7: Abnormal Molecular Mass

Correct answer: A. solute is dissociating

Explanation: Colligative effects depend on the number of effective solute particles. During dissociation, one formula unit separates into two or more ions, increasing the particle count. The observed vapour-pressure lowering, boiling-point elevation or freezing-point depression can therefore exceed the value calculated for undissociated molecules. Association has the opposite effect because it reduces the number of effective particles.

Correct answer: A. 2

Explanation: The van’t Hoff factor is the ratio of the actual number of particles in solution to the number expected without dissociation. One formula unit that completely produces two independent ions gives two effective particles, so i = 2 ideally. In real concentrated solutions, interionic interactions may make the observed value differ slightly.

Correct answer: A. 1

Explanation: Glucose dissolves in water as neutral molecules and does not normally dissociate into ions or associate significantly under the ideal dilute-solution assumption. Thus one glucose molecule contributes one effective solute particle, giving the van’t Hoff factor i = 1. A value of 2 or 3 would suggest particle multiplication, while i = 0 would incorrectly imply no solute particles.

Correct answer: A. Because the number of effective particles increases

Explanation: Dissociation splits one solute unit into two or more ions, increasing the number of independent particles. Since colligative effects are proportional to effective particle concentration, the observed effect becomes larger; for example, boiling-point elevation contains the factor i. Colour and solvent loss are unrelated explanations.

Correct answer: A. It decreases

Explanation: Association combines two or more solute particles into a larger species, reducing the number of independent particles in solution. Because colligative properties depend on that effective number, the observed effect decreases compared with the value calculated for unassociated particles. It does not become negative.

Correct answer: A. Dissociation of solute

Explanation: For a non-associating, non-dissociating solute, i is approximately 1. A value greater than 1 means that more effective particles are present than formula units initially introduced. The usual explanation is dissociation into ions or other species. Association would reduce the particle count and generally give i < 1.

Correct answer: A. Association of solute

Explanation: The van’t Hoff factor compares the actual number of dissolved particles with the number expected from the formula units. If i is less than 1, the solution contains fewer effective particles than expected. This commonly results when solute molecules associate into dimers or larger units, as in some weakly interacting systems. Dissociation generally gives i greater than 1, not less than 1.

Correct answer: A. 2

Explanation: The van’t Hoff factor is the ratio of particles after dissociation to the original formula units. If every formula unit produces exactly two ions and dissociation is complete, two particles result from one, so i = 2. This is an ideal value; interionic interactions or incomplete dissociation can make the measured value different.

Correct answer: A. 1

Explanation: A non-electrolyte remains essentially as intact molecules in solution and does not dissociate into ions. Therefore the actual number of particles is approximately the same as the number expected from the amount dissolved. Since the van't Hoff factor is actual particles divided by expected particles, i is normally 1. Values such as 2 are associated with particular dissociation cases, not ordinary non-electrolytes.

Correct answer: A. Particles have associated

Explanation: The van’t Hoff factor compares the actual effective number of particles with the number expected from the undissociated solute. An i value below one means that fewer particles are present than expected. This commonly results from association, in which molecules combine. Complete ionisation would instead increase the particle count and normally give i greater than one.

Correct answer: B. Two

Explanation: For ideal complete dissociation, the van’t Hoff factor equals the number of particles produced from one formula unit. Thus i = 2 indicates that one formula unit produces two ions, as happens for an electrolyte such as AB → A⁺ + B⁻. It does not mean two formula units; it refers to two resulting solute particles.

Correct answer: A. Particles are neither ionising nor associating

Explanation: The van’t Hoff factor i compares the effective number of particles in solution with the number expected from the formula units. i = 1 means there is no net change in particle number: the solute behaves approximately as an undissociated, non-associating species. Three-ion dissociation would give i near 3, association would give i below 1, and a solute-free solution is not represented by this comparison.

Correct answer: C. Greater

Explanation: For colligative properties, the electrolyte correction appears as a factor i: for example, ΔTf = iKf m, ΔTb = iKb m and π = iCRT. If i > 1, the solution contains more effective particles than the undissociated calculation assumes, usually because of dissociation. Multiplication by a factor greater than one makes the observed property greater than the expected value.

Correct answer: B. Lower

Explanation: In expressions such as ΔTf = iKf m and ΔTb = iKb m, the van’t Hoff factor multiplies the ideal particle-based result. When i < 1, fewer effective particles are present than expected, commonly because of association. Multiplication by a number below one reduces the observed colligative effect, so it is lower rather than equal to or greater than the expected value.

