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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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0.2 mole per litre
0.1 mole per litre
0.5 mole per litre
2.0 mole per litre
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solute particles may associate
solute is completely dissociating
solute is highly volatile
solvent is absent
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Greater lowering and lower molar mass
Smaller lowering and higher molar mass
Zero lowering and zero molar mass
Infinite lowering and infinite molar mass
Medium · Level 1View options
Lowering will be less than expected
Lowering will be more than expected
Lowering will always be zero
Vapour pressure will become infinite
Medium · Level 1View options
Lowering will increase
Lowering will decrease
No lowering will occur
Vapour pressure will become higher than pure solvent
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Because effective number of particles increases
Because solvent mass becomes zero
Because solute does not remain in vapour
Because external pressure is not constant
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Because effective number of particles decreases
Because solvent mole fraction becomes zero
Because liquid stops boiling
Because vapour pressure always increases
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Association
Dissociation
Loss of non-volatility
Becoming pure solvent
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Dissociation
Association
Complete vaporisation
More particles by complete ionisation
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To express the effect of actual number of particles
To tell the colour of solution
To tell the smell of solvent
To tell the volume of container
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Dissociation or association
Only container colour
Changing the solvent’s name
Length of scale
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It will be lower
It will be higher
It will always be zero
Change will be impossible
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Due to ionisation or association of solute
Due to colour of thermometer
Due to shape of glass
Due to name of solvent
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Less than actual
Greater than actual
Always zero
Always infinite
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Greater than actual
Less than actual
Always zero
Exactly same always
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ΔT_b = iK_bm
ΔT_b = K_b/m
ΔT_b = iRT
ΔT_b = K_fm
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ΔT_f = iK_fm
ΔT_f = K_bm
ΔT_f = i/K_f
ΔT_f = CRT
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When solute dissociates into ions
When solute particles associate
When solvent is pure
When solution has no solute
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When solute particles associate
When solute fully ionises
When particle number increases
When solution is highly dilute
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Association of solute
Ionisation of solute
Change in solvent colour
Solute not dissolving
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Ionisation of solute
Association of solute
Increase in particle number
Complete vaporisation
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During dissociation of solute
During association of solute
During colour change of solvent
During change in container shape
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2
1/2
1
4
Medium · Level 1View options
Association
Dissociation
Complete ionisation
Evaporation
Medium · Level 1View options
Solute particles are associating
The solute is completely dissociated
The solute is forming a gas
The solute must be ionic
Question 1MediumLevel 1
18 g glucose is dissolved in water to make 500 mL solution. Molar mass of glucose is 180 g/mol. What is the molarity?
Correct answer: A
Use the molarity relation M = n/V(solution). First find the amount of glucose: n = mass/molar mass = 18 g/(180 g mol⁻¹) = 0.10 mol. Convert the final volume: 500 mL = 0.500 L. Therefore M = 0.10 mol/0.500 L = 0.20 mol L⁻¹, so option A is correct. Option B is only the number of moles, not the concentration; C and D use incorrect scaling.
If observed molar mass of solute is greater than the actual value, what can it indicate about solute behaviour?
Correct answer: A
If solute particles associate, several original particles combine into fewer species. The measured colligative effect is then smaller than expected for the stated amount of solute. In calculations that assume independent particles, a smaller effect corresponds to a larger apparent or observed molar mass. Dissociation would increase the particle count and usually make the observed molar mass smaller, not larger.
If solute particles associate with each other, what may happen to lowering of vapour pressure and observed molar mass?
Correct answer: B
Association combines several solute particles into fewer larger species. The effective particle count therefore decreases, causing a smaller colligative effect and smaller lowering of vapour pressure than expected for non-associated particles. Since molar mass is calculated from that effect, the observed molar mass appears higher than the true monomeric value.
If a solute associates in the liquid to form larger particles, what happens to lowering of vapour pressure for the same mass?
Correct answer: A
Association joins two or more solute molecules into one larger effective particle. For the same mass of solute, the number of particles is therefore smaller than the number calculated by assuming no association. Because vapour-pressure lowering depends on particle number, the observed lowering is less than expected and the experimentally calculated molar mass appears unusually high.
If a solute dissociates in solution to produce more particles, what is the effect on lowering of vapour pressure?
Correct answer: A
Dissociation converts one formula unit into two or more solute particles. The effective particle concentration is therefore greater than the nominal concentration. This lowers the solvent mole fraction more strongly, so the relative lowering of vapour pressure increases. The solution pressure still remains below the pure-solvent pressure for a non-volatile solute; it does not become higher than it.
Why can lowering of vapour pressure become greater when solute particles dissociate in a solution?
Correct answer: A
Lowering of vapour pressure is a colligative property, so it depends on the number of dissolved particles. When one formula unit dissociates into two or more ions, the effective particle count increases. The solvent mole fraction consequently falls more than expected for undissociated solute, producing a greater lowering; this effect is represented by the van’t Hoff factor.
If solute particles associate, why is lowering of vapour pressure less than expected?
Correct answer: A
Association joins two or more dissolved particles into a larger species, so the number of independent solute particles becomes smaller. Since vapour-pressure lowering depends on particle number, the observed colligative effect is less than the value calculated by assuming no association. The solvent mole fraction does not become zero, and pressure does not automatically increase.
If observed molar mass of a solute is less than its actual value, which behaviour is likely?
Correct answer: B
Dissociation increases the number of effective solute particles. Consequently, a colligative effect such as vapour-pressure lowering is larger than expected from the formula mass of undissociated solute. If the calculation assumes fewer particles, the inferred number of moles is too large and the calculated molar mass appears too small. Association gives the opposite trend.
If observed molar mass of a solute is greater than its actual value, which behaviour is likely?
Correct answer: B
Association combines several solute molecules into fewer effective particles. The colligative effect is then smaller than expected for the same mass of unassociated solute. If this smaller effect is interpreted using the no-association formula, the calculated number of moles is too small, so the apparent molar mass becomes larger than the true value. Dissociation causes the reverse result.
The van’t Hoff factor accounts for the actual number of particles produced in solution. An electrolyte may dissociate into several ions, increasing the particle count, while some solutes associate into larger units, decreasing it. Consequently, colligative equations are written with i, such as ΔTb = iKb m and ΔTf = iKf m. It is unrelated to colour, smell, or container volume.
What can be one reason for abnormal molar mass obtained from colligative properties?
Correct answer: A
Molar mass calculated from a colligative effect assumes a particular number of independent solute particles. