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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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Partial association
Complete dissociation
Increase in particles
Formation of three ions
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A change in the number of solute particles from the expected value
A change in the colour of the glass container
A change in the size of the thermometer
Only shaking the solution
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KCl
Urea
Glucose
Sucrose
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Three
Two
One
Half
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Whether the solute associates or dissociates
The colour of the solution
The metal of the container
The colour used to print the question
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Different from the normal value
Always zero
Always double
Always unchanged
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The effective number of particles in solution
The colour of the solvent
The shape of the container
The sound during dissolving
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Dissociation or association
Only the name of the solvent
Thickness of the vessel
Direction of a clock
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Dissociation
Association
Only precipitation
No particles
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Association
Complete dissociation
Increase in particles
Only evaporation
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One
Zero
Two
Half
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Due to dissociation into ions
Due to joining of molecules
Because the solvent disappears
Because the salt remains solid
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Three times
Half
Zero
One third
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2
1
0.5
3
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3
2
1
0.33
Easy · Level 4View options
0.5
1
2
4
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Dissociation can make i greater than one
Association always makes i equal to two
If particles do not change, i is zero
The van’t Hoff factor is unrelated to colligative properties
Easy · Level 4View options
0.5
1
2
1.5
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i
K
M
T
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The fraction of solute that has dissociated
The total mass
Only the volume of solvent
The intensity of colour
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1
2
0
0.5
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2
1
0.5
3
Easy · Level 4View options
Whether the number of solute particles changes
The colour of the vessel
The size of the laboratory
Who prepared the solution
Easy · Level 4View options
Lower observed molar mass
Higher observed molar mass
Lower particle number
A van’t Hoff factor less than one
Easy · Level 4View options
Divide normal molar mass by observed molar mass
Add temperature to observed molar mass
Multiply pressure by colour
Change the solvent name
Question 1EasyLevel 4
If i = 0.80, what process may it indicate?
Correct answer: A
For a solute that neither associates nor dissociates, the van’t Hoff factor is 1. When i is below 1, the effective number of solute particles is lower than expected. This usually happens because two or more solute molecules combine to form larger associated species. Since i = 0.80 is below 1 but not necessarily the limiting value for complete association, it most appropriately indicates partial association. Dissociation would make i greater than 1.
Which option gives the correct reason for abnormal molar mass?
Correct answer: A
Abnormal molar mass arises when the solute does not remain as independent formula units in solution. Dissociation increases the particle count, whereas association decreases it. Since colligative measurements respond to particle number, the calculated molar mass differs from the normal value; container colour and shaking are irrelevant.
Which substance generally shows i > 1 in aqueous solution because of dissociation?
Correct answer: A
KCl is an ionic compound and, when dissolved in water, it separates into K⁺ and Cl⁻ ions. Thus one formula unit gives approximately two solute particles, making i greater than 1 in the ideal limiting case. Urea, glucose and sucrose are molecular nonelectrolytes and normally remain as individual molecules, so their i is close to 1.
If i = 3 and complete dissociation is assumed, how many particles are formed from one formula unit?
Correct answer: A
Under the ideal assumption of complete dissociation, one formula unit separates entirely into its constituent particles, and the van’t Hoff factor equals the number of particles produced. Therefore, i = 3 means that one formula unit gives three particles in solution. For example, a formula unit that produces three ions would have an ideal i value of 3. The values 1 and 2 would correspond to one or two particles, not three.
In an abnormal molar mass problem, what should be identified first?
Correct answer: A
The first step is to determine whether the solute particles split apart or join together in solution. Dissociation increases the effective particle number and usually gives i > 1, whereas association decreases it and gives i < 1. This decision tells you which relation and direction of molar-mass change to use.
When the number of solute particles changes from the expected value, how may the molar mass appear?
Correct answer: A
Colligative-property measurements use the effective number of dissolved particles. If dissociation increases that number, the calculated molar mass generally appears lower; if association decreases it, the calculated value appears higher. Therefore it can differ from the normal molar mass, but it is not always zero, double or unchanged.
The van’t Hoff factor is most helpful in explaining which change?
Correct answer: A
The van’t Hoff factor compares the actual effective particle count with the count expected if each formula unit remained unchanged. It therefore explains changes in colligative properties and abnormal molar mass caused by association or dissociation. It has no role in describing container shape, colour or sound.
Abnormal molar mass is a consequence of a changed number of dissolved particles. Dissociation breaks one species into several particles, while association combines several species into fewer particles. Because colligative properties depend on particle number, either process causes the molar mass calculated from such properties to differ from the normal value.
If the van’t Hoff factor is greater than one, what does it generally indicate?
Correct answer: A
i greater than 1 means that the solution contains more effective particles than expected from the undissociated solute. The usual reason is dissociation into ions or smaller molecules. Association would reduce the particle count and make i less than 1. The conclusion assumes no other unusual process dominates the measurement.
If the van’t Hoff factor is less than one, which process is generally indicated?
Correct answer: A
The van’t Hoff factor measures the effective change in the number of solute particles. If there is no association or dissociation, i equals 1. Association joins several solute molecules into fewer larger species, so the effective particle number decreases and i becomes less than 1. Dissociation has the opposite effect and generally makes i greater than 1. Therefore, i < 1 generally indicates association, although the exact value depends on its extent.
If a solute neither dissociates nor associates, what is the van’t Hoff factor?
