7: Abnormal Molecular Mass

Correct answer: A. 1.5

Explanation: For AB → A + B, complete dissociation would change one particle into two, so the increase is one particle per dissociated formula unit. The relation is i = 1 + α, where α is the fraction dissociated. Here, 50% means α = 0.50; therefore i = 1 + 0.50 = 1.50. The value 1.0 would represent no dissociation, not 50% dissociation.

Correct answer: A. 1.8

Explanation: For the dissociation AB2 → A + 2B, one formula unit produces three particles on complete dissociation. Therefore, for fractional dissociation α, the governing expression is i = 1 + (3 − 1)α = 1 + 2α. With α = 40/100 = 0.40, i = 1 + 2(0.40) = 1.80. Option B incorrectly adds α only once, so option A is correct.

Correct answer: A. 1.5

Explanation: The ideal dissociation is A₂B → 2A + B, so one formula unit gives three particles when dissociation is complete. For a compound producing n particles, i = 1 + (n − 1)α. Here n = 3 and α = 25/100 = 0.25, so i = 1 + 2(0.25) = 1.50. The other values result from miscounting particles or using the percentage incorrectly.

Correct answer: A. 25%

Explanation: For an AB electrolyte that dissociates into two ions, i = 1 + α, where α is the fraction dissociated. Rearranging gives α = i − 1 = 1.25 − 1 = 0.25. Multiplying by 100 converts the fraction to a percentage: 25%. A value of 125% is impossible because a degree of dissociation cannot exceed 100%.

Correct answer: A. 60%

Explanation: For AB₂ → A + 2B, three particles can form from one formula unit, so i = 1 + (3 − 1)α = 1 + 2α. Substituting i = 2.2 gives 2.2 = 1 + 2α, hence 2α = 1.2 and α = 0.6. Converting the fraction to a percentage gives 60%. The other percentages do not satisfy the equation.

Correct answer: A. 0.70

Explanation: In dimerization, two solute molecules combine to form one dimer: 2A → A₂. If α is the fraction of molecules undergoing dimerization, the number of particles decreases by α/2 per original molecule, giving i = 1 − α/2. For 60% dimerization, α = 0.60, so i = 1 − 0.60/2 = 1 − 0.30 = 0.70. Association therefore gives i below one.

Correct answer: A. i = 3, dissociation

Explanation: At fixed concentration, temperature, and gas constant, osmotic pressure is directly proportional to the van’t Hoff factor: π = iCRT. If the observed pressure is three times the expected value calculated with i = 1, then i must be 3. A factor greater than one means that the number of dissolved particles has increased, which indicates dissociation rather than association or no change.

Correct answer: A. 2.5

Explanation: The governing concept is the van’t Hoff factor, which represents the effective number of solute particles produced by dissociation. For an AB₃ electrolyte, the dissociation is AB₃ → A + 3B, so one formula unit can produce four particles when dissociation is complete. If α is the degree of dissociation, the factor is i = 1 + (ν − 1)α, where ν = 4. Thus i = 1 + 3α. For 50% dissociation, α = 0.50, so i = 1 + 3(0.50) = 2.5. Therefore option A is correct. The value 4.0 would require complete dissociation. The value 1.5 would correspond to an electrolyte producing only two particles at complete dissociation, and 3.5 results from an incorrect substitution or overestimating the extent of dissociation.

Correct answer: A. Whether the number of solute particles changed

Explanation: Colligative properties depend primarily on the number of solute particles, not on their individual chemical identity. Association decreases the number of particles, whereas dissociation increases it. Therefore, the first question is whether the solute particle count has changed. Colour, container material, and stirring do not explain abnormal molecular mass in this context.

Correct answer: A. Dissociation

Explanation: Calculate the factor from i = normal molecular mass / observed molecular mass = 98 / 49 = 2. A factor greater than one means that the number of solute particles has increased compared with the undissociated state. Such an increase is caused by dissociation, in which one solute unit splits into two or more particles. Association would give i less than one.

Correct answer: A. Solute particles have associated

Explanation: Using i = normal molecular mass / observed molecular mass, i = 60 / 120 = 0.5. A value below one indicates fewer particles than expected from the original solute units. The usual explanation is association; for example, two particles may combine to form one larger particle. Complete dissociation and doubled particle number would instead produce i greater than one.

Correct answer: A. It becomes lower than normal

Explanation: Dissociation increases the number of particles in solution. The resulting colligative effect is larger than expected for an undissociated solute. When molecular mass is calculated from that larger effect without correction, the value obtained is smaller than the normal molecular mass. Thus dissociation produces a lower observed molecular mass, while association produces a higher one.

Correct answer: A. Greater than normal

Explanation: When two molecules associate into one particle, the total number of solute particles decreases. Colligative effects consequently become smaller than expected for the original number of molecules. Calculation without an association correction then gives a molecular mass larger than the normal value. Therefore, association leads to an increased observed molecular mass.

