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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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Easy · Level 1View options
solute is dissociating
solute is associating
solute is absent
solute does not lower vapour pressure
Easy · Level 1View options
2
1
0
0.5
Easy · Level 1View options
1
2
3
0
Easy · Level 1View options
Because the number of effective particles increases
Because colour changes
Because solvent disappears
Because temperature always decreases
Easy · Level 1View options
It decreases
It greatly increases
It remains unchanged
It becomes negative
Easy · Level 1View options
Dissociation of solute
Association of solute
Colourless solution
Solvent stops freezing
Easy · Level 1View options
Association of solute
Complete dissociation of solute
Increase in number of ions
Boiling of solution stops
Easy · Level 1View options
2
1
0.5
0
Easy · Level 1View options
1
2
0
−1
Easy · Level 1View options
Particles have associated
Complete ionisation occurred
Particle number doubled
Solute is not non-volatile
Easy · Level 1View options
One
Two
Three
Four
Easy · Level 1View options
Particles are neither ionising nor associating
Particles fully split into three ions
Particles fully associate to half
Solution has no solute
Easy · Level 1View options
Lower
Same
Greater
Zero
Easy · Level 1View options
Greater
Lower
Always same
Infinite
Easy · Level 1View options
Because of dissociation or association of the solute
Because of the colour of the solvent
Because of the shape of the container
Because of the time shown by a clock
Easy · Level 1View options
It becomes less than the true value
It remains equal to the true value
It becomes greater than the true value
It always becomes zero
Easy · Level 1View options
It becomes less than the true value
It becomes greater than the true value
It always becomes zero
It always becomes infinite
Easy · Level 1View options
20 g mol⁻¹
30 g mol⁻¹
60 g mol⁻¹
120 g mol⁻¹
Easy · Level 1View options
50 g mol⁻¹
100 g mol⁻¹
150 g mol⁻¹
200 g mol⁻¹
Easy · Level 1View options
When the solute dissociates or associates in solution
When the solution is colourless
When the container is round
When the temperature is not recorded
Easy · Level 1View options
Van’t Hoff factor (वैन्ट हॉफ गुणक)
Colour factor (रंग गुणक)
Smell factor (गंध गुणक)
Shape factor (आकार गुणक)
Easy · Level 1View options
The calculated molar mass will be greater than the true molar mass (परिकलित मोलर द्रव्यमान वास्तविक मोलर द्रव्यमान से अधिक होगा)
The calculated molar mass will be less than the true molar mass (परिकलित मोलर द्रव्यमान वास्तविक मोलर द्रव्यमान से कम होगा)
The calculated molar mass will equal the true molar mass (परिकलित मोलर द्रव्यमान वास्तविक मोलर द्रव्यमान के बराबर होगा)
The calculated molar mass will be zero (परिकलित मोलर द्रव्यमान शून्य होगा)
Easy · Level 1View options
1
0
0.5
2
Easy · Level 1View options
0.5
1
2
150
Easy · Level 1View options
Association (संघटन)
Dissociation (वियोजन)
Complete ionisation (पूर्ण आयनीकरण)
Evaporation (वाष्पीकरण)
Question 1EasyLevel 1
If the observed colligative effect is greater than expected, what may it indicate about the solute?
Correct answer: A
Colligative effects depend on the number of effective solute particles. During dissociation, one formula unit separates into two or more ions, increasing the particle count. The observed vapour-pressure lowering, boiling-point elevation or freezing-point depression can therefore exceed the value calculated for undissociated molecules. Association has the opposite effect because it reduces the number of effective particles.
If a solute completely dissociates into two ions, what is the ideal value of i?
Correct answer: A
The van’t Hoff factor is the ratio of the actual number of particles in solution to the number expected without dissociation. One formula unit that completely produces two independent ions gives two effective particles, so i = 2 ideally. In real concentrated solutions, interionic interactions may make the observed value differ slightly.
What is the usual value of i for a non-electrolyte solute like glucose?
Correct answer: A
Glucose dissolves in water as neutral molecules and does not normally dissociate into ions or associate significantly under the ideal dilute-solution assumption. Thus one glucose molecule contributes one effective solute particle, giving the van’t Hoff factor i = 1. A value of 2 or 3 would suggest particle multiplication, while i = 0 would incorrectly imply no solute particles.
