7: Abnormal Molecular Mass

Correct answer: C. Double the normal molar mass

Explanation: Using i = Mnormal/Mobserved, substitute i = 0.5: 0.5 = Mnormal/Mobserved. Rearranging gives Mobserved = 2Mnormal. Thus the observed molar mass is twice the normal value. Such a result is consistent with strong association, for example complete dimerization, which reduces the number of effective particles.

Correct answer: B. Between 1 and 2

Explanation: If KCl does not dissociate, each formula unit remains one particle and i = 1. On complete dissociation, KCl → K⁺ + Cl⁻ gives two particles, so i = 2. Partial dissociation lies between these limiting cases; consequently, for a nonzero but incomplete degree of dissociation, 1 < i < 2. It cannot exceed the complete-dissociation limit of 2.

Correct answer: B. Between 0.5 and 1

Explanation: If no molecules associate, each remains separate and i = 1. If dimerization is complete, two molecules become one particle and i = 0.5. Partial dimerization gives an intermediate effective particle number, so the van’t Hoff factor lies between 0.5 and 1. It cannot exceed 1 for association alone.

Correct answer: A. Total particles formed from one formula unit

Explanation: In the dissociation equation, α is the fraction of solute units that dissociate, while n is the total number of particles produced from one dissociating unit. For NaCl, n = 2; for CaCl₂, n = 3. Substituting this value gives the increase in particle number at a chosen degree of dissociation. It is not a temperature, pressure, or solvent quantity.

Correct answer: C. 1.8

Explanation: The governing concept is the van’t Hoff factor for partial electrolyte dissociation. NaCl gives two ions, Na+ and Cl−, on complete dissociation, so n = 2. For degree of dissociation α, i = 1 + α(n − 1). Substitution gives i = 1 + 0.8(2 − 1) = 1.8. The limiting values are 1 for no dissociation and 2 for complete dissociation, so option C is correct.

Correct answer: B. 2.0

Explanation: CaCl2 dissociates as CaCl2 → Ca2+ + 2Cl−, producing n = 3 particles when complete. For partial dissociation, the governing equation is i = 1 + α(n − 1). With α = 0.5, i = 1 + 0.5(3 − 1) = 2.0. The value 1.5 uses an incomplete particle count, while 3.0 applies only to complete dissociation. Therefore, option B is correct.

Correct answer: C. 0.6

Explanation: AB dissociates into two ions, so n = 2. The dissociation relation becomes i = 1 + α(2 − 1) = 1 + α. With i = 1.6, α = 1.6 − 1 = 0.6. Thus 60% of the AB units are dissociated under the stated conditions. The value 1.6 is the particle factor, not the degree of dissociation.

Correct answer: A. Number of molecules combining to form one larger particle

Explanation: In association, n original solute molecules combine to form one associated particle. Thus n describes the size of the associated unit: n = 2 for a dimer and n = 3 for a trimer. The term (1 − 1/n) measures the fractional decrease in particle number on complete association. Temperature, pressure, and solution volume are not represented by n in this equation.

Correct answer: C. 0.75

Explanation: Dimerization is association in which two solute particles combine to form one particle. If α is the fraction dimerized, the particle factor is i = 1 − α(1 − 1/2) = 1 − α/2. For α = 50% = 0.5, i = 1 − 0.5/2 = 0.75. Since association lowers the number of particles, i must be below 1; hence option C is correct.

Correct answer: B. 0.4

Explanation: For dimer formation, two particles become one, so the van’t Hoff factor follows i = 1 − α/2. Using the given i value, 0.8 = 1 − α/2. Therefore α/2 = 0.2 and α = 0.4, meaning 40% association. The values 0.2 and 0.6 arise from incomplete rearrangement, while 0.8 incorrectly treats i as α. Thus, option B is correct.

Correct answer: A. Dissociation of the solute

Explanation: Freezing-point depression is a colligative property, so it increases with the number of dissolved particles. If the observed depression is greater than the value calculated for undissociated solute, more particles must be present than expected. Dissociation produces additional ions and gives i > 1, increasing ΔTf. Association would reduce the number of particles and make the depression smaller.

Correct answer: B. Partial association

Explanation: Elevation in boiling point depends on the number of dissolved particles. A value smaller than expected means that fewer effective particles are present than assumed in the normal calculation, so i < 1. Association combines solute molecules into larger units and reduces particle number. Partial association is therefore possible; complete dissociation or particle doubling would instead increase the boiling-point elevation.

