7: Abnormal Molecular Mass

Molality is defined as the number of moles of solute divided by the mass of solvent in kilograms. Laboratory data, however, are often recorded using grams of solvent. Since 1 kg equals 1000 g, conversion from grams to kilograms introduces the factor 1000 in formulas such as m = 1000w₂/(M₂w₁). The factor is therefore a unit-conversion factor, not a temperature, pressure, or fixed-solute-mass factor.

Association occurs when two or more solute molecules combine to form one larger associated particle. Consequently, the number of solute particles becomes smaller than expected. Colligative properties then indicate fewer particles, so the calculated or observed molar mass becomes greater than the normal molar mass. Therefore, option A is correct.

For abnormal colligative behaviour, the van’t Hoff factor is related to molar mass by i = Mnormal / Mobserved, or equivalently Mobserved = Mnormal / i. If dissociation occurs, i is greater than 1 and the observed molar mass is smaller. If association occurs, i is less than 1 and the observed molar mass is larger. Hence, option A is correct.

The relation between the observed and normal molar masses is Mobserved = Mnormal / i. Substituting i = 2 gives Mobserved = Mnormal / 2, so the observed molar mass is one-half of the normal molar mass. This situation is associated with an increase in particle number, commonly due to dissociation. Therefore, option A is correct.

Use the relation Mobserved = Mnormal / i. When i = 0.5, Mobserved = Mnormal / 0.5 = 2Mnormal. Thus, the observed molar mass is double the normal molar mass. A value of i less than 1 indicates association, because association decreases the number of particles in solution and makes the calculated molar mass appear larger. Hence, option C is correct.

The relation between normal and observed molar masses is i = Mnormal/Mobserved. Substituting the given values gives i = 120/60 = 2. A value greater than one indicates that the number of solute particles has increased, usually because the solute dissociates into two or more particles in solution. Therefore, option C is correct.

Use the relation i = Mnormal/Mobserved for abnormal molar mass. Here, i = 60/120 = 0.5. A van’t Hoff factor below one means that the effective number of solute particles is lower than expected, generally because solute molecules associate, such as dimerisation. Therefore, the correct answer is 0.5.

Abnormal molar mass is observed when the solute particles dissociate into more particles or associate to form fewer particles in solution. These changes alter the van’t Hoff factor from unity and therefore change the colligative effect used for calculation. If the solute neither dissociates nor associates, the number of particles remains unchanged, i = 1, so the experimentally observed molar mass equals the normal molar mass. The colour of a solution is irrelevant.

The freezing-point depression relation is ΔT_f = iK_fm, where i is the van’t Hoff factor, K_f is the cryoscopic constant, and m is molality. Rearranging gives i = ΔT_f/(K_fm). Substitution gives i = 0.186/(1.86 × 0.1) = 0.186/0.186 = 1. Thus the solute shows normal, non-dissociating behaviour under the stated conditions, so option B is correct.

For abnormal molar mass determined from a colligative property, the van’t Hoff factor is related by i = Mnormal/Mobserved. Rearranging gives Mobserved = Mnormal/i. Substituting the given values, Mobserved = 100/2 = 50 g mol⁻¹. Since i is greater than 1, dissociation has increased the number of particles and made the observed molar mass smaller. Thus, B is correct.

The van’t Hoff factor is related to normal and observed molar masses by i = normal molar mass ÷ observed molar mass. Let the normal molar mass be M. The observed value is M/2, so i = M ÷ (M/2) = 2. A value greater than one indicates an increase in the number of effective particles, commonly caused by dissociation. Therefore, option C is correct.

The van’t Hoff factor is calculated from i = normal molar mass ÷ observed molar mass. If the normal molar mass is M and the observed molar mass is 2M, then i = M ÷ 2M = 0.5. A value below one means that the effective number of solute particles has decreased, usually because solute molecules associate. Thus the correct answer is option A.

