7: Abnormal Molecular Mass

Correct answer: A. \(30\,\mathrm{g\,mol^{-1}}\)

Explanation: The van’t Hoff factor relates the normal and observed molar masses as \(i=M_{\text{normal}}/M_{\text{observed}}\). Therefore, \(M_{\text{observed}}=90/3=30\,\mathrm{g\,mol^{-1}}\). Since \(i>1\), the solute produces more particles, usually because of dissociation, so the colligative effect is larger and the calculated observed molar mass is smaller than the normal value. Hence, option A is correct.

Correct answer: A. Use the van’t Hoff factor

Explanation: Dissociation increases the number of particles in solution, so the measured colligative effect is larger than it would be for an equal amount of non-dissociating solute. This particle-number change is represented by the van’t Hoff factor \(i\). For an electrolyte, the appropriate factor is included in equations such as \(\Delta T_f=iK_fm\), \(\Delta T_b=iK_bm\), or \(\pi=iCRT\). Therefore, option A is correct.

Correct answer: C. 2

Explanation: The van’t Hoff factor is related to normal and observed molar masses by i = Mnormal/Mobserved. Therefore, i = 60/30 = 2. A factor greater than one means that the number of solute particles has increased relative to the undissociated formula units. This commonly occurs through dissociation, so the observed molar mass becomes smaller than the normal molar mass.

Correct answer: B. Dimer association

Explanation: The observed molar mass is twice the normal molar mass: 240/120 = 2. Since i = Mnormal/Mobserved, the van’t Hoff factor is i = 120/240 = 0.5. A value below one indicates that solute particles have associated, reducing the number of particles in solution. Organic acids commonly form hydrogen-bonded dimers in nonpolar benzene, so the effective particle mass doubles.

Correct answer: A. 32.5 g mol⁻¹

Explanation: For abnormal colligative behaviour, the van’t Hoff factor relates the normal and observed molar masses by i = Mnormal/Mobserved. Therefore, Mobserved = Mnormal/i. Using the given values, Mobserved = 58.5 ÷ 1.8 = 32.5 g mol⁻¹ approximately. Since i is greater than one because of dissociation, the observed molar mass is smaller than the normal value. Option A is correct.

Correct answer: C. 120 g mol⁻¹

Explanation: For association, the van’t Hoff factor is calculated from the number of solute particles after association. Consider one mole initially: 0.5 mol remains as individual molecules, while 0.5 mol molecules forms dimers and becomes 0.25 mol of dimers. Thus the total particles are 0.75 mol, so i = 0.75. Since observed molar mass = normal molar mass ÷ i, M observed = 90 ÷ 0.75 = 120 g mol⁻¹. Therefore, option C is correct.

Correct answer: B. 62.5 g mol⁻¹

Explanation: For AB dissociating into A⁺ and B⁻, the number of particles increases from one to two for each dissociated unit. The van’t Hoff factor is i = 1 + α(n − 1), where α = 0.60 and n = 2. Thus i = 1 + 0.60(1) = 1.60. The observed molar mass is Mnormal/i = 100/1.60 = 62.5 g mol⁻¹. Therefore, option B is correct.

Correct answer: B. The solute particles undergo dissociation

Explanation: A lower observed molar mass indicates that the solution contains more effective particles than expected from the undissociated solute. This occurs when one solute particle breaks into two or more particles, a process called dissociation. The van’t Hoff factor then becomes greater than one, and the observed molar mass is lower than the normal value. Hence, option B is correct.

Correct answer: B. Association of solute particles

Explanation: A higher observed molar mass means that the effective number of solute particles is lower than expected. Association causes two or more solute molecules to combine into a larger molecular unit, thereby reducing the number of independent particles. Consequently, the van’t Hoff factor becomes less than one and the calculated molar mass appears higher than the normal value. Thus, option B is correct.

Correct answer: B. 50 g mol⁻¹

Explanation: For abnormal molar mass, the relation between normal molar mass and observed molar mass is M_observed = M_normal/i. Substituting the given values, M_observed = 75 g mol⁻¹ ÷ 1.5 = 50 g mol⁻¹. Therefore, the observed molar mass is 50 g mol⁻¹, which is option B. The factor must be used as a divisor here, not as a multiplier.

Correct answer: A. 64 g mol⁻¹

Explanation: The van’t Hoff factor is related to the true or normal molar mass and the observed molar mass by i = M_normal/M_observed. Rearranging gives M_normal = i × M_observed. Substitution gives M_normal = 0.8 × 80 = 64 g mol⁻¹. Therefore, option A is correct. The factor below one indicates association, which reduces the number of particles and makes the experimentally observed molar mass appear larger than the normal value.

Correct answer: B. Dimerisation of benzoic acid in benzene

Explanation: Benzoic acid molecules associate in benzene and commonly form hydrogen-bonded dimers. Thus, two solute molecules behave as one effective particle, reducing the total number of solute particles. Since colligative properties depend on particle number, the observed effect becomes smaller than expected and the calculated molar mass becomes larger than the true value. In contrast, ionisation or dissociation increases particle number and gives a lower apparent molar mass.

