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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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25 questions
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Easy · Level 6View options
2
0.5
1
4
Easy · Level 6View options
0.5
2
1.5
3
Easy · Level 6View options
Partial dissociation may occur
Complete association is certain
No particles are present
Molecular mass is zero
Easy · Level 6View options
Partial association may occur
Complete dissociation is occurring
The number of particles is increasing
Three ions are forming
Easy · Level 6View options
2
1
3
0.5
Easy · Level 6View options
4
3
2
1
Easy · Level 6View options
4
3
2
5
Easy · Level 6View options
3
1
0.3
1.5
Easy · Level 6View options
0.5
2
1
3
Easy · Level 6View options
Association or dissociation of solute particles
The colour of the solution
The thickness of the glass
The light in the laboratory
Easy · Level 6View options
Dissociation can increase the value of colligative properties
Dissociation decreases particle number
Association increases particle number
i has no relation to particle number
Easy · Level 6View options
Association always makes i greater than 1
Dissociation can increase particle number
Dimer formation decreases particle number
i = 1 can show normal behaviour
Easy · Level 6View options
2
3
1
4
Easy · Level 6View options
Less than normal
Greater than normal
Equal
Infinite
Easy · Level 6View options
Greater than normal
Less than normal
Always zero
Always the same
Easy · Level 6View options
NaCl in water
Acetic acid in benzene
Glucose in water
Urea in water
Easy · Level 6View options
Acetic acid in benzene
KCl in water
NaCl in water
CaCl2 in water
Easy · Level 6View options
3
2
1
4
Easy · Level 6View options
Correction factor for particle number
Factor for colour change
Factor for solvent smell
Factor for container height
Easy · Level 6View options
It will be greater than normal
It will be less than normal
It will be zero
It will have no meaning
Easy · Level 6View options
Only solute particles being heavy
Association or dissociation of solute in solution
Change in solvent colour
Change in container shape
Easy · Level 6View options
It decreases
It increases
It becomes zero
It is unaffected
Easy · Level 6View options
Greater than normal
Less than normal
Equal to normal
Always double
Easy · Level 6View options
सामान्य मोलर द्रव्यमान / प्रेक्षित मोलर द्रव्यमान / Normal molar mass divided by observed molar mass
प्रेक्षित मोलर द्रव्यमान / सामान्य मोलर द्रव्यमान / Observed molar mass divided by normal molar mass
विलायक द्रव्यमान / विलेय द्रव्यमान / Solvent mass divided by solute mass
तापमान / दाब / Temperature divided by pressure
Easy · Level 6View options
When complete association occurs
When complete dissociation occurs
When neither association nor dissociation occurs
When the solute is volatile
Question 1EasyLevel 6
If a solute makes a colligative property twice the expected value, what is i?
Correct answer: A
At the same concentration and temperature, the van’t Hoff factor is the ratio of the observed colligative effect to the value expected without particle-number change. If the observed effect is twice the expected value, i = 2. This indicates twice as many effective particles and is commonly associated with dissociation.
If the observed colligative property is half the normal value, what is i?
Correct answer: A
The van’t Hoff factor compares the observed colligative effect with the normal effect for the same analytical concentration. Therefore i = observed effect ÷ normal effect = one-half ÷ one = 0.5. A value below one means fewer effective particles, which may result from association such as dimer formation.
The value 1.4 is greater than one, so the solution contains more effective particles than expected from the undissociated solute. Since it is below the ideal value for complete dissociation of many electrolytes, partial dissociation is a suitable conclusion. The number does not imply absence of particles or zero molecular mass.
Because 0.7 is less than one, the effective number of independent particles is lower than expected. Partial association, in which only some solute molecules combine, can produce this result. Complete association is not established by the value alone, while dissociation and formation of three ions would normally increase i above one.
If KCl completely dissociates in water, what will be the ideal value of i?
Correct answer: A
Complete dissociation of KCl is represented by KCl → K⁺ + Cl⁻. One formula unit therefore produces two independent ions. For ideal complete dissociation, the van’t Hoff factor equals this particle count, so i = 2. A value of 1 would mean no change in particle number, 3 overcounts the products, and 0.5 is not possible for complete dissociation of this 1:1 salt.
If FeCl₃ undergoes complete dissociation, what is the ideal value of i?
Correct answer: A
Complete dissociation of one formula unit of FeCl₃ gives one Fe³⁺ ion and three Cl⁻ ions: FeCl₃ → Fe³⁺ + 3Cl⁻. The total number of particles is therefore 1 + 3 = 4. Under the ideal complete-dissociation assumption, the van’t Hoff factor equals this number, so i = 4. The ionic charges affect electrical neutrality, but they do not change the count of particles.
If K₃PO₄ undergoes complete dissociation, what is the ideal value of i?
Correct answer: A
On complete dissociation, one formula unit of K₃PO₄ produces three potassium ions and one phosphate ion: K₃PO₄ → 3K⁺ + PO₄³⁻. The total number of particles is 3 + 1 = 4, so the ideal van’t Hoff factor is i = 4. The subscript 3 applies to potassium, while PO₄ is one polyatomic ion; its four atoms do not represent four separate solute particles.
If the number of particles in a solution becomes three times the expected number, what will be the value of i?
Correct answer: A
The van’t Hoff factor compares the effective or actual number of particles with the number expected from the undissociated solute: i = actual particles/expected particles. If the actual number is three times the expected number, then i = 3. A value of 1 means no particle-number change, while 0.3 would indicate fewer particles, not more.
If the actual number of particles becomes half the expected number, what will be the value of i?
