7: Abnormal Molecular Mass

Correct answer: A. Partial association

Explanation: For a solute that neither associates nor dissociates, the van’t Hoff factor is 1. When i is below 1, the effective number of solute particles is lower than expected. This usually happens because two or more solute molecules combine to form larger associated species. Since i = 0.80 is below 1 but not necessarily the limiting value for complete association, it most appropriately indicates partial association. Dissociation would make i greater than 1.

Correct answer: A. A change in the number of solute particles from the expected value

Explanation: Abnormal molar mass arises when the solute does not remain as independent formula units in solution. Dissociation increases the particle count, whereas association decreases it. Since colligative measurements respond to particle number, the calculated molar mass differs from the normal value; container colour and shaking are irrelevant.

Correct answer: A. KCl

Explanation: KCl is an ionic compound and, when dissolved in water, it separates into K⁺ and Cl⁻ ions. Thus one formula unit gives approximately two solute particles, making i greater than 1 in the ideal limiting case. Urea, glucose and sucrose are molecular nonelectrolytes and normally remain as individual molecules, so their i is close to 1.

Correct answer: A. Three

Explanation: Under the ideal assumption of complete dissociation, one formula unit separates entirely into its constituent particles, and the van’t Hoff factor equals the number of particles produced. Therefore, i = 3 means that one formula unit gives three particles in solution. For example, a formula unit that produces three ions would have an ideal i value of 3. The values 1 and 2 would correspond to one or two particles, not three.

Correct answer: A. Whether the solute associates or dissociates

Explanation: The first step is to determine whether the solute particles split apart or join together in solution. Dissociation increases the effective particle number and usually gives i > 1, whereas association decreases it and gives i < 1. This decision tells you which relation and direction of molar-mass change to use.

Correct answer: A. Different from the normal value

Explanation: Colligative-property measurements use the effective number of dissolved particles. If dissociation increases that number, the calculated molar mass generally appears lower; if association decreases it, the calculated value appears higher. Therefore it can differ from the normal molar mass, but it is not always zero, double or unchanged.

Correct answer: A. The effective number of particles in solution

Explanation: The van’t Hoff factor compares the actual effective particle count with the count expected if each formula unit remained unchanged. It therefore explains changes in colligative properties and abnormal molar mass caused by association or dissociation. It has no role in describing container shape, colour or sound.

Correct answer: A. Dissociation or association

Explanation: Abnormal molar mass is a consequence of a changed number of dissolved particles. Dissociation breaks one species into several particles, while association combines several species into fewer particles. Because colligative properties depend on particle number, either process causes the molar mass calculated from such properties to differ from the normal value.

Correct answer: A. Dissociation

Explanation: i greater than 1 means that the solution contains more effective particles than expected from the undissociated solute. The usual reason is dissociation into ions or smaller molecules. Association would reduce the particle count and make i less than 1. The conclusion assumes no other unusual process dominates the measurement.

Correct answer: A. Association

Explanation: The van’t Hoff factor measures the effective change in the number of solute particles. If there is no association or dissociation, i equals 1. Association joins several solute molecules into fewer larger species, so the effective particle number decreases and i becomes less than 1. Dissociation has the opposite effect and generally makes i greater than 1. Therefore, i < 1 generally indicates association, although the exact value depends on its extent.

Correct answer: A. One

Explanation: The van’t Hoff factor is defined as the ratio of the actual effective number of solute particles to the number expected from the original formula units. If the solute neither dissociates nor associates, every formula unit remains one particle in solution. Thus the numerator and denominator are equal, giving i = 1. A value above 1 suggests dissociation, while a value below 1 suggests association; neither occurs in this case.

Correct answer: A. Due to dissociation into ions

Explanation: Sodium chloride is an ionic compound. In water, polar water molecules separate and hydrate its ions, so the crystal units dissociate according to NaCl → Na⁺ + Cl⁻. One formula unit therefore gives two dissolved particles rather than one formula unit remaining as a single unit. This increases the effective particle number and can increase colligative effects. The solvent does not disappear, and undissolved solid salt would not produce this dissolved-particle increase.

