7: Abnormal Molecular Mass

Dimerisation is an association process in which two separate solute molecules combine to form one dimer particle. For example, two monomer units become one effective unit, so the number of independent particles is lower than before association. This reduces colligative effects and generally gives i < 1. It cannot double or become infinite; therefore option A is correct.

The depression in freezing point is proportional to the effective number of solute particles: ΔTf = iKf m. Dissociation produces more particles than the original solute units, so i becomes greater than one and the depression is larger than the normal value. Association would lower the effect because it reduces particles. Thus option A follows directly from the colligative-property relation.

Boiling-point elevation follows ΔTb = iKb m and therefore depends on the effective particle count. Association joins solute molecules, decreases the number of independent particles, and gives i < 1. At the same analytical concentration, the observed elevation is then smaller than the ideal or normal value. Dissociation and ionisation generally increase the elevation, so option A is correct.

For abnormal molar mass, the van’t Hoff factor is related to molecular masses by i = normal molar mass/observed molar mass. Substituting the given values gives i = 60/120 = 0.5. A value below one indicates association: several solute units behave as fewer particles in solution, which makes the observed molar mass larger than the normal value. Therefore 2 would be the inverse ratio and is not correct.

The van’t Hoff factor for abnormal molar mass is calculated as i = normal molar mass/observed molar mass. Hence i = 58.5/29.25 = 2. The value is greater than one because the solute produces more particles than expected, as happens during dissociation. Option B is the reciprocal and would correspond to interchanging the two masses; i = 1 would indicate no abnormality, while 4 is not supported by the given ratio.

An i value of one means that the observed colligative property equals the value predicted from the original number of solute units. Under the usual ideal treatment, the solute has neither appreciable association nor dissociation, so the effective particle count is unchanged. Complete association would make i smaller, and complete dissociation would make it larger. Thus A is correct.

NaCl dissociates into two particles, Na⁺ and Cl⁻, so n = 2. The degree of dissociation is 50% = 0.50. Using i = 1 + (n − 1)α gives i = 1 + (2 − 1)(0.50) = 1.50. This lies midway between 1, for no dissociation, and 2, for complete dissociation. Therefore option A is correct; 2.5 exceeds the complete-dissociation limit.

For dissociation into two particles, n = 2. Substituting this in i = 1 + (n − 1)α gives i = 1 + α. With i = 1.8, we obtain 1.8 = 1 + α, so α = 0.8. Thus the degree of dissociation is 0.8, or 80% if expressed as a percentage. The value 1.8 is the particle factor i, not the dissociation fraction itself.

The molecular-mass relation is i = Mnormal/Mobserved. Rearranging gives Mobserved = Mnormal/i. For i = 0.5, Mobserved = Mnormal/0.5 = 2Mnormal. The lower factor indicates fewer particles, as in association, so the ordinary colligative calculation reports a larger apparent mass. Therefore the observed molecular mass is double, making A correct.

Using i = Mnormal/Mobserved, rearrange to Mobserved = Mnormal/i. When i = 2, Mobserved = Mnormal/2, so the observed molecular mass is half the normal value. The factor above one indicates an increased number of effective particles, commonly due to dissociation. Thus option A is correct, while doubling would reverse the required ratio.

NaCl is an electrolyte and can dissociate in water into Na⁺ and Cl⁻. This increases the effective particle count above that predicted for one NaCl unit and can make the molecular mass obtained from a colligative property abnormal. Sugar, urea, and glucose are generally nonelectrolytes in water and remain as molecules under ordinary treatment, so they normally give i close to one. A is correct.

Glucose is treated as a nonelectrolyte in aqueous solution. Its molecules dissolve but do not normally dissociate into ions, so one glucose molecule contributes approximately one solute particle. Consequently the actual particle count is close to the ideal count and i ≈ 1. It does not form three ions or convert water into vapour; therefore option A is correct.

