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In Class 12 Chemistry, the topic Abnormal Molecular Mass explains why the molar mass calculated from colligative properties may differ from the actual molecular mass. Students learn how solute particles associate, such as through dimerisation, or dissociate into ions in solution, changing the number of particles present. They use the van’t Hoff factor to relate these changes to relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure, and calculate the degree of association or dissociation in solutions.
TOPIC PRACTICE
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Up to 25 questions from this page. Select your focus, then start.
25 questions
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Easy · Level 7View options
1
2
3
0.5
Easy · Level 7View options
1
2
3
4
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2
3
4
5
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2
1
0.5
3
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3
1
1/3
2/3
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To improve solution colour
To correct for actual particle number
To keep temperature constant
To convert grams into kilograms
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Structural properties
Colligative properties
Magnetic properties
Combustion properties
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The solute is associating
The solute is dissociating
The solute is unaffected
The solute is insoluble
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The solute is completely dissociated
The solute is partially associated
The solute shows ideal behaviour
The solute forms three ions
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Dimerization of benzoic acid in benzene
Association of ethanoic acid in a non-polar solvent
Dissociation of a salt in water
Molecules combining to form a larger particle
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Dissociation of KNO3 in water
Dissociation of NaCl in water
Association of benzoic acid in benzene
Dissociation of CaCl2 in water
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1
2
3
4
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2
3
4
0.5
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van’t Hoff factor
Density factor
Colour factor
Pressure factor
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Association: i > 1
Dissociation: i < 1
Association: observed molar mass is higher
Dissociation: observed molar mass is higher
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NaCl
CaCl₂
Benzoic acid in benzene
Glucose
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0.5
1
2
3
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Dissociation of particles
Association of particles
Increase in particle number
Complete ionization
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Slight or partial dissociation
Complete dimer association
No change
Complete trimer association
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It will be slightly greater than the normal value
It will be slightly less than the normal value
It will be double the normal value
It will be zero
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Checking the colour of the solute
Checking whether the solute can associate or dissociate
Checking the shape of the container
Shaking the solution
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It will decrease
It will increase
It will remain unchanged
It will become zero
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Greater than normal
Less than normal
Always zero
Always equal to normal
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The solute has dissociated
The solute has associated
The solute is insoluble
The solution is pure solvent
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Association
Complete dissociation
Evaporation of solvent
Precipitation of solute
Question 1EasyLevel 7
For complete dissociation of NaCl, what ideal value of the van’t Hoff factor may be taken?
Correct answer: B
In water, one NaCl formula unit dissociates ideally as NaCl → Na⁺ + Cl⁻. Therefore, one solute unit produces two particles. The ideal van’t Hoff factor is the number of particles formed per original unit, so i = 2. In real solutions, ion interactions can make the measured value slightly different, but the ideal complete-dissociation value is 2.
For complete dissociation of CaCl₂, what is the ideal value of the van’t Hoff factor, i?
Correct answer: C
One formula unit of calcium chloride dissociates completely as CaCl₂ → Ca²⁺ + 2Cl⁻. This produces one calcium ion and two chloride ions, giving three particles in total from one original unit. Therefore, the ideal van’t Hoff factor is i = 3. The subscript 2 on chlorine must be included; counting only the types of ions would incorrectly give 2.
For complete dissociation of AlCl₃, what is the ideal value of the van’t Hoff factor, i?
Correct answer: C
Aluminium chloride dissociates completely according to AlCl₃ → Al³⁺ + 3Cl⁻. Thus, one formula unit produces one aluminium ion and three chloride ions, for a total of four particles. The ideal van’t Hoff factor is therefore i = 4. The ionic charges identify the ions but are not added; the number of particles, including the coefficient 3, determines i.
If two molecules combine to form one dimer and association is complete, what is i?
Correct answer: C
In complete dimerization, every two original molecules become one effective particle. Thus the particle number becomes one-half of the number expected without association. Since i is the ratio of effective to expected particles, i = 1/2 = 0.5. This also causes the observed molar mass to become twice the normal value.
For complete association in which three molecules form one trimer, what is the value of the van’t Hoff factor?
Correct answer: C
In complete trimerization, three original solute molecules combine to form one larger particle: 3A → A₃. Thus the number of effective particles becomes one-third of the original number. Since the van’t Hoff factor is the ratio of actual to expected particle number, i = 1/3. In general, complete association of n molecules gives i = 1/n.
Why is the van’t Hoff factor used in colligative properties?
Correct answer: B
Colligative properties depend on the number of dissolved particles, not simply on the chemical formula of the solute. Association reduces the number and dissociation increases it. The van’t Hoff factor i incorporates this change by multiplying the ideal colligative expression, so the calculated effect matches the actual particle population.
Abnormal molar mass is most closely related to which type of properties?
Correct answer: B
Molar mass is commonly determined from osmotic pressure, elevation of boiling point, or depression of freezing point. These are colligative properties and depend on the number of solute particles. Association or dissociation changes that number, so using such measurements without the van’t Hoff correction can produce an abnormal molar mass.
If a solute has i = 2.5, which statement is correct?
Correct answer: B
A van’t Hoff factor greater than 1 means that the solution contains more effective particles than expected from the undissociated formula unit. This occurs when molecules or formula units split into smaller particles, which is dissociation. Association would give i less than 1, while i = 1 would indicate no change in particle number.
If a solute has i = 0.65, which conclusion is appropriate?
Correct answer: B
A van’t Hoff factor below 1 means that the effective number of solute particles has decreased. This is caused by association, in which two or more molecules combine. Since i = 0.65 is greater than the complete-dimerization limit of 0.5 but less than 1, it is consistent with partial dimerization under a simple model. It does not indicate dissociation or ideal behaviour.
