7: Abnormal Molecular Mass

Mass percentage is the solute mass divided by the total solution mass, multiplied by 100. With a solute-to-solvent ratio of 2:3, the total mass is 2 + 3 = 5 parts. Thus, solute mass percentage = (2 ÷ 5) × 100 = 40%. Option B is the solvent percentage. Option D comes from comparing solute only with solvent, not with the complete solution.

Dissociation means that one solute unit breaks into two or more particles in solution. The number of particles therefore becomes greater than the number expected from the undissociated formula. Since colligative properties depend on particle number, the calculated observed molar mass becomes lower than the normal molar mass. Thus, option A is correct.

The governing principle is that the ideal van’t Hoff factor equals the number of independent particles formed by one formula unit after complete dissociation. Barium chloride dissociates as BaCl₂ → Ba²⁺ + 2Cl⁻. One barium ion and two chloride ions are formed, so the total is 1 + 2 = 3 particles. Therefore i = 3. The answer is not 2, because both chloride ions must be counted separately.

On complete dissociation, one formula unit of A₃B separates into three A-type ions and one B-type ion. Hence, one original particle becomes four particles in solution. The van’t Hoff factor is the ratio of final effective particles to the original particles, so i = 4/1 = 4. Option B would count only the three A ions and omit the B ion, while i = 1 represents no change in particle number.

Sugar is a molecular, non-electrolyte solute. When it dissolves in water under ordinary dilute-solution conditions, its molecules remain essentially as individual molecules rather than splitting into ions or combining into larger groups. Thus, the number of solute particles remains nearly unchanged and i is approximately 1. It does not produce three ions, always form dimers, or change water into a gas.

Abnormal molar mass arises because association or dissociation changes the effective number of solute particles. Therefore, first determine whether the particle number has increased or decreased. An increase generally indicates dissociation and gives i greater than 1; a decrease indicates association and gives i less than 1. The other properties are irrelevant.

Complete dissociation of SrCl₂ is represented as SrCl₂ → Sr²⁺ + 2Cl⁻. One formula unit therefore produces one strontium ion and two chloride ions, giving three solute particles in total. Under the ideal complete-dissociation assumption, the van’t Hoff factor equals this number, so i = 3.

NH₄Cl is an ionic electrolyte and, on complete dissociation in water, gives NH₄⁺ and Cl⁻. These are two separate solute particles, so the effective particle number becomes twice the original formula-unit number. Therefore, the ideal van’t Hoff factor is i = 2. The polyatomic ion NH₄⁺ is counted as one particle; its four hydrogen atoms are not counted separately. Hence 3 and 4 are incorrect.

For a dilute solution, osmotic pressure is given by π = iCRT. At fixed concentration and temperature, the normal value corresponds to i = 1, so π_observed/π_normal = i. Since the observed pressure is 1.8 times the normal pressure, i = 1.8. This indicates an increase in effective particles, commonly due to dissociation.

Boiling-point elevation follows ΔT_b = iK_bm and depends on the effective number of solute particles. If the observed elevation is lower than the normal value, i is less than one. Association combines separate molecules into fewer particles, reducing the colligative effect. Dissociation or ionisation would increase i and therefore increase the elevation instead.

For a dilute solution, relative lowering of vapour pressure is proportional to the effective mole fraction, and therefore to the number of solute particles at fixed composition. A lower-than-expected lowering means fewer effective particles are present. Association, such as dimer formation, reduces particle number. Complete ionisation would increase the lowering rather than reduce it.

Because M_app = M_true/i, an apparent mass smaller than the true mass means i > 1. Therefore the effective number of solute particles increased relative to the original formula units. For a salt, this commonly results from dissociation into ions, although the observation alone establishes increased particle number rather than a particular ionisation mechanism.

The van’t Hoff factor compares the actual effective particle number with the number expected without association or dissociation. Since i = 0.70 is less than one, only 70% of the normal effective particle count is present. This indicates association or another particle-reducing process. It does not mean the colligative effect is zero; it is merely reduced.

A van’t Hoff factor of one means the effective number of solute particles is unchanged from the normal molecular state. Consequently, the colligative property has its normal value and M_app = M_true. This can occur for a non-electrolyte or when opposing effects happen to cancel, so the precise safe conclusion is no net effective association or dissociation.

The van’t Hoff factor compares the actual effective number of solute particles with the number expected without association or dissociation. Since i = 0.90 is slightly below 1, the particle number has decreased slightly. Association combines particles, whereas dissociation would make i greater than 1. Therefore, slight association is the best explanation.

For a solute that forms three ions on complete dissociation, the van’t Hoff relation is i = 1 + α(3 − 1) = 1 + 2α. Substituting i = 1.2 gives 1.2 = 1 + 2α, so 2α = 0.2 and α = 0.10. Expressed as a percentage, the degree of dissociation is 10%, so option A is correct. The small increase of i above one indicates limited dissociation.

The governing relation is M_app = M_true/i, because the van’t Hoff factor changes the colligative effect used to determine molar mass. Since i = 2.25 = 9/4, M_app/M_true = 1/2.25 = 4/9. Thus the required ratio is 4:9, so option A is correct. The AB3 information is unnecessary once i is supplied; 9:4 reverses the ratio, while 1:1 would imply i = 1.

The governing idea is partial dissociation and the van’t Hoff factor. One formula unit of AB2 produces three ions when completely dissociated, so n = 3. For dissociation fraction α, i = 1 + α(n − 1). Substitution gives i = 1 + 0.55(3 − 1) = 1 + 1.10 = 2.10. Thus option B is correct. A value of 1.55 would incorrectly use only α, while 3.00 represents complete dissociation.

In complete trimer association, every three original solute particles become one aggregate, so the effective particle number is one-third of the original number. Hence i = 1/3. Using M_app = M/i gives M_app = 3M, and therefore M_app : M = 3 : 1. Complete association increases apparent molar mass.

Using M_app = M/i, i = 150/300 = 0.50. Complete dimer association changes two original particles into one group, giving i = 1/2 = 0.50. Thus the result matches complete dimerisation. Complete dissociation would make i greater than one, normal behaviour would give i = 1, and complete trimerisation would give i = 1/3.

The apparent molar mass is M_app = M/i. Substituting i = 2.8 gives M_app = M/2.8 = 0.357M, which rounds to 0.36M. Since dissociation increases the effective number of particles, the apparent molar mass must be smaller than the true value; this rules out options above M.

Abnormal molar mass is related to the van’t Hoff factor by Mapp = M/i. A molar mass 60% greater than the true value means Mapp = 1.60M. Substituting gives 1.60M = M/i, so i = 1/1.60 = 0.625. Therefore option A is correct. The value below one is consistent with association, which reduces the number of particles; 1.60 is only the multiplier for the apparent mass, not i.

The van’t Hoff factor converts formal concentration, which counts formula units, into effective particle concentration, which includes particles created by dissociation. The governing relation is Ceff = iC. Substituting the values gives Ceff = 1.72 × 0.25 = 0.430 M. Therefore option C is correct. The formal value 0.250 M ignores the additional ions, while 1.720 M incorrectly treats i as a concentration.

Use M_app = M/i. Substitution gives M_app = 144/0.72 = 200 g mol⁻¹. Since i is less than one, the effective particle number is reduced by association and the apparent molar mass must be greater than the true mass. This also eliminates the smaller numerical options.

For a fixed solution, the van’t Hoff factor equals the observed colligative effect divided by the normal effect. Hence i = 0.60/0.24 = 2.50. A value greater than one indicates an increased number of particles, such as from dissociation. The given molality is consistent with the stated normal value but is not needed again. Thus C is correct.