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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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Medium · Level 7View options
1.5
0.5
2.5
1.0
Medium · Level 7View options
1.8
1.4
2.0
0.8
Medium · Level 7View options
1.5
1.25
2.25
0.75
Medium · Level 7View options
25%
50%
75%
125%
Medium · Level 7View options
60%
40%
80%
20%
Medium · Level 7View options
0.70
1.60
0.40
1.30
Medium · Level 7View options
i = 3, dissociation
i = 1/3, association
i = 1, no change
i = 0, precipitation
Medium · Level 7View options
2.5
1.5
3.5
4.0
Medium · Level 7View options
Whether the number of solute particles changed
Whether solvent colour changed
Whether the container is glass or metal
Whether the solution was stirred
Medium · Level 7View options
Dissociation
Association
No change
Crystallization
Medium · Level 7View options
Solute particles have associated
Solute has completely dissociated into ions
Particle number has doubled
Only solvent remains in solution
Medium · Level 7View options
It becomes lower than normal
It becomes higher than normal
It always remains normal
It becomes impossible to measure
Medium · Level 7View options
Greater than normal
Less than normal
Always zero
Independent of concentration
Medium · Level 7View options
i = normal molecular mass / observed molecular mass
i = observed molecular mass / normal molecular mass
i = product of the two masses
i = observed mass − normal mass
Medium · Level 7View options
Association through hydrogen bonding
Complete ionic dissociation
Formation of water
Formation of metal ions
Medium · Level 7View options
It dissociates into K⁺ and Cl⁻
It forms a dimer
It solidifies the solvent
It changes colour
Medium · Level 7View options
0.5
2
1
1.5
Medium · Level 7View options
5
3
2
6
Medium · Level 7View options
Between 1 and 2
Between 0 and 1
Exactly 2
Greater than 2
Medium · Level 7View options
1.50
1.25
2.25
0.75
Medium · Level 7View options
1.60
1.20
2.20
0.80
Medium · Level 7View options
30%
60%
20%
80%
Medium · Level 7View options
i = 1 − α/2
i = 1 + α/2
i = 1 + 2α
i = 2 + α
Medium · Level 7View options
Acetic acid in benzene
NaCl in water
KBr in water
MgCl₂ in water
Medium · Level 7View options
NaNO₃ in water
Acetic acid in benzene
Benzoic acid in benzene
Glucose in water
Question 1MediumLevel 7
If an AB-type electrolyte is 50% dissociated, what is the value of the van’t Hoff factor, i?
Correct answer: A
For AB → A + B, complete dissociation would change one particle into two, so the increase is one particle per dissociated formula unit. The relation is i = 1 + α, where α is the fraction dissociated. Here, 50% means α = 0.50; therefore i = 1 + 0.50 = 1.50. The value 1.0 would represent no dissociation, not 50% dissociation.
If an AB₂-type electrolyte is 40% dissociated, what is the value of the van’t Hoff factor, i?
Correct answer: A
For the dissociation AB2 → A + 2B, one formula unit produces three particles on complete dissociation. Therefore, for fractional dissociation α, the governing expression is i = 1 + (3 − 1)α = 1 + 2α. With α = 40/100 = 0.40, i = 1 + 2(0.40) = 1.80. Option B incorrectly adds α only once, so option A is correct.
If an A₂B-type salt is 25% dissociated, what is the value of the van’t Hoff factor, i?
Correct answer: A
The ideal dissociation is A₂B → 2A + B, so one formula unit gives three particles when dissociation is complete. For a compound producing n particles, i = 1 + (n − 1)α. Here n = 3 and α = 25/100 = 0.25, so i = 1 + 2(0.25) = 1.50. The other values result from miscounting particles or using the percentage incorrectly.
If i = 1.25 for an AB-type solute, what is the degree of dissociation?
Correct answer: A
For an AB electrolyte that dissociates into two ions, i = 1 + α, where α is the fraction dissociated. Rearranging gives α = i − 1 = 1.25 − 1 = 0.25. Multiplying by 100 converts the fraction to a percentage: 25%. A value of 125% is impossible because a degree of dissociation cannot exceed 100%.
For an AB₂ electrolyte, if the van’t Hoff factor is i = 2.2, what is the degree of dissociation?
