7: Abnormal Molecular Mass

Correct answer: A. 0.2 mole per litre

Explanation: Use the molarity relation M = n/V(solution). First find the amount of glucose: n = mass/molar mass = 18 g/(180 g mol⁻¹) = 0.10 mol. Convert the final volume: 500 mL = 0.500 L. Therefore M = 0.10 mol/0.500 L = 0.20 mol L⁻¹, so option A is correct. Option B is only the number of moles, not the concentration; C and D use incorrect scaling.

Correct answer: A. solute particles may associate

Explanation: If solute particles associate, several original particles combine into fewer species. The measured colligative effect is then smaller than expected for the stated amount of solute. In calculations that assume independent particles, a smaller effect corresponds to a larger apparent or observed molar mass. Dissociation would increase the particle count and usually make the observed molar mass smaller, not larger.

Correct answer: B. Smaller lowering and higher molar mass

Explanation: Association combines several solute particles into fewer larger species. The effective particle count therefore decreases, causing a smaller colligative effect and smaller lowering of vapour pressure than expected for non-associated particles. Since molar mass is calculated from that effect, the observed molar mass appears higher than the true monomeric value.

Correct answer: A. Lowering will be less than expected

Explanation: Association joins two or more solute molecules into one larger effective particle. For the same mass of solute, the number of particles is therefore smaller than the number calculated by assuming no association. Because vapour-pressure lowering depends on particle number, the observed lowering is less than expected and the experimentally calculated molar mass appears unusually high.

Correct answer: A. Lowering will increase

Explanation: Dissociation converts one formula unit into two or more solute particles. The effective particle concentration is therefore greater than the nominal concentration. This lowers the solvent mole fraction more strongly, so the relative lowering of vapour pressure increases. The solution pressure still remains below the pure-solvent pressure for a non-volatile solute; it does not become higher than it.

Correct answer: A. Because effective number of particles increases

Explanation: Lowering of vapour pressure is a colligative property, so it depends on the number of dissolved particles. When one formula unit dissociates into two or more ions, the effective particle count increases. The solvent mole fraction consequently falls more than expected for undissociated solute, producing a greater lowering; this effect is represented by the van’t Hoff factor.

Correct answer: A. Because effective number of particles decreases

Explanation: Association joins two or more dissolved particles into a larger species, so the number of independent solute particles becomes smaller. Since vapour-pressure lowering depends on particle number, the observed colligative effect is less than the value calculated by assuming no association. The solvent mole fraction does not become zero, and pressure does not automatically increase.

Correct answer: B. Dissociation

Explanation: Dissociation increases the number of effective solute particles. Consequently, a colligative effect such as vapour-pressure lowering is larger than expected from the formula mass of undissociated solute. If the calculation assumes fewer particles, the inferred number of moles is too large and the calculated molar mass appears too small. Association gives the opposite trend.

Correct answer: B. Association

Explanation: Association combines several solute molecules into fewer effective particles. The colligative effect is then smaller than expected for the same mass of unassociated solute. If this smaller effect is interpreted using the no-association formula, the calculated number of moles is too small, so the apparent molar mass becomes larger than the true value. Dissociation causes the reverse result.

Correct answer: A. To express the effect of actual number of particles

Explanation: The van’t Hoff factor accounts for the actual number of particles produced in solution. An electrolyte may dissociate into several ions, increasing the particle count, while some solutes associate into larger units, decreasing it. Consequently, colligative equations are written with i, such as ΔTb = iKb m and ΔTf = iKf m. It is unrelated to colour, smell, or container volume.

Correct answer: A. Dissociation or association

Explanation: Molar mass calculated from a colligative effect assumes a particular number of independent solute particles. Dissociation increases particle number and association decreases it, so the measured effect no longer matches the simple molecular formula. The resulting calculated molar mass may therefore appear abnormally low or high.

Correct answer: A. It will be lower

Explanation: Association combines two or more solute particles into a larger species. The number of independently moving particles therefore becomes smaller than the number calculated from the original formula units. Since colligative effects depend on particle number, the observed effect is lower than the ideal expected value, and the van’t Hoff factor is generally less than one.

Correct answer: A. Due to ionisation or association of solute

Explanation: Colligative measurements are interpreted through the number of dissolved particles. Ionisation produces more particles than the formula units suggest, whereas association produces fewer. If these changes are ignored and the ideal formula is used directly, the calculated molar mass differs from the true value. The van’t Hoff factor corrects this abnormality.