Correct answer: A
By definition, i = actual number of particles/expected number of particles. If the actual number is one-half of the expected number, then i = 1/2 = 0.5. This value below one indicates fewer independent particles, often because of association. The value 2 describes the opposite change, i = 1 means no change, and 3 means three times as many particles.
To understand abnormal molecular mass, what should be checked first?
Correct answer: A
The first diagnostic step is to identify whether solute particles associate or dissociate. Dissociation increases the number of independent particles and generally makes i greater than one, whereas association decreases the number and generally makes i less than one. These changes, not the colour or laboratory equipment, explain abnormal molecular mass.
Dissociation converts one solute unit into two or more independent particles. Since colligative properties depend on the number of dissolved particles, the corresponding effect can become larger. Association instead reduces independent particles, and i is specifically a measure related to the effective particle-number change.
Association joins solute molecules and therefore generally reduces the number of independent particles. Its usual effect is to make i less than one, not greater than one. Dissociation can increase particle number, dimerisation reduces it, and i equal to one represents no net change in effective particle number, so those statements are consistent.
What is the ideal i for complete dissociation of MgSO₄ in water?
Correct answer: A
Under the ideal complete-dissociation assumption, MgSO₄ → Mg²⁺ + SO₄²⁻. The sulfate group is one polyatomic ion, so one formula unit produces two independent particles: one magnesium ion and one sulfate ion. Therefore the ideal van’t Hoff factor is i = 2; it is not three because the atoms within sulfate are not counted separately.
When observed molecular mass is greater than normal, how is the observed colligative property?
Correct answer: A
The van’t Hoff factor is related to molar masses by i = normal molar mass/observed molar mass. If the observed molar mass is greater, i is less than 1. Since a colligative property is multiplied by i, its observed value is lower than the normal value. This generally indicates association of solute particles.
When the observed molar mass is less than the normal molar mass, how does the observed colligative property compare with the normal value?
Correct answer: A
The van’t Hoff factor is related to molar mass by i = Mnormal/Mobserved. Therefore, if the observed molar mass is smaller, i is greater than 1. Since colligative properties are multiplied by i, the observed boiling-point elevation, freezing-point depression, osmotic pressure, or relative lowering of vapour pressure becomes greater than the normal value. This usually indicates dissociation.
Which example can show abnormal molecular mass due to dissociation?
Correct answer: A
NaCl is an electrolyte and, in water, it can separate into Na+ and Cl− ions. One formula unit therefore produces two solute particles, increasing the particle number and giving i greater than 1. Acetic acid in benzene is mainly an association example, while glucose and urea generally remain undissociated.
Which example can show abnormal molecular mass due to association?
Correct answer: A
In benzene, acetic acid molecules can associate through hydrogen bonding and form dimers. Two original molecules then behave as one effective particle, so the particle number decreases, i becomes less than 1, and the observed molar mass becomes higher than the normal value. The salt examples generally show dissociation in water.
If an AB₂-type salt undergoes complete dissociation, what is the ideal value of the van’t Hoff factor, i?
Correct answer: A
One formula unit of an AB₂ salt produces one A ion and two B ions on complete dissociation. Thus, one original solute particle becomes three particles in solution. Since the ideal van’t Hoff factor is the ratio of particles after dissociation to the original number of particles, i = 3. The ionic charges do not change this particle count.
In abnormal molecular mass questions, what is the easiest way to understand i?
Correct answer: A
The van’t Hoff factor i compares the actual effective number of solute particles with the number expected from the formula. Association lowers this number and gives i less than 1; dissociation raises it and gives i greater than 1. Thus i is best understood as a particle-number correction factor in colligative-property equations.
A student finds that i is less than 1. What should the student write about observed molecular mass?
Correct answer: A
For molar-mass measurements, i = normal molar mass/observed molar mass. If i is less than 1, the denominator must be larger than the numerator; therefore the observed molar mass is greater than the normal molar mass. This usually results from association, which reduces the effective particle number.
Molar mass obtained from a colligative property assumes a certain number of independent solute particles. If solute molecules associate, the effective number decreases; if an electrolyte dissociates, it increases. The measured colligative effect then differs from the expected one, producing an abnormal observed molar mass.
If solute particles combine in solution to form larger particles, what happens to the observed molar mass?
Correct answer: B
Combining solute particles is association. For example, two molecules may form one dimer, reducing the number of particles that contribute to a colligative property. The observed effect is therefore smaller, so the calculation gives i less than 1 and an observed molar mass greater than the normal molar mass.
If a solute breaks into ions in solution, what will be the observed molar mass?
Correct answer: B
Breaking a solute into ions is dissociation. It increases the number of particles and therefore increases the colligative effect. In the relation i = normal molar mass/observed molar mass, dissociation gives i greater than 1, so the observed molar mass must be lower than the normal molar mass.
Which statement correctly defines the van’t Hoff factor in terms of molar mass?
Correct answer: A
The van’t Hoff factor compares the actual number of solute particles with the number expected from the formula. When expressed through molar mass, it is i = Mnormal/Mobserved. Dissociation lowers the observed molar mass and gives i > 1, whereas association raises it and gives i < 1. Therefore the reverse ratio in option B is incorrect.
In which situation is the van’t Hoff factor, i, equal to 1?
Correct answer: C
The van’t Hoff factor compares the actual number of solute particles in solution with the number expected if each formula unit remained as one particle. If solute particles neither combine nor split, the particle number is unchanged, so i = 1. Complete association usually gives i below 1, whereas dissociation increases i above 1; volatility alone does not define i.
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