Correct answer: A
For AB₂ → A + 2B, three particles can form from one formula unit, so i = 1 + (3 − 1)α = 1 + 2α. Substituting i = 2.2 gives 2.2 = 1 + 2α, hence 2α = 1.2 and α = 0.6. Converting the fraction to a percentage gives 60%. The other percentages do not satisfy the equation.
If a substance undergoes 60% dimerization, what is the van’t Hoff factor, i?
Correct answer: A
In dimerization, two solute molecules combine to form one dimer: 2A → A₂. If α is the fraction of molecules undergoing dimerization, the number of particles decreases by α/2 per original molecule, giving i = 1 − α/2. For 60% dimerization, α = 0.60, so i = 1 − 0.60/2 = 1 − 0.30 = 0.70. Association therefore gives i below one.
If the osmotic pressure of a solution is three times the expected value, what are i and the process?
Correct answer: A
At fixed concentration, temperature, and gas constant, osmotic pressure is directly proportional to the van’t Hoff factor: π = iCRT. If the observed pressure is three times the expected value calculated with i = 1, then i must be 3. A factor greater than one means that the number of dissolved particles has increased, which indicates dissociation rather than association or no change.
If an AB₃-type electrolyte is 50% dissociated, what is the van’t Hoff factor (i)?
Correct answer: A
The governing concept is the van’t Hoff factor, which represents the effective number of solute particles produced by dissociation. For an AB₃ electrolyte, the dissociation is AB₃ → A + 3B, so one formula unit can produce four particles when dissociation is complete. If α is the degree of dissociation, the factor is i = 1 + (ν − 1)α, where ν = 4. Thus i = 1 + 3α. For 50% dissociation, α = 0.50, so i = 1 + 3(0.50) = 2.5. Therefore option A is correct. The value 4.0 would require complete dissociation. The value 1.5 would correspond to an electrolyte producing only two particles at complete dissociation, and 3.5 results from an incorrect substitution or overestimating the extent of dissociation.
What should be checked first when explaining abnormal molecular mass?
Correct answer: A
Colligative properties depend primarily on the number of solute particles, not on their individual chemical identity. Association decreases the number of particles, whereas dissociation increases it. Therefore, the first question is whether the solute particle count has changed. Colour, container material, and stirring do not explain abnormal molecular mass in this context.
A solute has a normal molecular mass of 98 and an observed molecular mass of 49. Which process may be occurring?
Correct answer: A
Calculate the factor from i = normal molecular mass / observed molecular mass = 98 / 49 = 2. A factor greater than one means that the number of solute particles has increased compared with the undissociated state. Such an increase is caused by dissociation, in which one solute unit splits into two or more particles. Association would give i less than one.
A solute has a normal molecular mass of 60 and an observed molecular mass of 120. What is the most suitable explanation?
Correct answer: A
Using i = normal molecular mass / observed molecular mass, i = 60 / 120 = 0.5. A value below one indicates fewer particles than expected from the original solute units. The usual explanation is association; for example, two particles may combine to form one larger particle. Complete dissociation and doubled particle number would instead produce i greater than one.
When solute particles break into ions, how does the observed molecular mass change?
Correct answer: A
Dissociation increases the number of particles in solution. The resulting colligative effect is larger than expected for an undissociated solute. When molecular mass is calculated from that larger effect without correction, the value obtained is smaller than the normal molecular mass. Thus dissociation produces a lower observed molecular mass, while association produces a higher one.
If two solute molecules combine to form one larger particle, how does observed molecular mass behave?
Correct answer: A
When two molecules associate into one particle, the total number of solute particles decreases. Colligative effects consequently become smaller than expected for the original number of molecules. Calculation without an association correction then gives a molecular mass larger than the normal value. Therefore, association leads to an increased observed molecular mass.
Which formula correctly relates i to normal and observed molecular masses?
Correct answer: A
For a colligative-property measurement, the van’t Hoff factor is defined as the ratio of the normal molecular mass to the molecular mass calculated from the observed effect. Hence i = M normal / M observed. This relation correctly gives i greater than one for dissociation and less than one for association; reversing the ratio would give the opposite interpretation.
Why can benzoic acid in benzene show a molecular mass greater than normal?
Correct answer: A
Benzoic acid molecules can form hydrogen-bonded dimers in a non-polar solvent such as benzene. Two molecules then behave as one solute particle, reducing the particle number responsible for colligative effects. The reduced effect gives an observed molecular mass greater than the normal monomeric mass. Ionic dissociation and metal-ion formation are not involved.
