7: Abnormal Molecular Mass

For a given solvent, solvent mass, and solute amount, the normal colligative effect is multiplied by the van’t Hoff factor i. Thus i = observed depression/normal depression = 0.558/0.372. Dividing both values by 0.372 gives i = 1.50. A value greater than one indicates that the solute produces more particles than expected, usually because of dissociation. Therefore, option C is correct.

The relation between molar masses and the van’t Hoff factor is i = Mnormal/Mobserved. Given that Mobserved = 1.25 Mnormal, substitute this into the relation: i = Mnormal/(1.25 Mnormal) = 1/1.25 = 0.80. A value below one indicates association of solute particles, so the correct answer is option A.

The van’t Hoff factor relation is i = Mnormal/Mobserved. Rearranging gives Mobserved = Mnormal/i. Substitution gives Mobserved = 170/0.85 = 200 g mol⁻¹. Association reduces the number of solute particles, so i is less than one and the observed molar mass becomes greater than the normal value. Therefore, option C is correct.

A colligative method measures the effective number of solute particles. The van’t Hoff factor relates the observed molar mass to the true molar mass through Mobserved = Mtrue/i. Substituting the given values gives Mobserved = 180/1.8 = 100 g mol⁻¹. The concentration confirms the solution context but is not needed separately because the particle-number correction is already represented by i. Therefore, option C is correct.

For dimerization, two monomer particles combine to form one dimer, so the van’t Hoff factor is i = 1 − α/2. With α = 60% = 0.60, i = 1 − 0.60/2 = 0.70. Since the observed molar mass is related by Mobserved = Mtrue/i, the true molar mass is Mtrue = i × Mobserved = 0.70 × 200 = 140 g mol⁻¹. Hence, option C is correct.

An AB solute dissociates into two ions, so for a degree of dissociation α, the van’t Hoff factor is i = 1 + α. With α = 0.60, i = 1.60. The relation between observed and true molar mass is Mobserved = Mtrue/i, or Mtrue = i × Mobserved. Therefore, Mtrue = 1.60 × 75 = 120 g mol⁻¹. Hence, option C is the unambiguous correct answer.

For a solute that produces n particles on complete dissociation, the van’t Hoff factor is given by i = 1 + α(n − 1), where α is the degree of dissociation. Here, n = 5 and i = 3.4. Therefore, 3.4 = 1 + α(5 − 1) = 1 + 4α. Thus, α = 2.4/4 = 0.60, or 60%. Hence, option B is correct.

A colligative property depends on the number of solute particles rather than merely on the mass of solute. If two acid molecules associate to form one dimer in benzene, the number of particles becomes smaller than expected. The observed freezing-point depression is therefore smaller, and calculation gives an apparent molar mass nearly twice the true molar mass. This is evidence of association, commonly dimerisation through hydrogen bonding.

For abnormal colligative behaviour, the van’t Hoff factor is related to true and apparent molar masses by i = M_true/M_apparent. Substitution gives i = 60/30 = 2. Thus the solution contains twice as many effective solute particles as expected from undissociated molecules. This indicates dissociation, or another process that increases particle number. Therefore, the unambiguous numerical answer is option C.

Complete dissociation of one formula unit into three ions makes the van’t Hoff factor i = 3, assuming ideal complete dissociation and negligible ion pairing. Since the apparent molar mass obtained from a colligative property is M_apparent = M_true/i, it becomes M_true/3. Therefore, the apparent molar mass is one-third of the true molar mass. The increased particle count makes the colligative effect three times larger.

The van’t Hoff factor is calculated from the true and apparent molar masses using i = M_true/M_apparent. Here, i = 90/180 = 0.5. A value below 1 is consistent with association because several solute molecules combine to form fewer effective particles. The reduced particle number produces a smaller colligative effect and therefore an apparent molar mass larger than the true value. Hence option C is correct.