Why do colligative effects generally increase when a solute dissociates?
Correct answer: A
Dissociation splits one solute unit into two or more ions, increasing the number of independent particles. Since colligative effects are proportional to effective particle concentration, the observed effect becomes larger; for example, boiling-point elevation contains the factor i. Colour and solvent loss are unrelated explanations.
What usually happens to the observed colligative effect during association?
Correct answer: A
Association combines two or more solute particles into a larger species, reducing the number of independent particles in solution. Because colligative properties depend on that effective number, the observed effect decreases compared with the value calculated for unassociated particles. It does not become negative.
For a non-associating, non-dissociating solute, i is approximately 1. A value greater than 1 means that more effective particles are present than formula units initially introduced. The usual explanation is dissociation into ions or other species. Association would reduce the particle count and generally give i < 1.
The van’t Hoff factor compares the actual number of dissolved particles with the number expected from the formula units. If i is less than 1, the solution contains fewer effective particles than expected. This commonly results when solute molecules associate into dimers or larger units, as in some weakly interacting systems. Dissociation generally gives i greater than 1, not less than 1.
If an electrolyte completely dissociates into two ions, what can be the ideal value of i?
Correct answer: A
The van’t Hoff factor is the ratio of particles after dissociation to the original formula units. If every formula unit produces exactly two ions and dissociation is complete, two particles result from one, so i = 2. This is an ideal value; interionic interactions or incomplete dissociation can make the measured value different.
What is the usual value of i for a non-electrolyte solute?
Correct answer: A
A non-electrolyte remains essentially as intact molecules in solution and does not dissociate into ions. Therefore the actual number of particles is approximately the same as the number expected from the amount dissolved. Since the van't Hoff factor is actual particles divided by expected particles, i is normally 1. Values such as 2 are associated with particular dissociation cases, not ordinary non-electrolytes.
If a solute has i = 0.5, what is the most suitable meaning?
Correct answer: A
The van’t Hoff factor compares the actual effective number of particles with the number expected from the undissociated solute. An i value below one means that fewer particles are present than expected. This commonly results from association, in which molecules combine. Complete ionisation would instead increase the particle count and normally give i greater than one.
If an electrolyte ideally has i = 2, it indicates formation of how many ions from one formula unit?
Correct answer: B
For ideal complete dissociation, the van’t Hoff factor equals the number of particles produced from one formula unit. Thus i = 2 indicates that one formula unit produces two ions, as happens for an electrolyte such as AB → A⁺ + B⁻. It does not mean two formula units; it refers to two resulting solute particles.
If a solution has i = 1, it is closest to which behaviour?
Correct answer: A
The van’t Hoff factor i compares the effective number of particles in solution with the number expected from the formula units. i = 1 means there is no net change in particle number: the solute behaves approximately as an undissociated, non-associating species. Three-ion dissociation would give i near 3, association would give i below 1, and a solute-free solution is not represented by this comparison.
If i is greater than 1, how will the observed colligative property compare with the expected value?
Correct answer: C
For colligative properties, the electrolyte correction appears as a factor i: for example, ΔTf = iKf m, ΔTb = iKb m and π = iCRT. If i > 1, the solution contains more effective particles than the undissociated calculation assumes, usually because of dissociation. Multiplication by a factor greater than one makes the observed property greater than the expected value.
If i is less than 1, how will the observed colligative property compare with the expected value?
Correct answer: B
In expressions such as ΔTf = iKf m and ΔTb = iKb m, the van’t Hoff factor multiplies the ideal particle-based result. When i < 1, fewer effective particles are present than expected, commonly because of association. Multiplication by a number below one reduces the observed colligative effect, so it is lower rather than equal to or greater than the expected value.
Why can an abnormal molar mass be obtained for a solute in solution?
Correct answer: A
Abnormal molar mass is obtained when the solute does not remain as the expected number of independent particles in solution. Dissociation breaks one solute unit into two or more particles, increasing the particle number, whereas association combines particles and decreases their number. Since colligative properties depend on the number of particles, the molar mass calculated from them differs from the true value. Therefore, option A is correct.
What happens to the observed molar mass when dissociation occurs?
