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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 9View options
200
50
100
150
Easy · Level 9View options
The number of solute particles changes
The solvent colour changes
The container shape changes
The solution is filtered
Easy · Level 9View options
3
2
4
1
Easy · Level 9View options
1.30
0.30
1.60
2.30
Easy · Level 9View options
75%
25%
50%
100%
Easy · Level 9View options
0.80
1.40
0.60
1.20
Easy · Level 9View options
Greater than 1
Less than 1
Exactly 1
Exactly 0
Easy · Level 9View options
i < 1 and association is possible
i > 1 and dissociation is certain
i = 1 and behavior is normal
i = 0 and no particles exist
Easy · Level 9View options
They depend on the number of particles
They depend only on molecular colour
They depend only on manometer size
They depend only on solvent smell
Easy · Level 9View options
When i = 1
When i = 0.5
When i = 2
When i = 3
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One-fourth
Four times
Half
Equal
Easy · Level 9View options
It will be greater than normal
It will be less than normal
It will be equal to normal
It will be zero
Easy · Level 9View options
Dimer
Trimer
Tetramer
Pentamer
Easy · Level 9View options
The actual number of solute particles changes by association or dissociation
The solvent is always non-volatile
Temperature is never measured
Molar mass has no meaning
Easy · Level 9View options
i = 3, dissociation
i = 1/3, association
i = 1, normal behaviour
i = 0, no particles
Easy · Level 9View options
108 g mol−1
180 g mol−1
300 g mol−1
60 g mol−1
Easy · Level 9View options
1:5
5:1
1:1
2:5
Easy · Level 9View options
Complete dimer association
Complete dissociation into two ions
No change
Complete dissociation into three ions
Easy · Level 9View options
0.42M
0.58M
1.40M
2.40M
Easy · Level 9View options
0.80
1.25
0.25
1.80
Easy · Level 9View options
0.035 M
0.100 M
0.135 M
1.350 M
Easy · Level 9View options
60.5 g mol⁻¹
110 g mol⁻¹
165 g mol⁻¹
200 g mol⁻¹
Easy · Level 9View options
0.75M
1.00M
1.33M
2.00M
Easy · Level 9View options
i = 2, particles doubled
i = 0.5, particles halved
i = 1, no change
i = 4, particles quadrupled
Easy · Level 9View options
16.7%
33.3%
50.0%
66.7%
Question 1EasyLevel 9
If the normal molecular mass is 100 and i = 0.5, what will be the observed molecular mass?
Correct answer: A
Use i = normal molecular mass / observed molecular mass. On rearrangement, observed molecular mass = normal molecular mass / i. Therefore the observed value is 100 / 0.5 = 200. The factor is below one because association decreases the number of particles; accordingly, the observed molecular mass becomes larger than the normal molecular mass rather than smaller.
Why can molecular mass calculated from colligative properties become abnormal?
Correct answer: A
Colligative properties such as osmotic pressure and freezing-point depression depend on the number of dissolved particles. If solute molecules associate, the number decreases; if they dissociate, it increases. The measured property then differs from the value expected for unchanged molecules, producing an abnormal calculated molecular mass. Colour and container shape are irrelevant to this effect.
What is the ideal value of (i) for Na₂SO₄ on complete dissociation?
Correct answer: A
On complete dissociation, sodium sulfate gives Na₂SO₄ → 2Na⁺ + SO₄²⁻. Thus, one formula unit produces two sodium ions and one sulfate ion, making three particles in solution. For an ideal completely dissociated electrolyte, the van’t Hoff factor equals this particle count, so i = 3. The sulfate ion remains one particle despite containing several atoms.
If an AB-type solute is 30% dissociated, what is the value of (i)?
Correct answer: A
For the dissociation AB → A + B, the van’t Hoff relation is i = 1 + α, because one original particle becomes two and the particle count increases by one for each dissociated fraction. A 30% dissociation means α = 30/100 = 0.30. Substitution gives i = 1 + 0.30 = 1.30. Thus option A is correct; 0.30 is the degree of dissociation itself, not the van’t Hoff factor.
If i = 1.75 for an AB-type solute, what is its degree of dissociation?