Correct answer: A. Less than actual

Explanation: Dissociation increases the number of particles and therefore makes the observed colligative effect larger than the ideal molecular calculation predicts. If that enhanced effect is interpreted without applying the van’t Hoff factor, the calculation appears to involve more moles of solute than are actually present. Since molar mass equals mass divided by moles, the apparent molar mass becomes lower than the actual value.

Correct answer: A. Greater than actual

Explanation: Association joins several solute molecules into fewer larger particles, reducing the observed colligative effect. If the reduced effect is analysed using the non-associating formula, the calculated number of moles appears smaller than the true number. For a fixed sample mass, mass divided by this underestimated mole value gives an apparent molar mass greater than the actual molar mass.

Correct answer: A. ΔT_b = iK_bm

Explanation: For a non-electrolyte, the boiling point elevation is ΔT_b = K_bm. If a solute dissociates or associates, the effective number of particles differs from the simple formula-unit count, so the van't Hoff factor i is included: ΔT_b = iK_bm. The expression K_b/m has the wrong dependence, iRT is not the boiling-point relation, and K_fm describes freezing point depression.

Correct answer: A. ΔT_f = iK_fm

Explanation: For a non-electrolyte, freezing point depression is ΔT_f = K_fm. Dissociation increases the number of particles and association decreases it, so the actual effect is corrected by the van't Hoff factor i. The complete relation is therefore ΔT_f = iK_fm. K_bm belongs to boiling point elevation, i/K_f has the wrong form, and CRT is associated with the dilute osmotic-pressure equation.

Correct answer: A. When solute dissociates into ions

Explanation: Dissociation changes one formula unit into two or more particles, increasing the effective particle count. Since colligative effects become larger, a calculation that assumes no dissociation interprets the effect as coming from more solute moles. The calculated apparent molar mass, which is inversely related to the inferred amount, therefore becomes lower than the true molar mass.

Correct answer: A. When solute particles associate

Explanation: Association combines several solute particles into fewer units, so the effective number of particles becomes smaller than expected. The observed colligative effect is consequently reduced. If the calculation ignores association, it infers fewer solute moles than are actually present; dividing the solute mass by this underestimated amount gives an apparent molar mass greater than the true value.

Correct answer: B. Ionisation of solute

Explanation: The freezing-point depression is ΔTf = iKf m. If the experimentally observed value is larger than the value calculated with i = 1, the effective particle factor i is greater than one. Ionisation or dissociation produces additional particles and raises i, thereby increasing the depression. Association would lower the number of particles and reduce ΔTf instead.

Correct answer: B. Association of solute

Explanation: Boiling-point elevation follows ΔTb = iKb m. A value lower than the ideal non-electrolyte prediction indicates i < 1, meaning fewer effective particles than expected. Association combines solute particles and causes this decrease in particle number. Ionisation and an increase in particles would make i greater than one and would produce a larger, not smaller, elevation.

Correct answer: A. During dissociation of solute

Explanation: Dissociation breaks solute molecules into two or more particles, increasing the number of effective particles in solution. Consequently, the observed colligative effect is larger than expected for the undissociated solute. If the ordinary formula is used without correcting for this increase, the calculated or observed molar mass becomes smaller than the actual value. Association generally causes the opposite error.

Correct answer: A. 2

Explanation: For abnormal molar-mass behaviour, the van't Hoff factor is related to molar masses by i = normal molar mass/observed molar mass. Substituting the given values gives i = M/(M/2) = 2. An observed molar mass smaller than the normal value means that the solute has produced more effective particles, usually through dissociation. Therefore, the correct answer is 2.

Correct answer: A. Association

Explanation: A greater observed molar mass corresponds to a van't Hoff factor less than one. This happens when two or more solute molecules join to form larger associated species, thereby reducing the number of independently behaving particles. The reduced particle count produces a smaller colligative effect and makes the calculated molar mass appear larger. Dissociation and ionisation would increase particle number and generally lower the observed molar mass.

Correct answer: A. Solute particles are associating

Explanation: The van’t Hoff factor i compares the actual number of solute particles in solution with the number expected if no association or dissociation occurred. When i is less than one, individual solute molecules combine to form larger associated units, such as dimers. The total number of particles consequently decreases. Because colligative effects depend on particle number, the observed effect is smaller than the normal calculated value, and the experimentally determined molar mass appears larger.