7: Abnormal Molecular Mass

Correct answer: A. Lower

Explanation: For complete dissociation, every formula unit breaks into the maximum possible number of particles. In partial dissociation, only some units break, so the effective particle increase is smaller. Consequently, i is lower than the complete-dissociation value, although it can still be greater than one for a dissociating solute.

Correct answer: A. Greater than for complete association

Explanation: Complete association produces the greatest reduction in the number of independent solute particles for the stated association pattern. Partial association leaves some molecules unassociated, so the particle count is higher than in the complete case. Therefore its i value is greater than the complete-association value, though it generally remains below one.

Correct answer: A. Dissociation and i > 1

Explanation: The van’t Hoff factor i compares the actual number of solute particles with the number expected without association or dissociation. During dissociation, one solute unit breaks into two or more particles, so the particle count increases and i becomes greater than 1. Association combines particles and generally gives i < 1. Therefore, dissociation with i > 1 is the only correct matching; the other pairs reverse or misstate the effect.

Correct answer: A. It is explained by a change in particle number

Explanation: Molar mass is often calculated from a colligative property, and colligative properties depend on the number of dissolved particles. Dissociation increases that number, while association decreases it; either process can make the calculated molar mass abnormal. Colour, physical container and solid state are not the governing explanation.

Correct answer: A. Change in the number of solute particles in solution

Explanation: Molecular mass may be inferred from colligative properties such as osmotic pressure or boiling-point elevation. These properties depend on the number of solute particles, not simply on the chemical formula written before dissolution. Association and dissociation alter that number and therefore produce an abnormal calculated value.

Correct answer: A. Less than normal

Explanation: Dissociation produces more independent particles from the original solute units. The resulting colligative effect is larger than expected for an undissociated solute. Since i = normal molar mass ÷ observed molar mass and i is greater than one, the observed molar mass must be less than the normal value.

Correct answer: A. Greater than normal

Explanation: Association combines two or more solute molecules into fewer independent particles. This lowers the colligative effect at a given analytical concentration. Therefore i becomes less than one, and from i = normal molar mass ÷ observed molar mass, the observed molecular mass is greater than the normal molecular mass.

Correct answer: A. When the solute dissociates

Explanation: The van’t Hoff factor compares the actual effective number of particles with the number expected from the undissociated solute. Dissociation creates additional particles, so this ratio becomes greater than one. Dimer formation and other association processes reduce particle number and instead tend to make i less than one.

Correct answer: A. Association

Explanation: When i < 1, the number of independent solute particles is lower than the number expected from the formula units initially dissolved. The usual cause is association, where two or more molecules combine into one larger species. Dissociation or ionisation increases the particle count and gives i > 1. Thus association is the correct interpretation.

Correct answer: A. The solute is behaving normally

Explanation: A van’t Hoff factor of one means that the effective number of particles is the same as the number expected from the solute formula. Thus no net association or dissociation is affecting the measurement, and the solute behaves normally for that condition. It does not mean that the solute has zero mass or is fully ionised.

Correct answer: A. 2

Explanation: Use i = normal molecular mass/observed molecular mass. Substitution gives i = 80/40 = 2. Thus the effective number of particles is twice the normal expected number. Such a value is consistent with dissociation, although the numerical calculation itself only establishes the factor. Option B is the reciprocal, option C represents no abnormality, and option D is not a ratio.

Correct answer: A. 2/3

Explanation: For abnormal molar mass, the van’t Hoff factor is related to molar masses by i = normal molar mass / observed molar mass. Substituting the given values gives i = 60/90 = 2/3. Since this value is less than 1, it is consistent with association, in which several solute particles combine into fewer particles. The value 3/2 would result from reversing the ratio.

Correct answer: A. Two

Explanation: On complete dissociation in water, one formula unit of NaCl separates into one sodium ion and one chloride ion: NaCl → Na⁺ + Cl⁻. Thus, one solute unit produces two particles. The charges show that the ions are different, but charge does not add extra particles. Therefore, the ideal van’t Hoff factor for complete dissociation of NaCl is i = 2, not 1, 3, or 4.

