7: Abnormal Molecular Mass

Correct answer: A. 2

Explanation: At the same concentration and temperature, the van’t Hoff factor is the ratio of the observed colligative effect to the value expected without particle-number change. If the observed effect is twice the expected value, i = 2. This indicates twice as many effective particles and is commonly associated with dissociation.

Correct answer: A. 0.5

Explanation: The van’t Hoff factor compares the observed colligative effect with the normal effect for the same analytical concentration. Therefore i = observed effect ÷ normal effect = one-half ÷ one = 0.5. A value below one means fewer effective particles, which may result from association such as dimer formation.

Correct answer: A. Partial dissociation may occur

Explanation: The value 1.4 is greater than one, so the solution contains more effective particles than expected from the undissociated solute. Since it is below the ideal value for complete dissociation of many electrolytes, partial dissociation is a suitable conclusion. The number does not imply absence of particles or zero molecular mass.

Correct answer: A. Partial association may occur

Explanation: Because 0.7 is less than one, the effective number of independent particles is lower than expected. Partial association, in which only some solute molecules combine, can produce this result. Complete association is not established by the value alone, while dissociation and formation of three ions would normally increase i above one.

Correct answer: A. 2

Explanation: Complete dissociation of KCl is represented by KCl → K⁺ + Cl⁻. One formula unit therefore produces two independent ions. For ideal complete dissociation, the van’t Hoff factor equals this particle count, so i = 2. A value of 1 would mean no change in particle number, 3 overcounts the products, and 0.5 is not possible for complete dissociation of this 1:1 salt.

Correct answer: A. 4

Explanation: Complete dissociation of one formula unit of FeCl₃ gives one Fe³⁺ ion and three Cl⁻ ions: FeCl₃ → Fe³⁺ + 3Cl⁻. The total number of particles is therefore 1 + 3 = 4. Under the ideal complete-dissociation assumption, the van’t Hoff factor equals this number, so i = 4. The ionic charges affect electrical neutrality, but they do not change the count of particles.

Correct answer: A. 4

Explanation: On complete dissociation, one formula unit of K₃PO₄ produces three potassium ions and one phosphate ion: K₃PO₄ → 3K⁺ + PO₄³⁻. The total number of particles is 3 + 1 = 4, so the ideal van’t Hoff factor is i = 4. The subscript 3 applies to potassium, while PO₄ is one polyatomic ion; its four atoms do not represent four separate solute particles.

Correct answer: A. 3

Explanation: The van’t Hoff factor compares the effective or actual number of particles with the number expected from the undissociated solute: i = actual particles/expected particles. If the actual number is three times the expected number, then i = 3. A value of 1 means no particle-number change, while 0.3 would indicate fewer particles, not more.

Correct answer: A. 0.5

Explanation: By definition, i = actual number of particles/expected number of particles. If the actual number is one-half of the expected number, then i = 1/2 = 0.5. This value below one indicates fewer independent particles, often because of association. The value 2 describes the opposite change, i = 1 means no change, and 3 means three times as many particles.

Correct answer: A. Association or dissociation of solute particles

Explanation: The first diagnostic step is to identify whether solute particles associate or dissociate. Dissociation increases the number of independent particles and generally makes i greater than one, whereas association decreases the number and generally makes i less than one. These changes, not the colour or laboratory equipment, explain abnormal molecular mass.

Correct answer: A. Dissociation can increase the value of colligative properties

Explanation: Dissociation converts one solute unit into two or more independent particles. Since colligative properties depend on the number of dissolved particles, the corresponding effect can become larger. Association instead reduces independent particles, and i is specifically a measure related to the effective particle-number change.

Correct answer: A. Association always makes i greater than 1

Explanation: Association joins solute molecules and therefore generally reduces the number of independent particles. Its usual effect is to make i less than one, not greater than one. Dissociation can increase particle number, dimerisation reduces it, and i equal to one represents no net change in effective particle number, so those statements are consistent.