Correct answer: A
The van’t Hoff factor is defined as the ratio of the actual effective number of solute particles to the number expected from the original formula units. If the solute neither dissociates nor associates, every formula unit remains one particle in solution. Thus the numerator and denominator are equal, giving i = 1. A value above 1 suggests dissociation, while a value below 1 suggests association; neither occurs in this case.
Why does the number of particles generally increase when sodium chloride dissolves in water?
Correct answer: A
Sodium chloride is an ionic compound. In water, polar water molecules separate and hydrate its ions, so the crystal units dissociate according to NaCl → Na⁺ + Cl⁻. One formula unit therefore gives two dissolved particles rather than one formula unit remaining as a single unit. This increases the effective particle number and can increase colligative effects. The solvent does not disappear, and undissolved solid salt would not produce this dissolved-particle increase.
If i = 3, how does the effective particle number compare with the normal expectation?
Correct answer: A
The van’t Hoff factor is the ratio of the effective number of particles in solution to the number expected from the original solute units. Thus i = 3 means effective particle number divided by normal particle number equals 3. For the same amount of solute and comparable conditions, the solution behaves as though it contains three times as many particles. This can result from complete dissociation into three particles, although the numerical value alone does not prove the exact mechanism.
What is the ideal van’t Hoff factor for a completely dissociating solute that gives two ions?
Correct answer: A
Under complete dissociation, every formula unit separates into all of its constituent ions. If the substance gives two ions, the number of particles changes from one formula unit to two ions. Hence the ideal van’t Hoff factor is i = 2. The value 1 applies to a non-dissociating solute, while 0.5 applies to complete dimerisation.
What is the ideal van’t Hoff factor for a completely dissociating solute that gives three ions?
Correct answer: A
For complete dissociation, every formula unit separates into all of the particles represented by its dissociation equation. If one formula unit produces three ions, the number of particles changes from one to three. The ideal van’t Hoff factor is therefore i = 3. A value of 2 would apply to complete dissociation into two ions, i = 1 to no change in particle number, and 0.33 would represent a reduction rather than three-particle dissociation.
When two molecules combine to form one dimer, what is i for complete association?
Correct answer: A
In complete dimerisation, two original solute molecules behave as one particle. Thus the number of effective particles becomes one-half of the initial number. Since i is the effective-to-original particle ratio, i = 1/2 = 0.5. Partial dimerisation would give a value between 0.5 and 1.
Dissociation converts one solute unit into two or more independent particles, so the effective particle ratio can exceed one. Association instead lowers the ratio, and no change gives i = 1. The van’t Hoff factor is directly used in colligative-property equations, so the other statements contradict the definition.
If the effective number of particles in a solution becomes half, what will be the value of i?
Correct answer: A
The van’t Hoff factor is the ratio of the effective particle number after association or dissociation to the normal particle number before that change. If the effective number becomes half of the normal number, this ratio is 1/2. Therefore, i = 0.5. Such a reduction is consistent with association, in which several original molecules combine into fewer larger species. i = 1 would mean no change, whereas i = 2 would mean a doubling of particles.
Which symbol is used to represent the van’t Hoff factor?
Correct answer: A
The van’t Hoff factor is conventionally represented by the lowercase italic symbol i. It measures the change in the effective number of solute particles caused by dissociation or association. M is commonly used for molar mass, T for temperature, and K often denotes a constant, so those symbols do not represent the van’t Hoff factor in this context.
The degree of dissociation, α, is the fraction of the original solute units that have separated into smaller particles or ions. It has no unit and normally lies between 0 and 1. For example, 25% dissociation means α = 25/100 = 0.25. It is used in expressions such as i = 1 + (n − 1)α.
For dissociation into two ions, if α = 0, what is i?
Correct answer: A
For a solute that can dissociate into two ions, the relation is i = 1 + α. Substituting α = 0 gives i = 1 + 0 = 1. This means that no solute unit has dissociated, so the particle number remains at the normal value. The result is not zero because the original undissociated particles are still present.
For complete dissociation into two ions, what is i when α = 1?
Correct answer: A
For dissociation into two ions, the van’t Hoff factor is i = 1 + α. Here α is the fraction of solute units that dissociate. Complete dissociation means every unit dissociates, so α = 1. Substitution gives i = 1 + 1 = 2. This agrees with the particle count: one original formula unit produces two ions. A value of 1 would indicate no dissociation, while 0.5 would indicate fewer effective particles.
To understand abnormal molar mass, what should be checked first?
Correct answer: A
Abnormal molar mass is obtained when a colligative-property measurement is affected by a change in the number of solute particles. Therefore, first check whether the solute dissociates into smaller particles or associates into larger units. The vessel colour, laboratory size and operator do not determine this molecular effect.
What result is expected from dissociation of an ionic substance such as potassium chloride in solution?
Correct answer: A
In water, KCl can produce K⁺ and Cl⁻ ions, increasing the number of solute particles. The larger particle count increases the colligative effect, so the molar mass calculated from that effect becomes lower than the normal value. Thus dissociation gives i greater than one and a lower observed molar mass.
What is a simple way to find the van’t Hoff factor i in an abnormal molar-mass problem?
Correct answer: A
For abnormal molar mass, i is calculated as normal molar mass divided by observed molar mass: i = Mnormal/Mobserved. Both masses must be expressed in the same units. The factor compares the expected number of particles with the effective number indicated by the colligative property; temperature, colour, pressure, or solvent naming is not part of this direct relation.
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