Correct answer: A. 3

Explanation: Complete dissociation of one formula unit follows BaCl₂ → Ba²⁺ + 2Cl⁻. There is one barium ion and two chloride ions, making three independent particles in total. Hence the ideal van’t Hoff factor equals the number of particles, i = 3. The charge values do not add to the particle count; the stoichiometric numbers do.

Correct answer: A. Five

Explanation: The formula Al₂(SO₄)₃ contains two aluminium atoms and three sulfate groups. On complete dissociation, it produces 2Al³⁺ + 3SO₄²⁻. Counting the ions, not the total charge, gives 2 + 3 = 5 ions per formula unit. The subscripts outside and inside the parentheses are both important; counting only one type of ion would incorrectly give 2 or 3.

Correct answer: A. Below 1

Explanation: Dimerisation is an association process: two original molecules behave as one independent solute particle. The number of particles therefore becomes smaller than expected for the same amount of solute. Since i measures the effective particle ratio, dimer formation generally makes i less than one; its exact value depends on the extent of association.

Correct answer: A. 1.25

Explanation: For an AB-type electrolyte, dissociation produces two ions, so the relation is i = 1 + α. Convert the percentage into a fraction: α = 25/100 = 0.25. Substitution gives i = 1 + 0.25 = 1.25. The value 0.25 is only the degree of dissociation, not the van’t Hoff factor; 1.75 and 2.25 do not follow from the stated percentage.

Correct answer: A. 3

Explanation: Under complete dissociation, CaCl₂ → Ca²⁺ + 2Cl⁻. One formula unit therefore produces one calcium ion and two chloride ions, making three independent particles. The ideal van’t Hoff factor is consequently i = 3. In a real solution, ion interactions or incomplete dissociation may make the measured value lower.

Correct answer: A. Depression in freezing point

Explanation: Depression in freezing point is a colligative property, so its magnitude depends on the number of dissolved particles. Dissociation produces more particles from each solute unit, increasing the effective van’t Hoff factor and therefore increasing ΔTf under comparable conditions. Colour, smell, and container weight are not colligative properties and are not determined by particle count in this way.

Correct answer: A. Lower

Explanation: Osmotic pressure at a given temperature and analytical concentration is proportional to the effective number of solute particles. Association joins molecules into fewer independent units, so the observed osmotic pressure can be lower than the value calculated for non-associating particles. Dissociation would produce the opposite tendency.

Correct answer: A. Greater than normal

Explanation: Elevation in boiling point is a colligative effect and increases with the number of dissolved particles at fixed conditions. Dissociation creates additional ions from each original solute unit, so the observed boiling-point elevation can be greater than the value for an undissociated solute. Association would reduce the effect instead.

Correct answer: A. Less than normal

Explanation: Relative lowering of vapour pressure is a colligative property and is proportional to the effective number of solute particles at a given composition. Association joins molecules into larger units, reducing the number of independent particles. Hence the lowering can be smaller than the value calculated without association. It is not necessarily zero or exactly double, so A is correct.

Correct answer: A. Electrolytes

Explanation: Electrolytes can produce ions when dissolved, especially in a suitable polar solvent such as water. The ions increase the effective particle count and alter colligative properties, which can lead to an abnormal calculated molecular mass. A non-electrolyte that remains molecular generally gives i close to one under the stated conditions.

Correct answer: A. Because glucose generally does not dissociate into ions

Explanation: Glucose is treated as a nonelectrolyte in dilute aqueous solution. It dissolves as intact glucose molecules rather than splitting into ions, and no substantial association is assumed in the elementary model. Thus the effective particle count is approximately the expected count and i is taken as about one.

Correct answer: A. Because urea generally does not dissociate into ions

Explanation: Urea is treated as a nonelectrolyte in dilute water. Its molecules dissolve without producing ions, and the usual school-level treatment does not assign significant association. Consequently, the number of effective particles remains close to the expected number, i is approximately one, and the molecular mass calculated from colligative properties is normal.