In which case will the observed molar mass be less than the normal molar mass?
Correct answer: C
The observed molar mass is lower when i is greater than 1. Dissociation of a salt in water increases the number of particles because each formula unit can produce several ions, so the colligative effect is larger and the calculated molar mass becomes smaller. All other options describe association, which gives i less than 1.
A value i less than 1 means that the effective number of solute particles has decreased. Benzoic acid can form hydrogen-bonded dimers in benzene, so two molecules behave as one particle. The aqueous salts listed generally dissociate into ions and therefore tend to give i greater than 1, not less than 1.
If an AB2-type electrolyte dissociates completely, what will be i?
Correct answer: C
Complete dissociation of AB2 gives one A ion and two B ions, for example A2+ + 2B−. Thus one original formula unit produces three particles. Under the ideal complete-dissociation assumption, the van’t Hoff factor is the number of particles formed, so i = 3. The subscripts must be included when counting ions.
If an A₂B-type solute completely dissociates into 2A⁺ and B²⁻, what is the van’t Hoff factor, i?
Correct answer: B
The stated dissociation is A₂B → 2A⁺ + B²⁻. One formula unit therefore produces two A⁺ ions and one B²⁻ ion, making three particles in total. For complete dissociation, the ideal van’t Hoff factor equals this number of product particles per original unit, so i = 3. Ionic charge magnitudes are not added; only the number of particles is counted.
Which factor is most useful for explaining abnormal molar mass?
Correct answer: A
Abnormal molar mass arises because association or dissociation changes the effective number of solute particles. The van’t Hoff factor i measures this change and enters the colligative-property equations directly. Values below 1 indicate association, values above 1 indicate dissociation, and i = 1 indicates no change in particle number.
Association joins two or more solute particles, so the total number of particles decreases. Colligative effects then become smaller than expected, and the molar mass calculated from that effect appears higher than the normal molar mass. Therefore, the association–higher observed molar mass pair is correct; dissociation gives i > 1 and a lower observed molar mass.
If i = 3 and the solute is completely dissociated, which solute may be a suitable example?
Correct answer: B
The van’t Hoff factor i represents the number of particles formed from one formula unit, relative to the undissociated solute. On complete dissociation, CaCl₂ produces one Ca²⁺ ion and two Cl⁻ ions, giving three particles; therefore i = 3. NaCl gives only two ions, while glucose is a non-electrolyte and benzoic acid in benzene commonly associates.
For an uncharged and non-associated solute like glucose, what is usually taken as i?
Correct answer: B
Glucose is a molecular, non-electrolyte solute in ordinary dilute solution. It dissolves as whole glucose molecules rather than splitting into ions, and it does not normally form dimers or larger associated particles under these conditions. Thus the number of solute particles remains unchanged, so the van’t Hoff factor is i = 1. Values above one indicate dissociation, whereas values below one indicate association.
In which case will the observed osmotic pressure be less than the normal value?
Correct answer: B
Osmotic pressure depends on the number of solute particles and is represented by π = iCRT. Association combines particles, so i becomes less than one and the effective particle concentration decreases. Consequently, the observed osmotic pressure is lower than the value calculated for unchanged solute molecules. Dissociation or ionization would instead increase it.
If the experimental value of i for a solute is 1.25, what type of behavior is possible?
Correct answer: A
For a solute that neither associates nor dissociates, i is approximately 1. Here i = 1.25, which is greater than one, so the number of particles has increased. Because the value is well below the ideal value for complete dissociation into several particles, partial or slight dissociation is the suitable interpretation. Association would instead give i less than one.
If i = 0.9, which statement about the observed value of a colligative property is correct?
Correct answer: B
For an abnormal solution, the measured colligative effect equals i times the value calculated for an ideal, non-associating and non-dissociating solute. With i = 0.9, the effect is 0.9 of the normal value, so it is slightly smaller, not zero. A value below one usually reflects association, which reduces the number of solute particles.
What is the safest first check in questions related to abnormal molar mass?
Correct answer: B
Abnormal molar mass arises when the number of solute particles differs from the number expected from the formula. The main causes in this chapter are association, which reduces particles, and dissociation, which increases them. Therefore, first check whether either process can occur; then use i and the appropriate colligative-property relation. Colour, vessel shape, and shaking are irrelevant.
If the number of particles in an electrolyte solution is found to be more than expected, what will happen to its observed molecular mass?
Correct answer: A
A larger number of particles produces a larger colligative effect at the same amount of solute. If that effect is interpreted using the unchanged-molecule formula, the calculated molar mass must appear smaller, because the effect is inversely related to molar mass in such calculations. Thus dissociation generally lowers observed molecular mass; it does not make it zero.
Due to association, the number of solute particles decreases. How will the observed molecular mass change?
Correct answer: A
Association combines separate solute molecules into larger units, so fewer particles are present than the formula suggests. The resulting colligative effect is smaller. When the smaller effect is used to calculate molar mass, the value appears larger than the normal molar mass. Therefore association gives a higher observed molecular mass, while dissociation gives a lower one.
If i > 1, what is the correct conclusion about the solute?
Correct answer: A
The van't Hoff factor compares the actual number of solute particles with the number expected if each formula unit remained intact. When i > 1, more particles are present than expected. The usual cause is dissociation into ions or smaller species. Association would reduce the particle count and give i < 1, so option A is the only suitable conclusion.
If i < 1, which process is most likely occurring in the solution?
Correct answer: A
The van’t Hoff factor compares the actual number of dissolved particles with the number expected from the formula units. When i is less than one, fewer particles are present than expected. This occurs when two or more solute molecules combine to form larger associated species, such as dimers. Dissociation would increase particles and give i greater than one.
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