Correct answer: A
For an AB-type electrolyte, the relation between the van’t Hoff factor and degree of dissociation is i = 1 + α, because one formula unit can produce two particles. Rearranging gives α = i − 1 = 1.75 − 1 = 0.75. Multiplying by 100 converts this fraction to 75%. Thus, the solute is 75% dissociated.
If 40% dimerization occurs, what is the value of the van’t Hoff factor (i)?
Correct answer: A
The governing concept is the van’t Hoff factor for association. During dimerization, two solute molecules combine to form one particle, so for an association fraction α, i = 1 − α/2. Here α = 40/100 = 0.40. Therefore, i = 1 − 0.40/2 = 1 − 0.20 = 0.80. Option A is correct. The value is below one because dimer formation decreases the total number of solute particles; values above one would indicate dissociation.
If observed molecular mass is 25% lower than normal, how will i behave?
Correct answer: A
The factor is i = M normal / M observed. If the observed molecular mass is lower than the normal mass, the denominator is smaller than the numerator, so the ratio is greater than one. This indicates an increased number of particles, usually because of dissociation. The exact percentage is not needed to decide the direction of i.
If observed molecular mass is greater than normal, which statement is correct?
Correct answer: A
Since i = M normal / M observed, an observed mass greater than the normal mass makes i less than one. A value below one means fewer particles are present than expected, which is consistent with association. The observation alone does not prove a particular mechanism with absolute certainty, so “association is possible” is the appropriately cautious statement.
Why are colligative properties important for studying abnormal molecular mass?
Correct answer: A
Colligative properties depend on the number of dissolved particles rather than directly on their chemical identity. Association and dissociation change that particle number, so measurements such as osmotic pressure can differ from the value expected for unchanged molecules. This difference leads to an abnormal calculated molecular mass. Colour, smell, and instrument size do not govern the colligative relation.
In which situation will solute behaviour be considered normal?
Correct answer: A
The van’t Hoff factor compares the actual number of solute particles with the number expected if each formula unit remained unchanged. Normal behaviour means there is neither dissociation nor association, so the actual and expected particle numbers are equal. Their ratio is therefore i = 1. Values below one indicate association, while values above one usually indicate dissociation.
If i = 4, the observed molecular mass will be what fraction of the normal molecular mass?
Correct answer: A
The relation is i = normal molecular mass / observed molecular mass. Rearranging gives observed molecular mass = normal molecular mass / i. For i = 4, the observed mass is normal mass divided by four, or one-fourth of it. A larger i means more effective particles and therefore a smaller observed molecular mass in this relation.
If i = 0.75, which statement about the observed molecular mass is correct?
Correct answer: A
Since i = normal molecular mass / observed molecular mass, a value of i less than one means the numerator is smaller than the denominator. Therefore the observed molecular mass must be greater than the normal molecular mass. This commonly results from association, where several solute units combine and the number of particles decreases. It cannot make the mass zero.
A solution has i = 0.25. If the solute undergoes complete association, what type of group is formed?
Correct answer: C
In complete association, n original particles combine to form one group, so the van’t Hoff factor is i = 1/n. Since 0.25 = 1/4, n equals 4. Thus four solute particles form one associated group, called a tetramer. A dimer, trimer, or pentamer would give factors 0.50, 0.333, and 0.20 respectively.
Why is molar mass obtained from a colligative property called abnormal in some cases?
Correct answer: A
Colligative properties depend on the number of dissolved particles, not directly on their chemical identity. Association joins particles and lowers their number, whereas dissociation separates them and increases their number. If this change is ignored, the calculated molar mass differs from the true value; this is called abnormal molar mass.
If the apparent molar mass of a solute is M/3, what is the value of i and what does it indicate?
Correct answer: A
The governing relation for abnormal molar mass is M_app = M/i, where M is the normal molar mass. Given M_app = M/3, equating M/i with M/3 gives i = 3. A van’t Hoff factor greater than one means that the number of solute particles has increased, which occurs through dissociation. Thus option A is correct. Association would produce i below one, normal behaviour gives i = 1, and i = 0 is physically inappropriate for a dissolved solute.
The true molar mass of a solute is 180 g mol−1 and i = 0.6. What is its apparent molar mass?