Correct answer: A. Because of dissociation or association of the solute

Explanation: Abnormal molar mass is obtained when the solute does not remain as the expected number of independent particles in solution. Dissociation breaks one solute unit into two or more particles, increasing the particle number, whereas association combines particles and decreases their number. Since colligative properties depend on the number of particles, the molar mass calculated from them differs from the true value. Therefore, option A is correct.

Correct answer: A. It becomes less than the true value

Explanation: During dissociation, one solute unit produces two or more particles, so the number of particles in solution increases. Colligative effects therefore become larger than expected for the undissociated solute. If molar mass is calculated from this enhanced effect without correcting for the van’t Hoff factor, the calculated or observed molar mass is lower than the true molar mass. Thus, option A is correct.

Correct answer: B. It becomes greater than the true value

Explanation: Association occurs when two or more solute particles combine to form a larger associated species. Consequently, the number of independent particles decreases and the colligative effect becomes smaller than expected. If the molar mass is calculated without correcting for this association, the observed molar mass is greater than the true molar mass. In this case the van’t Hoff factor is less than one, so option B is correct.

Correct answer: B. 30 g mol⁻¹

Explanation: The relation between true and observed molar mass is M_observed = M_true/i. This follows because the van’t Hoff factor represents the change in the effective number of solute particles. Substituting the values gives M_observed = 60 g mol⁻¹ ÷ 2 = 30 g mol⁻¹. Since i is greater than one, the solute behaves as if more particles are present and the observed molar mass is lower. Therefore, option B is correct.

Correct answer: D. 200 g mol⁻¹

Explanation: For abnormal molar mass, the observed value is calculated from M_observed = M_true/i. Therefore, M_observed = 100 g mol⁻¹ ÷ 0.5 = 200 g mol⁻¹. Dividing by one-half is the same as multiplying by two. Because i is less than one, association has reduced the number of independent particles, so the observed molar mass is greater than the true value. Hence, option D is correct.

Correct answer: A. When the solute dissociates or associates in solution

Explanation: The van’t Hoff factor i is required when the solute does not behave as an unchanged set of particles in solution. Dissociation produces more particles than expected, whereas association produces fewer particles. These changes alter colligative properties and cause the experimentally calculated molar mass to differ from the true molar mass. Including i corrects the particle count and gives a reliable molar-mass value.

Correct answer: A. Van’t Hoff factor (वैन्ट हॉफ गुणक)

Explanation: Abnormal molar mass is observed when a solute associates or dissociates in solution, changing the actual number of solute particles. The Van’t Hoff factor, represented by i, accounts for this change in particle number in colligative-property equations. For dissociation, i is generally greater than 1, while for association it is generally less than 1. Therefore, option A is correct.

Correct answer: A. The calculated molar mass will be greater than the true molar mass (परिकलित मोलर द्रव्यमान वास्तविक मोलर द्रव्यमान से अधिक होगा)

Explanation: During dimerisation, two solute molecules combine to form one larger particle. Thus, the number of dissolved particles decreases, and the osmotic pressure becomes lower than expected for the same mass of solute. In the osmotic-pressure relation M = wRT/(πV), a smaller π produces a larger apparent or calculated molar mass. Hence, association by dimerisation makes the calculated molar mass greater than the true value, so option A is correct.

Correct answer: A. 1

Explanation: The Van’t Hoff factor is calculated from the relation i = normal molar mass / observed molar mass. Substituting the given values gives i = 180/180 = 1. A value of i equal to 1 means that the effective number of solute particles has not changed. Therefore, under the ideal interpretation, neither association nor dissociation is occurring, and option A is correct.

Correct answer: C. 2

Explanation: For abnormal molar mass, the Van’t Hoff factor is given by i = normal molar mass / observed molar mass. Therefore, i = 100/50 = 2. This value greater than 1 indicates that the effective number of solute particles has increased, as happens during dissociation or ionisation. Thus, the correct numerical answer is 2, which is option C.

Correct answer: A. Association (संघटन)

Explanation: The Van’t Hoff factor is i = normal molar mass / observed molar mass = 60/120 = 0.5. A value of i less than 1 shows that the number of solute particles has decreased. This occurs when solute molecules associate, such as two molecules forming a dimer. Therefore, association is the possible process and option A is correct.