7: Abnormal Molecular Mass

For dissociation of one solute unit into n particles, the van’t Hoff factor is i = 1 + (n − 1)α, where α is the degree of dissociation. If the solute dissociates into two ions, n = 2, so i = 1 + (2 − 1)α = 1 + α. This formula gives values between 1 and 2 as dissociation increases from zero to complete. Association would instead reduce i below 1 and requires a different relation.

For dissociation into two ions, the van’t Hoff relation is i = 1 + α, because each dissociated unit changes from one particle to two. Fifty percent dissociation means α = 50/100 = 0.5. Substituting gives i = 1 + 0.5 = 1.5. The value 1 would mean no dissociation, while 2 would represent complete dissociation into two ions. Thus 1.5 is the only consistent answer.

For a solute that dissociates into two ions, i = 1 + α. Rearranging gives α = i − 1. Substituting i = 1.2 gives α = 1.2 − 1 = 0.2. Converting this fraction to a percentage, 0.2 × 100 = 20%. Therefore, the degree of dissociation is 0.2 or 20%. A value such as 1.2 or 120% is impossible here because α cannot exceed 1 or 100%.

The relation i = Mactual/Mobserved shows that an observed molar mass smaller than the actual value gives i greater than 1. This means that the effective number of particles has increased. Dissociation produces additional particles, such as ions, and is therefore the most likely process. Association would instead make the observed molar mass larger.

Using i = Mactual/Mobserved, if the denominator is greater than the numerator, the ratio must be less than one. Therefore i < 1. This commonly indicates association, because joining solute particles reduces the effective particle number. The factor is not necessarily exactly 0.5 or zero; its value depends on the extent and type of association.

Carboxylic acid molecules can form hydrogen-bonded dimers in benzene and some other nonpolar solvents. Two molecules then act as one effective solute particle, reducing particle number and producing i < 1. Salt splitting in water is dissociation and would increase particle number instead.

An ionic salt such as KCl can separate into hydrated ions in water. This increases the number of effective particles, so colligative measurements may give a lower observed molar mass than the formula-unit molar mass. Acid dimerisation in benzene is association and gives the opposite particle-number effect.

Start with i = Mnormal/Mobserved. Rearranging gives Mnormal = i × Mobserved. Substituting the values gives Mnormal = 2 × 40 = 80 g mol⁻¹. The answer is not 20 because that would divide by i in the wrong direction. Since i is greater than one, the normal molar mass is greater than the observed value.

For abnormal molar mass, the van’t Hoff factor is calculated from i = actual or normal molar mass / observed molar mass. Substituting the given values gives i = 90 / 45 = 2. Hence the solute produces twice the effective particle number compared with the normal, non-associated reference. This is consistent with dissociation, although the numerical value alone does not identify every possible cause. The answer is dimensionless because it is a ratio.

Use i = Mactual/Mobserved. Therefore i = 60/120 = 0.5. Because the observed molar mass is greater than the actual value, the factor is below one, indicating fewer effective particles than expected. This is commonly associated with association of solute molecules rather than dissociation.

Complete dissociation into three ions gives i = 3 because one formula unit produces three particles. Since i = Mactual/Mobserved, rearranging gives Mobserved = Mactual/i = Mactual/3. Therefore the observed molar mass is one third of the actual molar mass. The three-times option reverses the effect.

Suppose four original solute molecules combine completely to form one associated particle. Before association there are four particles; after association there is one particle. The effective particle ratio is therefore i = final particles / original particles = 1/4 = 0.25. The value 4 would describe an increase rather than a decrease. A value of 1 would indicate no association, and 2 would not match the stated four-to-one change.

The van’t Hoff factor is i = normal molar mass ÷ observed molar mass = 50/25 = 2. Thus the solute produces twice as many effective particles as expected. An increase in particle number indicates dissociation, in which one species separates into two or more species. Association would give i < 1, so option A is correct.

Calculate i = normal molar mass ÷ observed molar mass = 50/100 = 0.5. A value below one means that the solution contains fewer independent particles than expected. This commonly occurs when solute molecules combine to form larger associated species. Dissociation and ionisation increase particle number and generally give i > 1, so A is correct.

The correct relation is i = Mnormal/Mobserved. It follows because colligative-property equations use the effective number of particles, while the observed molar mass reflects that change. If particles increase by dissociation, the observed molar mass decreases and i becomes greater than one. Reversing the ratio would give the reciprocal and misidentify the behaviour.

In benzene, acetic acid molecules can associate through hydrogen bonding and form dimers, commonly represented as two acetic-acid molecules joined together. This reduces the number of independent solute particles, lowers the colligative effect and makes the observed molar mass higher than the normal value. No ions are required for this explanation.

In complete dimerisation, every two original solute molecules combine to form one dimer particle. Thus the number of independent particles becomes half of the number expected without association. The van’t Hoff factor is therefore i = 1/2 = 0.5. A value of one would represent no net particle-number change.

Use i = normal molecular mass ÷ observed molecular mass. With i = 0.5, 0.5 = normal mass ÷ observed mass, so observed mass = normal mass ÷ 0.5 = 2 × normal mass. The larger observed value is consistent with association, which reduces the number of independent particles.

Since i = normal molecular mass/observed molecular mass, putting i = 2 gives 2 = Mnormal/Mobserved. Rearranging, Mobserved = Mnormal/2. Therefore, the observed molecular mass is half the normal value. A doubled particle number commonly produces this result through dissociation; it does not mean that the molecular mass becomes double.

Let one mole of AB be considered. If the degree of dissociation is α, then (1 − α) mole remains as undissociated AB, while α mole produces 2α moles of ions, A⁺ and B⁻. The total particle amount is therefore (1 − α) + 2α = 1 + α. Since i is the ratio of actual particles to the original amount, i = 1 + α. This also gives i = 2 for complete dissociation.

NaCl dissociates according to NaCl → Na⁺ + Cl⁻, so it is an AB-type electrolyte and follows i = 1 + α. Using the given value, 1.6 = 1 + α, hence α = 1.6 − 1 = 0.6. Thus the degree of dissociation is 0.6, or 60% if expressed as a percentage. The value 1.6 is i itself, not α.

For a solute that can dissociate into two ions, the limiting value of i is 1 when no dissociation occurs and 2 when dissociation is complete. Therefore, a value strictly between 1 and 2 means that only some solute units have dissociated. This is partial dissociation; complete dissociation would give i = 2, while association would generally make i less than 1.

For dimerization, two original solute particles combine to form one particle. If dimerization is complete, the particle ratio becomes 1/2, so i = 0.5. If no association occurs, i = 1. A value between 0.5 and 1 therefore shows that only part of the solute has formed dimers, which is partial association.

Benzoic acid molecules can form hydrogen-bonded dimers in the relatively non-polar solvent benzene. Two molecules then behave as one effective solute particle, reducing the particle number and making i less than 1. Consequently, the colligative effect is smaller and the observed molar mass appears nearly doubled.

Use the molar-mass relation i = normal molar mass/observed molar mass. Substitution gives i = 60/30 = 2. Thus the effective particle number is twice the expected number, which is consistent with dissociation into two particles under ideal conditions. The unit cancels in the ratio.