Correct answer: C
Use the relation Mapparent = Mtrue/i. Substitution gives Mapparent = 180/0.6 = 300 g mol−1. Since i is less than one, the effective number of particles has decreased because of association, so the apparent molar mass must be greater than the true molar mass. This also rules out 108 and 60.
A solute forms pentamers completely. What is the ratio of apparent molar mass to true molar mass?
Correct answer: B
Complete pentamerisation means five original solute molecules combine into one effective particle. Therefore the van’t Hoff factor is i = 1/5. Apparent molar mass varies inversely with i, so M_app = M/i = M/(1/5) = 5M. Hence M_app : M = 5 : 1, making option B correct. The ratio 1:5 reverses the inverse relation and would incorrectly treat the apparent mass as smaller during association.
A colligative experiment gives an apparent molar mass of 240 g mol⁻¹, while the normal molar mass is 120 g mol⁻¹. Which conclusion is most correct?
Correct answer: A
The van’t Hoff factor is i = M/M_app = 120/240 = 0.50. Thus the effective number of particles has become half the original number. Complete dimer association converts every pair of particles into one particle and gives i = 1/2. Dissociation would produce i greater than one, while no change would give i = 1.
A salt MCl₂ has i = 2.4. If complete dissociation gives three ions, what is the apparent molar mass as a fraction of the true molar mass?
Correct answer: A
The apparent molar mass is related to the true mass by M_app = M/i. Substituting i = 2.4 gives M_app = M/2.4 = 0.4167M, which rounds to 0.42M. The statement about three ions confirms that i can arise from partial dissociation, but once i is given, the mass calculation uses only the inverse relation.
If the apparent molar mass is 25% greater than the true molar mass, what is the approximate value of i?
Correct answer: A
A 25% increase means M_app = 1.25M. Since M_app = M/i, we have 1.25M = M/i. Cancelling M gives i = 1/1.25 = 0.80. A value below one indicates association or some other process that reduces the effective particle count; dissociation would instead produce an apparent mass below M.
An XY electrolyte has i = 1.35. If its formal concentration is 0.10 M, what is the effective particle concentration?
Correct answer: C
The van’t Hoff factor converts formal concentration into the effective concentration of solute particles: Ceffective = iC. Substitution gives Ceffective = 1.35 × 0.10 = 0.135 M. Thus option C is correct. The value exceeds 0.10 M because partial dissociation of XY creates additional particles; 0.035 M would represent only the increase, not the total effective concentration.
A solute has i = 0.55 and a true molar mass of 110 g mol⁻¹. What is its apparent molar mass?
Correct answer: D
Use M_app = M/i. Substitution gives M_app = 110/0.55 = 200 g mol⁻¹. Because i is less than one, the effective number of particles is reduced, so the apparent molar mass is greater than the true mass. Multiplying 110 by 0.55 would give the wrong direction and does not satisfy the defining relation.
The elevation in boiling point is 25% lower than expected. If the cause is dimer association, what is the apparent molar mass relative to the true mass?
Correct answer: C
A 25% lower colligative property means i = 0.75. Since apparent molar mass is inversely related to i, M_app = M/i = M/0.75 = 1.333M, approximately 1.33M. The reduction in boiling-point elevation reflects fewer effective particles due to association. It is therefore incorrect to multiply M by 0.75.
For a 0.10 m solution, the normal freezing-point depression is 0.186 K and the observed depression is 0.372 K. Which conclusion is correct?
Correct answer: A
The van’t Hoff factor is the ratio of observed to normal colligative property: i = 0.372/0.186 = 2. Thus the effective particle concentration is twice the normal value. This is consistent with dissociation or another particle-increasing process. A factor of 0.5 would correspond to a depression smaller than normal, while i = 1 would indicate no abnormality.
For a solute, i = 4/3. If it dissociates into two ions, what is the degree of dissociation?
Correct answer: B
The governing relation for dissociation into two ions is i = 1 + α, because each dissociated solute unit changes from one particle into two, producing one additional particle. Substituting i = 4/3 gives α = 4/3 − 1 = 1/3. Converting this fraction to a percentage gives 33.3%. Thus option B is correct. Option A is half of the required value, while option D is the undissociated fraction, not the degree of dissociation.
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