Correct answer: A. The observed molar mass is greater than the actual molar mass

Explanation: Association means that two or more solute molecules combine to form fewer effective particles. Because colligative properties depend on the number of dissolved particles, the measured colligative effect becomes smaller than the value expected for non-associating molecules. When the usual formula is applied, this smaller effect gives an apparently larger molar mass. Therefore, association produces an observed molar mass greater than the actual molar mass and a van’t Hoff factor less than one.

Correct answer: A. Association of molecules in the solution

Explanation: When solute molecules associate, two or more molecules combine to form fewer effective particles. Since colligative properties depend on the number of particles, the measured effect becomes smaller than the ideal expected effect. Substitution into the ordinary molar-mass formula then gives a value greater than the true molar mass. Dissociation and ionisation increase particle number and generally produce the opposite result, namely a lower observed molar mass.

Correct answer: D. 120 g mol⁻¹

Explanation: For abnormal molar mass caused by dissociation or association, the van’t Hoff factor is related to molar masses by i = Mnormal/Mobserved. Therefore, Mnormal = i × Mobserved = 3 × 40 = 120 g mol⁻¹. The observed molar mass is smaller because dissociation produces more particles than expected from the original solute units. Thus, 40 g mol⁻¹ is not the normal value; option D is correct.

Correct answer: A. It will be lower than the normal molar mass

Explanation: For a colligative-property measurement, the observed effect is proportional to \(i/M\). If the electrolyte has \(i=2\) but the calculation incorrectly uses \(i=1\), the extra particles produced by dissociation are ignored. The measured effect is then interpreted as arising from twice as many solute moles, so the calculated molar mass becomes half the correct value, or lower than normal. Option A is correct.

Correct answer: A. The solute dissociated and doubled the number of effective particles

Explanation: For abnormal molar mass, the van’t Hoff factor is \(i=M_{\text{normal}}/M_{\text{observed}}=98/49=2\). Thus, the solution contains twice as many effective solute particles as expected from an undissociated solute. Dissociation, such as an ionic compound splitting into ions, produces more particles and makes the experimentally calculated molar mass smaller than the normal value. Therefore, option A is correct; dimerisation would instead give \(i<1\) and a larger observed molar mass.

Correct answer: C. 2

Explanation: The apparent molar mass obtained from a colligative property is related to the true molar mass by \(M_{\text{app}}=M_{\text{true}}/i\). Substitution gives \(90=180/i\), so \(i=180/90=2\). A value greater than one means that the number of effective particles has increased, usually because the solute dissociates into two or more particles. Therefore, option C is correct.

Correct answer: B. The solute is forming dimers

Explanation: The relation is \(M_{\text{observed}}=M_{\text{true}}/i\). Therefore, \(240=120/i\), which gives \(i=0.5\). A van’t Hoff factor below one means that the number of independent solute particles has decreased because particles associate. Formation of dimers combines two solute molecules into one effective particle, giving approximately \(i=1/2\) and doubling the apparent molar mass. Hence, option B is correct.

Correct answer: C. 50%

Explanation: For association, the van’t Hoff factor is i = Mtrue ÷ Mobserved = 60 ÷ 80 = 0.75. When monomers form dimers, the relation is i = 1 − α/2, because two original particles become one associated particle. Thus, 0.75 = 1 − α/2, so α/2 = 0.25 and α = 0.50. Therefore, the degree of association is 50%, making option C correct.

Correct answer: C. 1.25

Explanation: The relation between observed and true molar mass is Mobserved = Mtrue/i. The given condition is 0.8Mtrue = Mtrue/i. Cancelling Mtrue gives i = 1/0.8 = 1.25. Since i is greater than 1, the result also indicates an increase in the number of solute particles, such as dissociation. Hence option C is correct.

Correct answer: A. 0.625

Explanation: Use the relation Mobserved = Mtrue/i. The stated value gives 1.6Mtrue = Mtrue/i. After cancelling the true molar mass, i = 1/1.6 = 0.625. A van’t Hoff factor below unity means that association has reduced the number of independent solute particles. Therefore, option A is the correct value.

Correct answer: A. i = 2; dissociation is occurring

Explanation: For abnormal colligative behaviour, the van’t Hoff factor is i = Mtrue/Mobserved. Thus i = 100/50 = 2. A value greater than one means that the number of solute particles has increased compared with the undissociated state. This increase is caused by dissociation, so option A is correct.

Correct answer: C. 1.5

Explanation: For abnormal molar mass caused by association or dissociation, the van’t Hoff factor is related to molar masses by i = normal molar mass ÷ observed molar mass. Thus, i = 90 ÷ 60 = 1.5. Since i is greater than 1, the solute produces more particles in solution than expected, generally because of dissociation or ionisation. Hence, option C, 1.5, is the unambiguous correct answer.

Correct answer: C. 7.38 atm

Explanation: The osmotic-pressure equation is π = iCRT. At the same temperature, R and T are constant, so osmotic pressure is proportional to iC. For the first nonelectrolyte solution, iC = 1 × 0.2 = 0.2 M. For the second solution, iC = 3 × 0.1 = 0.3 M. Therefore, π₂/π₁ = 0.3/0.2 = 1.5, and π₂ = 1.5 × 4.92 = 7.38 atm. Hence option C is correct.