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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 6View options
Half the normal molar mass
Equal to the normal molar mass
Double the normal molar mass
Zero
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Between 0 and 1
Between 1 and 2
Between 2 and 3
Always greater than 2
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Between 1 and 2
Between 0.5 and 1
Between 2 and 3
Always above 1
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Total particles formed from one formula unit
Number of solvent molecules
Temperature of the solution
Value of pressure
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1.2
1.5
1.8
2.8
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1.5
2.0
2.5
3.0
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0.3
0.4
0.6
1.6
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Number of molecules combining to form one larger particle
Number of solvent molecules breaking
Volume of the solution
Number of temperatures
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0.25
0.50
0.75
1.50
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0.2
0.4
0.6
0.8
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Dissociation of the solute
Association of the solute
The solvent stops freezing
The solute being coloured
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Complete dissociation
Partial association
Increase in the number of ions
Doubling of particles
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π = iCRT
π = CRT/i
π = i + CRT
π = C + R + T + i
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ΔTf = Kf m
ΔTf = iKf m
ΔTf = Kf/m
ΔTf = i + Kf + m
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Complete dimer association
Complete dissociation
Half the mass dissolving
Evaporation of solvent
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60 g mol−1
80 g mol−1
120 g mol−1
180 g mol−1
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67.5 g mol−1
90 g mol−1
120 g mol−1
135 g mol−1
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Dissociation is incomplete
Association is complete
The solute did not dissolve at all
The solvent dissociated into ions
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Molar mass will always be correct
An abnormal or incorrect molar mass may be obtained
Temperature will become zero
Solvent mass will change
Medium · Level 6View options
The solute is completely associated
The solute is partially dissociated
The solute is ideal
The solute gives no ions at all
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Dissociation
Dimer association
Complete ionization
Breaking into three ions
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1/3
1
3
0.5
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i = normal molar mass / observed molar mass
i = observed molar mass / normal molar mass
i = normal molar mass + observed molar mass
i = normal molar mass − observed molar mass
Medium · Level 6View options
Association due to hydrogen bonding
Complete ionization
Formation of salt
Freezing of the solvent
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Between 1 and 2
Exactly 0
Exactly 3
Less than 1
Question 1MediumLevel 6
If a solute has i = 0.5, which statement about its observed molar mass is correct?
Correct answer: C
Using i = Mnormal/Mobserved, substitute i = 0.5: 0.5 = Mnormal/Mobserved. Rearranging gives Mobserved = 2Mnormal. Thus the observed molar mass is twice the normal value. Such a result is consistent with strong association, for example complete dimerization, which reduces the number of effective particles.
If KCl is partially dissociated, in which range will the van’t Hoff factor, i, lie?
Correct answer: B
If KCl does not dissociate, each formula unit remains one particle and i = 1. On complete dissociation, KCl → K⁺ + Cl⁻ gives two particles, so i = 2. Partial dissociation lies between these limiting cases; consequently, for a nonzero but incomplete degree of dissociation, 1 < i < 2. It cannot exceed the complete-dissociation limit of 2.
If a substance partially forms dimers, in which range will i lie?
Correct answer: B
If no molecules associate, each remains separate and i = 1. If dimerization is complete, two molecules become one particle and i = 0.5. Partial dimerization gives an intermediate effective particle number, so the van’t Hoff factor lies between 0.5 and 1. It cannot exceed 1 for association alone.
In the dissociation relation i = 1 + α(n − 1), what does n represent?
Correct answer: A
In the dissociation equation, α is the fraction of solute units that dissociate, while n is the total number of particles produced from one dissociating unit. For NaCl, n = 2; for CaCl₂, n = 3. Substituting this value gives the increase in particle number at a chosen degree of dissociation. It is not a temperature, pressure, or solvent quantity.
If the degree of dissociation of NaCl is α = 0.8, what is the van’t Hoff factor, i?