Correct answer: A. i = normal molecular mass / observed molecular mass

Explanation: For a colligative-property measurement, the van’t Hoff factor is defined as the ratio of the normal molecular mass to the molecular mass calculated from the observed effect. Hence i = M normal / M observed. This relation correctly gives i greater than one for dissociation and less than one for association; reversing the ratio would give the opposite interpretation.

Correct answer: A. Association through hydrogen bonding

Explanation: Benzoic acid molecules can form hydrogen-bonded dimers in a non-polar solvent such as benzene. Two molecules then behave as one solute particle, reducing the particle number responsible for colligative effects. The reduced effect gives an observed molecular mass greater than the normal monomeric mass. Ionic dissociation and metal-ion formation are not involved.

Correct answer: A. It dissociates into K⁺ and Cl⁻

Explanation: KCl is an ionic electrolyte and, in water, it separates into K⁺ and Cl⁻ ions. Thus one formula unit produces two solute particles instead of one, increasing the colligative effect. If this enhanced effect is interpreted without correcting for dissociation, the calculated molecular mass is lower than the normal value. Dimer formation would have the opposite effect.

Correct answer: A. 0.5

Explanation: The governing concept is association and its effect on the number of solute particles. In complete dimerization, two monomer molecules combine according to 2A → A₂. Consequently, two original particles become one dimer particle, so the final particle count is half the ideal non-associating count. Hence i = actual particles/expected particles = 1/2 = 0.5. Values above one describe dissociation, not complete association.

Correct answer: A. 5

Explanation: The governing concept is stoichiometric particle counting during complete dissociation. Aluminium sulfate separates according to Al₂(SO₄)₃ → 2Al³⁺ + 3SO₄²⁻. The subscript 2 gives two aluminium ions, and the coefficient 3 before sulfate gives three sulfate ions. Thus the total number of particles is 2 + 3 = 5, so the ideal van’t Hoff factor is i = 5. The charge values do not replace the stoichiometric coefficients.

Correct answer: A. Between 1 and 2

Explanation: For an AB solute, the governing relation for dissociation is i = 1 + α, because each dissociated AB unit changes from one particle into two. In incomplete dissociation, the degree of dissociation satisfies 0 < α < 1. Therefore 1 < i < 2. The value i = 2 applies only when dissociation is complete, while a value below one is associated with particle association rather than dissociation.

Correct answer: A. 1.50

Explanation: The governing formula for a solute producing n particles on complete dissociation is i = 1 + (n − 1)α. For AB₂ → A + 2B, n = 3, so i = 1 + 2α. The stated 25% dissociation gives α = 0.25. Therefore i = 1 + 2(0.25) = 1.50. The factor 2 represents the increase from one original particle to three particles, not the percentage itself.

Correct answer: A. 1.60

Explanation: For AB₃ → A + 3B, complete dissociation changes one solute particle into four, so the increase in particle number is 3. The appropriate relation is i = 1 + 3α. With 20% dissociation, α = 0.20, giving i = 1 + 3 × 0.20 = 1.60. The value 1.20 would incorrectly count only one additional particle, while 4 would represent complete rather than partial dissociation.

Correct answer: A. 30%

Explanation: For AB₂ → A + 2B, complete dissociation produces three particles from one, so the van’t Hoff relation is i = 1 + 2α. Substituting the given value gives 1.6 = 1 + 2α. Hence 2α = 0.6 and α = 0.3. Converting the decimal fraction to a percentage, α × 100 = 30%. Therefore option A is correct; 60% incorrectly omits the factor of two in the relation.

Correct answer: A. i = 1 − α/2

Explanation: The governing concept is particle reduction by association. For dimerization, 2A → A₂. Consider one initial mole: α moles of monomer participate, forming α/2 moles of dimers, while 1 − α moles remain as monomers. The total final particles are (1 − α) + α/2 = 1 − α/2 moles. Dividing by the initial one-mole particle count gives i = 1 − α/2, so option A is correct.

Correct answer: A. Acetic acid in benzene

Explanation: Acetic acid molecules can form hydrogen-bonded dimers in a non-polar solvent such as benzene. Dimer formation reduces the number of independent solute particles and can therefore produce an observed molecular mass higher than the normal value. The aqueous salts listed are ordinarily discussed as electrolytes that dissociate into ions, giving i greater than one rather than association.

Correct answer: A. NaNO₃ in water

Explanation: NaNO₃ is an ionic electrolyte and in water it dissociates approximately into Na⁺ and NO₃⁻ ions. This increases the number of solute particles, so i is greater than one, especially under the ideal assumption. Acetic and benzoic acids in benzene can associate, while glucose is a non-electrolyte and normally has i close to one.