed molecular mass, rearrange to obtain observed molecular mass = normal molecular mass/i. With i = 3, the observed value is M/3, or one-third of the normal value. The high i indicates increased particle number, as in dissociation, which makes the calculated observed mass smaller rather than larger.

Correct answer: A. It will be greater

Explanation: Using i = Mnormal/Mobserved, rearrange to Mobserved = Mnormal/i. Since i = 0.8 is less than one, dividing by it makes the observed mass larger than the normal mass: Mobserved = Mnormal/0.8 = 1.25Mnormal. This direction is characteristic of association, which reduces the effective number of particles.

Correct answer: A. KCl in water

Explanation: KCl is an electrolyte and, in water under the usual ideal approximation, dissociates into K⁺ and Cl⁻ ions. One formula unit therefore gives two particles, so i is greater than one. Acetic acid in benzene tends to associate, while sugar and a non-reacting molecular solute do not normally change particle number appreciably and have i close to one.

Correct answer: A. Benzoic acid in benzene

Explanation: Benzoic acid molecules associate through hydrogen bonding in benzene and commonly form dimers. Two original molecules then behave as one particle, reducing the effective particle count and making i less than one. The aqueous salts are electrolytes; their dissociation generally increases the particle number and gives i greater than one in the ideal treatment.

Correct answer: A. 5

Explanation: Complete dissociation means every formula unit separates into all its constituent ions. A₂B₃ gives two A ions and three B ions, so the total number of particles is 2 + 3 = 5. Hence the ideal van’t Hoff factor is i = 5. The values 2 and 3 count only one type of ion, while 6 is not the particle count.

Correct answer: A. 1

Explanation: The van’t Hoff factor compares the actual number of solute particles with the number expected if each solute unit remained unchanged. For a non-electrolyte with no association or dissociation, the particle number does not change. Thus the ratio is one, i = 1. Values above or below one indicate dissociation or association, respectively.

Correct answer: A. 3

Explanation: One formula unit of MgCl₂ dissociates as MgCl₂ → Mg²⁺ + 2Cl⁻. Thus, one solute unit becomes three independent particles in solution. For complete dissociation, the van’t Hoff factor equals the total number of particles formed, so i = 3. The charge on Mg²⁺ does not add another particle; it only indicates the ion’s charge.

Correct answer: A. 4

Explanation: The governing concept is particle counting in complete electrolyte dissociation. FeCl₃ separates as FeCl₃ → Fe³⁺ + 3Cl⁻. Thus, one formula unit produces one iron(III) ion and three chloride ions, giving 1 + 3 = 4 solute particles. Therefore, the ideal van’t Hoff factor, i, is 4. The ionic charge describes charge balance, whereas the coefficient determines how many particles are counted.

Correct answer: A. 360

Explanation: The relation between the van’t Hoff factor and molecular masses is i = normal molecular mass / observed molecular mass. Rearranging gives observed molecular mass = normal molecular mass / i. Substitution gives 180 / 0.5 = 360. Since i is less than one, association has reduced the number of particles and consequently increased the observed molecular mass.

Correct answer: A. i > 1

Explanation: The van’t Hoff factor compares the actual particle effect with the ideal effect calculated for unchanged solute units. If i is greater than one, the actual number of particles is greater than expected. This commonly results from dissociation into ions. An i value below one indicates association, while zero or negative values are not physically appropriate here.

Correct answer: A. Association of solute

Explanation: An i value of 0.6 is less than one, meaning that the actual number of solute particles is lower than the number expected from the original formula units. Particle reduction occurs when molecules associate or combine. Complete dissociation and increasing ion number would produce i greater than one, so they are unsuitable explanations.

Correct answer: A. 2

Explanation: The van’t Hoff factor is calculated by i = normal molecular mass / observed molecular mass. Substituting the given values gives i = 90 / 45 = 2. This means that the effective number of solute particles is twice the number expected without abnormal behaviour, usually because the solute dissociates. The mass value 45 is not itself the factor.