Correct answer: A. Three times

Explanation: The van’t Hoff factor is the ratio of the effective number of particles in solution to the number expected from the original solute units. Thus i = 3 means effective particle number divided by normal particle number equals 3. For the same amount of solute and comparable conditions, the solution behaves as though it contains three times as many particles. This can result from complete dissociation into three particles, although the numerical value alone does not prove the exact mechanism.

Correct answer: A. 2

Explanation: Under complete dissociation, every formula unit separates into all of its constituent ions. If the substance gives two ions, the number of particles changes from one formula unit to two ions. Hence the ideal van’t Hoff factor is i = 2. The value 1 applies to a non-dissociating solute, while 0.5 applies to complete dimerisation.

Correct answer: A. 3

Explanation: For complete dissociation, every formula unit separates into all of the particles represented by its dissociation equation. If one formula unit produces three ions, the number of particles changes from one to three. The ideal van’t Hoff factor is therefore i = 3. A value of 2 would apply to complete dissociation into two ions, i = 1 to no change in particle number, and 0.33 would represent a reduction rather than three-particle dissociation.

Correct answer: A. 0.5

Explanation: In complete dimerisation, two original solute molecules behave as one particle. Thus the number of effective particles becomes one-half of the initial number. Since i is the effective-to-original particle ratio, i = 1/2 = 0.5. Partial dimerisation would give a value between 0.5 and 1.

Correct answer: A. Dissociation can make i greater than one

Explanation: Dissociation converts one solute unit into two or more independent particles, so the effective particle ratio can exceed one. Association instead lowers the ratio, and no change gives i = 1. The van’t Hoff factor is directly used in colligative-property equations, so the other statements contradict the definition.

Correct answer: A. 0.5

Explanation: The van’t Hoff factor is the ratio of the effective particle number after association or dissociation to the normal particle number before that change. If the effective number becomes half of the normal number, this ratio is 1/2. Therefore, i = 0.5. Such a reduction is consistent with association, in which several original molecules combine into fewer larger species. i = 1 would mean no change, whereas i = 2 would mean a doubling of particles.

Correct answer: A. i

Explanation: The van’t Hoff factor is conventionally represented by the lowercase italic symbol i. It measures the change in the effective number of solute particles caused by dissociation or association. M is commonly used for molar mass, T for temperature, and K often denotes a constant, so those symbols do not represent the van’t Hoff factor in this context.

Correct answer: A. The fraction of solute that has dissociated

Explanation: The degree of dissociation, α, is the fraction of the original solute units that have separated into smaller particles or ions. It has no unit and normally lies between 0 and 1. For example, 25% dissociation means α = 25/100 = 0.25. It is used in expressions such as i = 1 + (n − 1)α.

Correct answer: A. 1

Explanation: For a solute that can dissociate into two ions, the relation is i = 1 + α. Substituting α = 0 gives i = 1 + 0 = 1. This means that no solute unit has dissociated, so the particle number remains at the normal value. The result is not zero because the original undissociated particles are still present.

Correct answer: A. 2

Explanation: For dissociation into two ions, the van’t Hoff factor is i = 1 + α. Here α is the fraction of solute units that dissociate. Complete dissociation means every unit dissociates, so α = 1. Substitution gives i = 1 + 1 = 2. This agrees with the particle count: one original formula unit produces two ions. A value of 1 would indicate no dissociation, while 0.5 would indicate fewer effective particles.

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

Explanation: Abnormal molar mass is obtained when a colligative-property measurement is affected by a change in the number of solute particles. Therefore, first check whether the solute dissociates into smaller particles or associates into larger units. The vessel colour, laboratory size and operator do not determine this molecular effect.

Correct answer: A. Lower observed molar mass

Explanation: In water, KCl can produce K⁺ and Cl⁻ ions, increasing the number of solute particles. The larger particle count increases the colligative effect, so the molar mass calculated from that effect becomes lower than the normal value. Thus dissociation gives i greater than one and a lower observed molar mass.

Correct answer: A. Divide normal molar mass by observed molar mass

Explanation: For abnormal molar mass, i is calculated as normal molar mass divided by observed molar mass: i = Mnormal/Mobserved. Both masses must be expressed in the same units. The factor compares the expected number of particles with the effective number indicated by the colligative property; temperature, colour, pressure, or solvent naming is not part of this direct relation.