Correct answer: C
The governing concept is the van’t Hoff factor for partial electrolyte dissociation. NaCl gives two ions, Na+ and Cl−, on complete dissociation, so n = 2. For degree of dissociation α, i = 1 + α(n − 1). Substitution gives i = 1 + 0.8(2 − 1) = 1.8. The limiting values are 1 for no dissociation and 2 for complete dissociation, so option C is correct.
If the degree of dissociation of CaCl₂ is α = 0.5, what is the van’t Hoff factor, i?
Correct answer: B
CaCl2 dissociates as CaCl2 → Ca2+ + 2Cl−, producing n = 3 particles when complete. For partial dissociation, the governing equation is i = 1 + α(n − 1). With α = 0.5, i = 1 + 0.5(3 − 1) = 2.0. The value 1.5 uses an incomplete particle count, while 3.0 applies only to complete dissociation. Therefore, option B is correct.
If i = 1.6 and an AB-type solute dissociates into A⁺ and B⁻, what is the degree of dissociation α?
Correct answer: C
AB dissociates into two ions, so n = 2. The dissociation relation becomes i = 1 + α(2 − 1) = 1 + α. With i = 1.6, α = 1.6 − 1 = 0.6. Thus 60% of the AB units are dissociated under the stated conditions. The value 1.6 is the particle factor, not the degree of dissociation.
In the association relation i = 1 − α(1 − 1/n), what does n mean?
Correct answer: A
In association, n original solute molecules combine to form one associated particle. Thus n describes the size of the associated unit: n = 2 for a dimer and n = 3 for a trimer. The term (1 − 1/n) measures the fractional decrease in particle number on complete association. Temperature, pressure, and solution volume are not represented by n in this equation.
If a substance undergoes 50% dimerization, what is the van’t Hoff factor, i?
Correct answer: C
Dimerization is association in which two solute particles combine to form one particle. If α is the fraction dimerized, the particle factor is i = 1 − α(1 − 1/2) = 1 − α/2. For α = 50% = 0.5, i = 1 − 0.5/2 = 0.75. Since association lowers the number of particles, i must be below 1; hence option C is correct.
If i = 0.8 for a solute that forms dimers, what is the degree of association, α?
Correct answer: B
For dimer formation, two particles become one, so the van’t Hoff factor follows i = 1 − α/2. Using the given i value, 0.8 = 1 − α/2. Therefore α/2 = 0.2 and α = 0.4, meaning 40% association. The values 0.2 and 0.6 arise from incomplete rearrangement, while 0.8 incorrectly treats i as α. Thus, option B is correct.
If the observed depression in freezing point is greater than the theoretical value, what is the most likely reason?
Correct answer: A
Freezing-point depression is a colligative property, so it increases with the number of dissolved particles. If the observed depression is greater than the value calculated for undissociated solute, more particles must be present than expected. Dissociation produces additional ions and gives i > 1, increasing ΔTf. Association would reduce the number of particles and make the depression smaller.
If the observed elevation in boiling point is less than the expected value, which process is possible?
Correct answer: B
Elevation in boiling point depends on the number of dissolved particles. A value smaller than expected means that fewer effective particles are present than assumed in the normal calculation, so i < 1. Association combines solute molecules into larger units and reduces particle number. Partial association is therefore possible; complete dissociation or particle doubling would instead increase the boiling-point elevation.
Which relation is correct for abnormal behavior in osmotic pressure?
Correct answer: A
For a dilute solution, the ideal osmotic-pressure equation is π = CRT. If association or dissociation changes the effective number of solute particles, the van’t Hoff factor corrects the equation by multiplication: π = iCRT. It is not placed in the denominator or added, because particle concentration directly scales the pressure.
Which relation is correct for freezing-point depression in an abnormal solution?
Correct answer: B
For an ideal non-associating, non-dissociating solute, ΔTf = Kf m. When the number of particles is abnormal, the ideal expression is multiplied by the van’t Hoff factor, giving ΔTf = iKf m. Association makes i less than 1, while dissociation makes i greater than 1, changing the magnitude accordingly.
When the observed molar mass of an acid in benzene is nearly double, what may be the reason?
