7: Abnormal Molecular Mass

Correct answer: B. Association of solute molecules

Explanation: For a dilute solution, osmotic pressure is related to the number of solute particles by π = iCRT. Association joins two or more solute molecules into larger units, decreasing the number of particles and lowering the observed osmotic pressure. If the ordinary formula is used, this smaller pressure gives an apparent molar mass larger than the true value. Therefore, the observation indicates association.

Correct answer: C. 1.5

Explanation: For abnormal molar-mass calculations, the van’t Hoff factor is defined as \(i=\frac{M_{\text{true}}}{M_{\text{observed}}}\), or equivalently \(M_{\text{observed}}=\frac{M_{\text{true}}}{i}\). Since the observed molar mass is \(\frac{2}{3}M_{\text{true}}\), \(i=\frac{M_{\text{true}}}{(2/3)M_{\text{true}}}=\frac{3}{2}=1.5\). Thus, option C is correct. A value greater than one indicates effective dissociation of KCl particles in solution.

Correct answer: C. 2

Explanation: For abnormal molar mass, the van’t Hoff factor is related to the normal and observed molar masses by i = Mnormal/Mobserved. Substitution gives i = 120/60 = 2. A value greater than one means that the number of solute particles is greater than expected for undissociated molecules. This result is consistent with dissociation, although the numerical question only asks for the factor.

Correct answer: A. Association of solute molecules

Explanation: Osmotic pressure depends on the number of dissolved particles. When solute molecules associate, several original molecules combine into fewer particles, reducing the observed osmotic pressure. If the usual formula is then used without correcting for association, the calculated or apparent molar mass becomes larger than the true molar mass. Dissociation would increase the particle count and produce a lower apparent molar mass, so association is indicated.

Correct answer: A. The observed molar mass will be greater than the normal molar mass

Explanation: Dimerisation combines two original solute molecules into one effective particle. Consequently, the number of particles and the colligative effect become smaller than expected for the same mass of solute. The calculation therefore interprets the sample as containing fewer moles, so the observed molar mass is higher than the true molar mass.

Correct answer: C. Less than the normal molar mass

Explanation: Dissociation changes one solute unit into two or more particles, so the number of effective particles increases. Colligative effects therefore become larger than expected for the undissociated substance. When the larger effect is used in the usual formula, the calculated number of moles is too high and the observed molar mass becomes lower than the true value.

Correct answer: C. Benzoic acid in benzene

Explanation: Benzoic acid associates through hydrogen bonding and commonly forms dimers in non-polar benzene. This lowers the number of effective solute particles, so the van’t Hoff factor is less than one. Because the colligative effect is smaller, the calculated apparent molar mass becomes greater than the true molar mass. Therefore option C is correct.

Correct answer: B. The substance forms dimers

Explanation: For abnormal molar mass caused by association, the van’t Hoff factor is related to molar masses by i = normal molar mass/observed molar mass. If the observed molar mass is twice the normal value, i = 1/2. Complete dimerisation changes two original molecules into one associated particle, giving i = 0.5. Thus the substance forms dimers, so option B is correct.

Correct answer: C. 2

Explanation: For abnormal molar masses, the relation between observed molar mass and true molar mass is Mobserved = Mtrue/i, where i is the van’t Hoff factor. The question states Mobserved = Mtrue/2. Comparing the two expressions gives Mtrue/i = Mtrue/2, so i = 2. A value greater than one is consistent with dissociation, because dissociation increases the number of particles in solution and makes the observed molar mass smaller.

Correct answer: A. 0.5

Explanation: The relation for abnormal molar mass is Mobserved = Mtrue/i. Here, Mobserved = 2Mtrue. Substitution gives 2Mtrue = Mtrue/i. Cancelling the non-zero true molar mass gives 2 = 1/i, so i = 1/2 = 0.5. A value below one indicates association, in which solute particles combine and the effective number of particles becomes smaller than expected, producing an observed molar mass greater than the true value.

Correct answer: A. When the observed value differs from the normal value

Explanation: The normal molar mass is the value expected from the molecular formula of a substance. In colligative-property measurements, association or dissociation can change the effective number of particles in solution. As a result, the molar mass calculated from the observed colligative effect may differ from the theoretical value. Such a deviation is called abnormal molar mass; it is not defined by the unit, solvent, or room temperature.

Correct answer: B. 60 g mol^-1

Explanation: For abnormal molar mass caused by dissociation or association, the observed molar mass is related to the true molar mass by M_observed = M_true/i. Since i = 2, the solute produces twice as many particles as expected for an undissociated solute. Therefore, M_observed = 120/2 = 60 g mol^-1. The observed value is lower because dissociation increases the number of particles. Hence, option B is correct.