At fixed concentration and temperature, osmotic pressure is proportional to the number of independent solute particles. A lower-than-normal value means that fewer effective particles are present, corresponding to i < 1. Association combines solute molecules and produces this reduction. Dissociation, ionisation, or breaking a salt into ions generally increases particle number and raises pressure, so A is correct.

Colligative properties are determined primarily by the number of dissolved particles, not by their chemical identity, colour, smell, or the shape of the container. The four standard examples are relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure. Association and dissociation matter because they change particle number. Hence option A states the governing principle.

The freezing-point depression relation is ΔTf = iKf m. Dissociation converts one solute unit into more independent particles, making i greater than one. At the same molality, the product iKf m is therefore larger than the normal value calculated with i = 1. The depression increases, not decreases or vanishes. Thus option A is correct, provided the solution remains within the assumed dilute-solution conditions.

For freezing-point depression, ΔTf = iKf m. Association joins separate solute molecules, reducing the number of independent particles and making i less than one. Consequently the observed depression is smaller than the value predicted for the same concentration of non-associating solute. The exact decrease depends on the extent of association; it is not necessarily exactly half or double. Therefore A is correct.

The relation is i = Mnormal/Mobserved. If the observed molecular mass is smaller than the normal molecular mass, the numerator is larger than the denominator, so the ratio must exceed one. This corresponds to an increased effective particle count, usually caused by dissociation. A van’t Hoff factor cannot be negative in this context, and zero would not represent a dissolved solute. Hence A is correct.

From i = Mnormal/Mobserved, a larger observed molecular mass makes the denominator larger than the numerator. Therefore i is less than one. This means fewer effective particles are present than expected, as occurs when molecules associate into dimers or larger groups. The exact value is not always 2 or 3; it depends on the extent of association. Thus option A is correct.

Acetic acid molecules can form hydrogen-bonded dimers. In a non-polar solvent such as benzene, the solvent does not strongly separate or solvate the polar acid molecules, so intermolecular association can be appreciable. The resulting dimers reduce the effective particle count and produce a higher apparent molecular mass. Association is not restricted to metallic solvents or every solid, so A is correct.

Electrolytes can dissociate in solution into ions or other charged species. This changes the number of independent particles relative to the original formula units, altering colligative properties such as osmotic pressure or freezing-point depression. If the ideal formula is used without the van’t Hoff correction, the calculated molecular mass becomes abnormal. Colour, solvent freezing, and glass reaction are not the general explanation, so A is correct.

A nonelectrolyte dissolves without producing ions, and the stated absence of association means its molecules also remain separate. Thus each original solute molecule contributes one effective particle, so the actual-to-ideal particle ratio is i = 1. Values above one would indicate particle production through dissociation, while values below one would indicate association. Hence option A is correct.

Complete dissociation of aluminium chloride is AlCl₃ → Al³⁺ + 3Cl⁻. One formula unit therefore produces one aluminium ion and three chloride ions, giving four independent particles. The ideal van’t Hoff factor is consequently i = 4. The ionic charges do not add extra particles; the coefficients in the dissociation equation provide the count. Thus option A is correct.

The complete dissociation equation is MgCl₂ → Mg²⁺ + 2Cl⁻. It produces one magnesium ion and two chloride ions, so each formula unit gives three independent particles. Under the ideal complete-dissociation assumption, i equals this particle count and is therefore 3. The 2+ charge does not mean two magnesium particles; it is still one ion. Hence option A is correct.

The van’t Hoff factor compares the effective number of solute particles with the number expected from undissociated formula units. One formula unit produces two ions, A⁺ and B⁻, on complete dissociation. Therefore, the particle ratio is 2 and i = 2. A value of 1 would mean no change in particle number.

The van’t Hoff factor compares the effective number of solute particles with the number expected without association or dissociation. For an unchanged solute, i = 1. A value greater than 1 means that the particle count has increased, which generally occurs when molecules dissociate into smaller particles or ions. Since i = 1.25, the increase is incomplete, so partial dissociation is the best choice; complete association would give i less than 1.