Correct answer: A
If the observed molar mass is about twice the normal value, i = normal/observed is about 1/2. Complete dimerization gives i = 0.5 because two acid molecules behave as one particle. Carboxylic acids such as benzoic acid commonly form hydrogen-bonded dimers in non-polar benzene, explaining the observation.
If normal molar mass is 120 g mol−1 and i = 1.5, what is the observed molar mass?
Correct answer: B
Rearrange i = normal molar mass/observed molar mass to get observed molar mass = normal molar mass/i. Therefore, Mobserved = 120/1.5 = 80 g mol−1. Since i is greater than 1, the result is appropriately lower than the normal molar mass, indicating an increase in particles such as by dissociation.
If normal molar mass is 90 g mol−1 and i = 0.75, what is the observed molar mass?
Correct answer: C
Use Mobserved = Mnormal/i. Substitution gives Mobserved = 90/0.75 = 120 g mol−1. Dividing by a number below 1 increases the result, which agrees with the general rule that i less than 1 produces a higher observed molar mass. The lower i indicates association of solute particles.
If the i value of an electrolyte is less than its ideal complete-dissociation value, what does it mean?
Correct answer: A
For an electrolyte producing n ions, ideal complete dissociation gives i = n. If the measured i is lower than n but generally above 1, only a fraction of the formula units has dissociated, so dissociation is incomplete. The lower value does not by itself prove complete association or absence of dissolution.
What error may occur if i is ignored while calculating molar mass from colligative properties?
Correct answer: B
Colligative properties depend on the actual number of dissolved particles. If a solute associates or dissociates and the calculation assumes i = 1, the measured effect is interpreted incorrectly. The resulting molar mass may be too high for association or too low for dissociation, so ignoring i can produce an abnormal value.
If i lies between 1 and 3 for an AB2 solute, which conclusion is correct?
Correct answer: B
For AB2, no dissociation gives one effective particle and i = 1. Complete ideal dissociation gives three ions and i = 3. Therefore, a value strictly between 1 and 3 indicates that some, but not all, formula units have dissociated. This is partial dissociation; it is not complete association or absence of ion formation.
If a substance has observed molar mass 200 g mol⁻¹ and normal molar mass 100 g mol⁻¹, which process is possible?
Correct answer: B
The relation is i = normal molar mass / observed molar mass. Here i = 100/200 = 0.5. For complete dimerization, two original molecules combine to form one associated particle, giving i = 1/2. Thus the doubled observed molar mass is consistent with dimer association, whereas dissociation would make the observed mass smaller.
If the observed molar mass is one-third of the normal molar mass, what is i?
Correct answer: C
For abnormal molar mass, the relation is i = normal molar mass / observed molar mass. Let the normal molar mass be M. The observed value is M/3, so i = M ÷ (M/3) = 3. A value greater than one means that the effective number of solute particles has increased, as happens during dissociation.
Which relation correctly represents the van't Hoff factor i?
Correct answer: A
When a colligative property is used to calculate molar mass without considering association or dissociation, the result is the observed molar mass. The correction factor is i = Mnormal/Mobserved. Thus i greater than one corresponds to a smaller observed mass, as in dissociation, while i below one corresponds to a larger observed mass, as in association.
The molecular mass of acetic acid in benzene is found to be higher than normal. What is the main reason?
Correct answer: A
Acetic acid molecules can form hydrogen-bonded dimers in the relatively non-polar solvent benzene. Two molecules then behave as one associated particle, reducing the number of particles in solution. The smaller colligative effect leads to a molar mass that appears higher than the normal value. Ionization and salt formation are not the relevant processes here.
For NaCl under incomplete dissociation, how will the value of i behave?
Correct answer: A
NaCl gives two ions, Na⁺ and Cl⁻, when dissociation is complete, so the limiting value is i = 2. If dissociation is incomplete, some NaCl remains as undissociated units while the rest forms ions. The particle number therefore increases above one but does not reach two, giving 1 < i < 2. It is not less than one because association is not being described.
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