Why can KCl in water show an observed molecular mass lower than normal?
Correct answer: A
KCl is an ionic electrolyte and, in water, it separates into K⁺ and Cl⁻ ions. Thus one formula unit produces two solute particles instead of one, increasing the colligative effect. If this enhanced effect is interpreted without correcting for dissociation, the calculated molecular mass is lower than the normal value. Dimer formation would have the opposite effect.
If a substance undergoes complete dimerization, what is the value of the van’t Hoff factor (i)?
Correct answer: A
The governing concept is association and its effect on the number of solute particles. In complete dimerization, two monomer molecules combine according to 2A → A₂. Consequently, two original particles become one dimer particle, so the final particle count is half the ideal non-associating count. Hence i = actual particles/expected particles = 1/2 = 0.5. Values above one describe dissociation, not complete association.
What is the ideal van’t Hoff factor (i) for Al₂(SO₄)₃ on complete dissociation?
Correct answer: A
The governing concept is stoichiometric particle counting during complete dissociation. Aluminium sulfate separates according to Al₂(SO₄)₃ → 2Al³⁺ + 3SO₄²⁻. The subscript 2 gives two aluminium ions, and the coefficient 3 before sulfate gives three sulfate ions. Thus the total number of particles is 2 + 3 = 5, so the ideal van’t Hoff factor is i = 5. The charge values do not replace the stoichiometric coefficients.
For incomplete dissociation of an AB-type solute, in which range will (i) lie?
Correct answer: A
For an AB solute, the governing relation for dissociation is i = 1 + α, because each dissociated AB unit changes from one particle into two. In incomplete dissociation, the degree of dissociation satisfies 0 < α < 1. Therefore 1 < i < 2. The value i = 2 applies only when dissociation is complete, while a value below one is associated with particle association rather than dissociation.
If an AB₂-type solute is 25% dissociated, what is the value of (i)?
Correct answer: A
The governing formula for a solute producing n particles on complete dissociation is i = 1 + (n − 1)α. For AB₂ → A + 2B, n = 3, so i = 1 + 2α. The stated 25% dissociation gives α = 0.25. Therefore i = 1 + 2(0.25) = 1.50. The factor 2 represents the increase from one original particle to three particles, not the percentage itself.
If an AB₃-type solute is 20% dissociated, what is the value of (i)?
Correct answer: A
For AB₃ → A + 3B, complete dissociation changes one solute particle into four, so the increase in particle number is 3. The appropriate relation is i = 1 + 3α. With 20% dissociation, α = 0.20, giving i = 1 + 3 × 0.20 = 1.60. The value 1.20 would incorrectly count only one additional particle, while 4 would represent complete rather than partial dissociation.
If i = 1.6 for AB₂, what is the degree of dissociation?
Correct answer: A
For AB₂ → A + 2B, complete dissociation produces three particles from one, so the van’t Hoff relation is i = 1 + 2α. Substituting the given value gives 1.6 = 1 + 2α. Hence 2α = 0.6 and α = 0.3. Converting the decimal fraction to a percentage, α × 100 = 30%. Therefore option A is correct; 60% incorrectly omits the factor of two in the relation.
If the degree of dimerization is α, which relation for the van’t Hoff factor (i) is correct?
Correct answer: A
The governing concept is particle reduction by association. For dimerization, 2A → A₂. Consider one initial mole: α moles of monomer participate, forming α/2 moles of dimers, while 1 − α moles remain as monomers. The total final particles are (1 − α) + α/2 = 1 − α/2 moles. Dividing by the initial one-mole particle count gives i = 1 − α/2, so option A is correct.
Which example can show abnormal molecular mass due to association?
Correct answer: A
Acetic acid molecules can form hydrogen-bonded dimers in a non-polar solvent such as benzene. Dimer formation reduces the number of independent solute particles and can therefore produce an observed molecular mass higher than the normal value. The aqueous salts listed are ordinarily discussed as electrolytes that dissociate into ions, giving i greater than one rather than association.
In which solution is i most likely to be greater than 1?
Correct answer: A
NaNO₃ is an ionic electrolyte and in water it dissociates approximately into Na⁺ and NO₃⁻ ions. This increases the number of solute particles, so i is greater than one, especially under the ideal assumption. Acetic and benzoic acids in benzene can associate, while glucose is a non-electrolyte and normally has i close to one.
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