Correct answer: C. 2

Explanation: The relation between true and observed molar mass is M_observed = M_true/i. Substituting the given values, 30 = 60/i. Multiplying by i gives 30i = 60, so i = 2. A value greater than one indicates that the solute produces more particles than expected, usually because of dissociation. Therefore, option C is correct.

Correct answer: A. 29.25 g mol^-1

Explanation: Complete dissociation of NaCl produces two particles, Na+ and Cl-, from one formula unit. Therefore, the ideal van't Hoff factor is i = 2. The observed molar mass obtained from a colligative property is M_observed = M_true/i. Thus, M_observed = 58.5/2 = 29.25 g mol^-1. The lower apparent molar mass reflects the increased number of particles. Hence, option A is correct.

Correct answer: A. Greater than the true molar mass

Explanation: Complete dimerisation means that two solute molecules combine to form one particle in solution. Therefore, the number of solute particles becomes half of the expected number, and the colligative effect, including lowering of vapour pressure, becomes smaller. For complete association into dimers, the van’t Hoff factor is i = 1/2. Since M observed = M true/i, the observed molar mass is 2M true, so it is greater than the true molar mass. This is why association produces an abnormally high observed molar mass.

Correct answer: C. 3/2

Explanation: The relation between observed molar mass and true molar mass is M observed = M true/i, where i is the van’t Hoff factor. The question gives M observed = (2/3)M true. Substitution gives (2/3)M true = M true/i. After cancelling M true, 2/3 = 1/i, so i = 3/2 = 1.5. Thus, option C is correct. The ratio 2/3 is the observed-to-true molar-mass ratio, not the van’t Hoff factor itself.

Correct answer: A. 0.80

Explanation: For abnormal molar masses, the observed molar mass is related to the true molar mass by M observed = M true/i. Here M observed = 1.25M true. Therefore, 1.25M true = M true/i, and after cancelling M true, i = 1/1.25 = 0.80. Thus, option A is correct. A value of i below one indicates association, because association decreases the number of solute particles and makes the observed molar mass greater than the true value.

Correct answer: A. Lower than the true value

Explanation: Dissociation breaks solute molecules or formula units into two or more particles, so the number of particles becomes larger than expected. Consequently, the observed colligative effect is greater. If the van’t Hoff factor is ignored, the calculation attributes this larger effect to more solute moles, and molar mass, calculated as mass divided by moles, appears lower than the true value. Therefore, option A is correct.

Correct answer: B. Higher than the true value

Explanation: Association combines two or more solute particles into a larger associated species, so the number of independent particles decreases. The measured colligative effect therefore becomes smaller than the effect expected for non-associated particles. If this reduction is ignored, the calculated number of solute moles is too small. Since molar mass equals mass divided by moles, the apparent molar mass becomes higher than the true value. Thus, option B is correct.

Correct answer: A. 1/2

Explanation: Complete dissociation of one AB formula unit produces two particles, A⁺ and B⁻. Therefore, the van’t Hoff factor is i = 2. For a colligative-property measurement, the observed molar mass is related to the true molar mass by M_observed = M_true/i. Hence M_observed = M_true/2, or one-half of the true value. Option A is correct.

Correct answer: C. Double

Explanation: In complete dimerisation, two solute molecules associate to form one dimer particle. Thus the number of particles becomes half the original number, giving a van’t Hoff factor i = 1/2. Since the observed molar mass obtained from colligative properties is M_observed = M_true/i, it becomes M_true/(1/2) = 2M_true. Therefore, it is double the true value, so option C is correct.

Correct answer: C. 2

Explanation: The relation between normal and observed molar masses is i = normal molar mass/observed molar mass. Therefore, i = 120/60 = 2. A van’t Hoff factor greater than one indicates that the number of solute particles has increased, usually because the solute dissociates or ionises in solution. Thus, option C is correct.

Correct answer: A. Association of solute particles

Explanation: For abnormal molar masses, i = normal molar mass/observed molar mass. If association occurs, several solute molecules combine to form larger units, so the number of particles decreases. The colligative effect becomes smaller than expected, giving i less than one and an observed molar mass greater than the normal value. Hence association is correct.

Correct answer: A. Increase in the number of particles due to dissociation

Explanation: Dissociation of an electrolyte produces two or more ions from each formula unit, thereby increasing the effective number of solute particles. Since colligative properties depend on particle number, the observed effect becomes larger than expected for the undissociated substance. When the ordinary formula is used to calculate molar mass, this larger effect gives a lower observed molar mass. Hence, option A is correct.

Correct answer: A. Because the number of effective solute particles decreases

Explanation: When two solute molecules associate to form one dimer, the number of independently acting solute particles becomes smaller. Colligative properties depend on the number of particles, so the observed effect becomes less than expected for the unassociated solute. If this smaller effect is interpreted using the ordinary formula, the calculated number of moles appears smaller and the molar mass, mass divided by moles, appears larger. This is called an abnormal molar mass due to association.