Correct answer: A
During dissociation, one solute unit produces two or more particles, so the number of particles in solution increases. Colligative effects therefore become larger than expected for the undissociated solute. If molar mass is calculated from this enhanced effect without correcting for the van’t Hoff factor, the calculated or observed molar mass is lower than the true molar mass. Thus, option A is correct.
What happens to the observed molar mass when association occurs?
Correct answer: B
Association occurs when two or more solute particles combine to form a larger associated species. Consequently, the number of independent particles decreases and the colligative effect becomes smaller than expected. If the molar mass is calculated without correcting for this association, the observed molar mass is greater than the true molar mass. In this case the van’t Hoff factor is less than one, so option B is correct.
If the true molar mass is 60 g mol⁻¹ and the van’t Hoff factor is i = 2, what is the observed molar mass?
Correct answer: B
The relation between true and observed molar mass is M_observed = M_true/i. This follows because the van’t Hoff factor represents the change in the effective number of solute particles. Substituting the values gives M_observed = 60 g mol⁻¹ ÷ 2 = 30 g mol⁻¹. Since i is greater than one, the solute behaves as if more particles are present and the observed molar mass is lower. Therefore, option B is correct.
If the true molar mass is 100 g mol⁻¹ and the van’t Hoff factor is i = 0.5, what is the observed molar mass?
Correct answer: D
For abnormal molar mass, the observed value is calculated from M_observed = M_true/i. Therefore, M_observed = 100 g mol⁻¹ ÷ 0.5 = 200 g mol⁻¹. Dividing by one-half is the same as multiplying by two. Because i is less than one, association has reduced the number of independent particles, so the observed molar mass is greater than the true value. Hence, option D is correct.
When is it necessary to use the van’t Hoff factor i in molar-mass determination?
Correct answer: A
The van’t Hoff factor i is required when the solute does not behave as an unchanged set of particles in solution. Dissociation produces more particles than expected, whereas association produces fewer particles. These changes alter colligative properties and cause the experimentally calculated molar mass to differ from the true molar mass. Including i corrects the particle count and gives a reliable molar-mass value.
Which factor is used to explain abnormal molar mass properly?
Correct answer: A
Abnormal molar mass is observed when a solute associates or dissociates in solution, changing the actual number of solute particles. The Van’t Hoff factor, represented by i, accounts for this change in particle number in colligative-property equations. For dissociation, i is generally greater than 1, while for association it is generally less than 1. Therefore, option A is correct.
A student determines the molar mass of a solute by the osmotic pressure method. If the solute undergoes dimerisation in solution, which conclusion about the calculated molar mass is correct?
Correct answer: A
During dimerisation, two solute molecules combine to form one larger particle. Thus, the number of dissolved particles decreases, and the osmotic pressure becomes lower than expected for the same mass of solute. In the osmotic-pressure relation M = wRT/(πV), a smaller π produces a larger apparent or calculated molar mass. Hence, association by dimerisation makes the calculated molar mass greater than the true value, so option A is correct.
If the normal molar mass is 180 g mol⁻¹ and the observed molar mass is 180 g mol⁻¹, what is the Van’t Hoff factor i?
Correct answer: A
The Van’t Hoff factor is calculated from the relation i = normal molar mass / observed molar mass. Substituting the given values gives i = 180/180 = 1. A value of i equal to 1 means that the effective number of solute particles has not changed. Therefore, under the ideal interpretation, neither association nor dissociation is occurring, and option A is correct.
If the normal molar mass is 100 g mol⁻¹ and the observed molar mass is 50 g mol⁻¹, what is the Van’t Hoff factor i?
Correct answer: C
For abnormal molar mass, the Van’t Hoff factor is given by i = normal molar mass / observed molar mass. Therefore, i = 100/50 = 2. This value greater than 1 indicates that the effective number of solute particles has increased, as happens during dissociation or ionisation. Thus, the correct numerical answer is 2, which is option C.
If the normal molar mass is 60 g mol⁻¹ and the observed molar mass is 120 g mol⁻¹, which process is possible?
Correct answer: A
The Van’t Hoff factor is i = normal molar mass / observed molar mass = 60/120 = 0.5. A value of i less than 1 shows that the number of solute particles has decreased. This occurs when solute molecules associate, such as two molecules forming a dimer. Therefore, association is the possible process and option A is correct.
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