For the same solute and the same mass of solvent, the molality is directly proportional to the mass of solute used. The solute mass changes from 3 g to 6 g, so it doubles. Freezing-point depression is a colligative property and therefore also doubles: ΔTf = 2 × 0.3 K = 0.6 K. Hence option A is correct.

Cryoscopic molar mass is inferred from freezing-point depression, which depends on the number of independent solute particles. Association combines molecules and reduces that number, producing a smaller-than-expected depression and therefore a higher apparent molar mass. A repeated high value should prompt investigation of association and solvent effects. Colour, vessel mass, and name length are irrelevant.

The apparent molecular mass obtained without accounting for abnormal particle behaviour is related to the normal mass by i = Mnormal/Mobserved. If dissociation increases particles, the observed mass becomes smaller and the ratio exceeds one; if association decreases particles, the observed mass becomes larger and the ratio is below one. Thus the numerator and denominator in A are correctly placed.

Consider one initial solute particle. If a fraction α dissociates, the undissociated fraction contributes 1 − α particle, while the dissociated fraction contributes nα particles. The total relative particle count is (1 − α) + nα = 1 + (n − 1)α, which is the van’t Hoff factor i. The plus sign is essential because dissociation increases particle number; the other expressions either subtract this increase or use the wrong dependence.

In complete dimerisation, two original solute molecules combine to form one effective particle. Thus, if two moles of monomer units are considered, they produce one mole of dimers, so the particle number becomes one-half of the original count. Since i is the ratio of actual to expected particles, i = 1/2 = 0.5. Hence option A is correct.

For a dimer-forming solute, no association gives i = 1, while complete association gives i = 0.5 because two monomer particles become one dimer. The value 0.75 lies between these limits, so only part of the solute has associated and the association is partial. It cannot represent complete association or dissociation. Therefore option A is correct.

In dimerisation, two solute molecules combine to form one dimer. If association is absent, i is 1; if dimerisation is complete, two original particles become one and i is 0.5. Partial dimerisation must therefore give an intermediate value, strictly between 0.5 and 1, depending on the extent of association.

Dissociation produces more particles than the original formula units. Since colligative effects become larger, a calculation that ignores dissociation attributes the effect to more moles of solute and gives a smaller molar mass. Thus observed molar mass is generally below the actual value; association produces the opposite trend.

Association combines several solute particles into fewer larger units. The effective particle number falls, so colligative effects are smaller than expected for the same mass of solute. If this reduction is ignored, the calculated or observed molar mass appears higher than the true molar mass. The exact increase depends on the extent of association.

Abnormal molar mass results because association or dissociation changes the number of particles measured by a colligative-property method. The van’t Hoff factor is defined as the ratio of the normal molar mass to the observed molar mass: i = Mnormal / Mobserved. Reversing the ratio would give the reciprocal, not i.

For abnormal molar mass, the van’t Hoff factor is related to molar masses by i = normal molar mass / observed molar mass. Rearranging gives observed molar mass = normal molar mass / i. Substitution gives 60 / 2 = 30 g mol⁻¹. Thus the observed value is 30 g mol⁻¹. A factor greater than 1 means more effective particles than expected, so the experimentally calculated molar mass is smaller than the normal value.

The relation between the van’t Hoff factor and molar masses is i = actual or normal molar mass divided by observed molar mass. Let the actual molar mass be M. The observed molar mass is M/2, so i = M ÷ (M/2) = 2. Therefore, the effective particle concentration is twice the normal comparison value. A factor of 0.5 would result if the observed molar mass were twice the actual value, not half.

Let the actual molar mass be M. From i = Mactual / Mobserved, 0.5 = M/Mobserved. Therefore Mobserved = M/0.5 = 2M. The observed molar mass is double the actual value. This is expected when association lowers the number of effective particles and makes i less than one.

Benzoic acid molecules can form hydrogen-bonded dimers in a nonpolar solvent such as benzene. Two molecules then behave approximately as one solute particle, reducing the effective particle number. The smaller colligative effect leads to a calculated molar mass higher than the monomer’s actual molar mass.