Correct answer: A. π = iCRT

Explanation: For a dilute solution, the ideal osmotic-pressure equation is π = CRT. If association or dissociation changes the effective number of solute particles, the van’t Hoff factor corrects the equation by multiplication: π = iCRT. It is not placed in the denominator or added, because particle concentration directly scales the pressure.

Correct answer: B. ΔTf = iKf m

Explanation: For an ideal non-associating, non-dissociating solute, ΔTf = Kf m. When the number of particles is abnormal, the ideal expression is multiplied by the van’t Hoff factor, giving ΔTf = iKf m. Association makes i less than 1, while dissociation makes i greater than 1, changing the magnitude accordingly.

Correct answer: A. Complete dimer association

Explanation: If the observed molar mass is about twice the normal value, i = normal/observed is about 1/2. Complete dimerization gives i = 0.5 because two acid molecules behave as one particle. Carboxylic acids such as benzoic acid commonly form hydrogen-bonded dimers in non-polar benzene, explaining the observation.

Correct answer: B. 80 g mol−1

Explanation: Rearrange i = normal molar mass/observed molar mass to get observed molar mass = normal molar mass/i. Therefore, Mobserved = 120/1.5 = 80 g mol−1. Since i is greater than 1, the result is appropriately lower than the normal molar mass, indicating an increase in particles such as by dissociation.

Correct answer: C. 120 g mol−1

Explanation: Use Mobserved = Mnormal/i. Substitution gives Mobserved = 90/0.75 = 120 g mol−1. Dividing by a number below 1 increases the result, which agrees with the general rule that i less than 1 produces a higher observed molar mass. The lower i indicates association of solute particles.

Correct answer: A. Dissociation is incomplete

Explanation: For an electrolyte producing n ions, ideal complete dissociation gives i = n. If the measured i is lower than n but generally above 1, only a fraction of the formula units has dissociated, so dissociation is incomplete. The lower value does not by itself prove complete association or absence of dissolution.

Correct answer: B. An abnormal or incorrect molar mass may be obtained

Explanation: Colligative properties depend on the actual number of dissolved particles. If a solute associates or dissociates and the calculation assumes i = 1, the measured effect is interpreted incorrectly. The resulting molar mass may be too high for association or too low for dissociation, so ignoring i can produce an abnormal value.

Correct answer: B. The solute is partially dissociated

Explanation: For AB2, no dissociation gives one effective particle and i = 1. Complete ideal dissociation gives three ions and i = 3. Therefore, a value strictly between 1 and 3 indicates that some, but not all, formula units have dissociated. This is partial dissociation; it is not complete association or absence of ion formation.

Correct answer: B. Dimer association

Explanation: The relation is i = normal molar mass / observed molar mass. Here i = 100/200 = 0.5. For complete dimerization, two original molecules combine to form one associated particle, giving i = 1/2. Thus the doubled observed molar mass is consistent with dimer association, whereas dissociation would make the observed mass smaller.

Correct answer: C. 3

Explanation: For abnormal molar mass, the relation is i = normal molar mass / observed molar mass. Let the normal molar mass be M. The observed value is M/3, so i = M ÷ (M/3) = 3. A value greater than one means that the effective number of solute particles has increased, as happens during dissociation.

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

Explanation: When a colligative property is used to calculate molar mass without considering association or dissociation, the result is the observed molar mass. The correction factor is i = Mnormal/Mobserved. Thus i greater than one corresponds to a smaller observed mass, as in dissociation, while i below one corresponds to a larger observed mass, as in association.

Correct answer: A. Association due to hydrogen bonding

Explanation: Acetic acid molecules can form hydrogen-bonded dimers in the relatively non-polar solvent benzene. Two molecules then behave as one associated particle, reducing the number of particles in solution. The smaller colligative effect leads to a molar mass that appears higher than the normal value. Ionization and salt formation are not the relevant processes here.

Correct answer: A. Between 1 and 2

Explanation: NaCl gives two ions, Na⁺ and Cl⁻, when dissociation is complete, so the limiting value is i = 2. If dissociation is incomplete, some NaCl remains as undissociated units while the rest forms ions. The particle number therefore increases above one but does not reach two, giving 1 < i < 2. It is not less than one because association is not being described.