Molality is defined as moles of solute divided by kilograms of solvent: \(m=n/\text{kg solvent}\). First convert the solvent mass: \(250\,\mathrm{g}=0.250\,\mathrm{kg}\). Hence, \(n=m\times\text{kg solvent}=0.5\times0.250=0.125\,\mathrm{mol}\). The mass of solvent, not the total mass of solution, is used in molality. Therefore, option B is correct.

For abnormal colligative behaviour, i = Mnormal/Mapparent, so Mapparent = Mnormal/i. Substitution gives Mapparent = 120/1.5 = 80 g mol−1. Because i is greater than one, the solution has more effective particles than expected, as in dissociation, and the apparent molar mass is consequently lower. Thus option A is correct.

Molecular mass obtained from a colligative property is based on the expected number of solute particles. Association joins particles and lowers their effective number, whereas dissociation separates a species into more particles. Either change makes the measured colligative effect differ from the ideal prediction, so the calculated mass becomes abnormal. Colour, vessel shape, and zero pressure are not the governing causes.

Dissociation increases the number of independent particles. Since a colligative effect is proportional to particle number, the observed effect becomes larger than expected for one undissociated molecule. When the ordinary formula is used without correction, this larger effect is interpreted as a larger amount of solute and therefore gives a smaller apparent molecular mass. Hence A is correct.

Association combines two or more solute molecules into fewer independent particles. The colligative effect therefore becomes smaller than the value expected from the original molecular formula. Using an ordinary formula to interpret that smaller effect makes the calculated or observed molecular mass larger than the normal molecular mass. The exact increase depends on the extent and type of association, so only option A is generally valid.

The van’t Hoff factor accounts for the actual number of particles produced or lost in solution through dissociation or association. Colligative equations are corrected by multiplying the ideal particle-dependent term by i. Thus i connects the observed property with abnormal molecular mass and particle count. It has no role in correcting colour, forcing density to zero, or converting a solvent into a gas.

For an unchanged solute formula, i = 1 represents the expected particle count. If i is greater than one, more particles are present than one formula unit would provide. Dissociation into ions or smaller species causes this increase; for example, one salt unit may produce two or more ions. Association usually gives i < 1, so A is correct.

An i value below one means that the solution contains fewer independent particles than predicted from the original solute formula. Association, such as dimer formation, combines molecules and produces this reduction. Dissociation and ionisation generally increase particle number and give i greater than one. Therefore option A is the appropriate interpretation, although the extent of association must be judged from the actual value.

For a given concentration and temperature, the ideal colligative value assumes the solute remains in its expected particle form. The van’t Hoff factor is i = observed colligative property/ideal or normal colligative property. It therefore measures the correction caused by the actual particle count. Colour ratios, sums of temperature and pressure, and mass–volume differences do not define i.

Complete dissociation of one NaCl formula unit is represented as NaCl → Na⁺ + Cl⁻. One initial unit therefore produces two independent solute particles. Under the ideal complete-dissociation assumption, the van’t Hoff factor equals the number of resulting particles, so i = 2. Partial dissociation would give a value between 1 and 2, but it is not the condition stated here.

The complete dissociation equation is CaCl₂ → Ca²⁺ + 2Cl⁻. Counting the ions gives one calcium ion plus two chloride ions, or three independent particles for each formula unit. Therefore the ideal van’t Hoff factor is i = 3. The charge on an ion does not add to the particle count; it is the number of separate species that matters.

Complete dissociation means that every formula unit of K₂SO₄ separates into its constituent ions: K₂SO₄ → 2K⁺ + SO₄²⁻. Thus one dissolved formula unit produces three solute particles in total. The van’t Hoff factor i is the ratio of the actual number of particles to the number expected without dissociation, so for complete dissociation its ideal value is 3. It is not 2, because two potassium ions and one sulfate ion must all be counted.

In a relatively non-polar solvent such as benzene, acetic acid molecules can associate through hydrogen bonding to form dimers, often represented as (CH₃COOH)₂. Two original molecules then behave as one effective particle, reducing the colligative effect. An ordinary calculation interprets this smaller effect as a larger molecular mass. Therefore dimer formation, option A, explains the observation.