Correct answer: A. 2

Explanation: Under the ideal complete-dissociation assumption, MgSO₄ → Mg²⁺ + SO₄²⁻. The sulfate group is one polyatomic ion, so one formula unit produces two independent particles: one magnesium ion and one sulfate ion. Therefore the ideal van’t Hoff factor is i = 2; it is not three because the atoms within sulfate are not counted separately.

Correct answer: A. Less than normal

Explanation: The van’t Hoff factor is related to molar masses by i = normal molar mass/observed molar mass. If the observed molar mass is greater, i is less than 1. Since a colligative property is multiplied by i, its observed value is lower than the normal value. This generally indicates association of solute particles.

Correct answer: A. Greater than normal

Explanation: The van’t Hoff factor is related to molar mass by i = Mnormal/Mobserved. Therefore, if the observed molar mass is smaller, i is greater than 1. Since colligative properties are multiplied by i, the observed boiling-point elevation, freezing-point depression, osmotic pressure, or relative lowering of vapour pressure becomes greater than the normal value. This usually indicates dissociation.

Correct answer: A. NaCl in water

Explanation: NaCl is an electrolyte and, in water, it can separate into Na+ and Cl− ions. One formula unit therefore produces two solute particles, increasing the particle number and giving i greater than 1. Acetic acid in benzene is mainly an association example, while glucose and urea generally remain undissociated.

Correct answer: A. Acetic acid in benzene

Explanation: In benzene, acetic acid molecules can associate through hydrogen bonding and form dimers. Two original molecules then behave as one effective particle, so the particle number decreases, i becomes less than 1, and the observed molar mass becomes higher than the normal value. The salt examples generally show dissociation in water.

Correct answer: A. 3

Explanation: One formula unit of an AB₂ salt produces one A ion and two B ions on complete dissociation. Thus, one original solute particle becomes three particles in solution. Since the ideal van’t Hoff factor is the ratio of particles after dissociation to the original number of particles, i = 3. The ionic charges do not change this particle count.

Correct answer: A. Correction factor for particle number

Explanation: The van’t Hoff factor i compares the actual effective number of solute particles with the number expected from the formula. Association lowers this number and gives i less than 1; dissociation raises it and gives i greater than 1. Thus i is best understood as a particle-number correction factor in colligative-property equations.

Correct answer: A. It will be greater than normal

Explanation: For molar-mass measurements, i = normal molar mass/observed molar mass. If i is less than 1, the denominator must be larger than the numerator; therefore the observed molar mass is greater than the normal molar mass. This usually results from association, which reduces the effective particle number.

Correct answer: B. Association or dissociation of solute in solution

Explanation: Molar mass obtained from a colligative property assumes a certain number of independent solute particles. If solute molecules associate, the effective number decreases; if an electrolyte dissociates, it increases. The measured colligative effect then differs from the expected one, producing an abnormal observed molar mass.

Correct answer: B. It increases

Explanation: Combining solute particles is association. For example, two molecules may form one dimer, reducing the number of particles that contribute to a colligative property. The observed effect is therefore smaller, so the calculation gives i less than 1 and an observed molar mass greater than the normal molar mass.

Correct answer: B. Less than normal

Explanation: Breaking a solute into ions is dissociation. It increases the number of particles and therefore increases the colligative effect. In the relation i = normal molar mass/observed molar mass, dissociation gives i greater than 1, so the observed molar mass must be lower than the normal molar mass.

Correct answer: A. सामान्य मोलर द्रव्यमान / प्रेक्षित मोलर द्रव्यमान / Normal molar mass divided by observed molar mass

Explanation: The van’t Hoff factor compares the actual number of solute particles with the number expected from the formula. When expressed through molar mass, it is i = Mnormal/Mobserved. Dissociation lowers the observed molar mass and gives i > 1, whereas association raises it and gives i < 1. Therefore the reverse ratio in option B is incorrect.

Correct answer: C. When neither association nor dissociation occurs

Explanation: The van’t Hoff factor compares the actual number of solute particles in solution with the number expected if each formula unit remained as one particle. If solute particles neither combine nor split, the particle number is unchanged, so i = 1. Complete association usually gives i below 1, whereas dissociation increases i above 1; volatility alone does not define i.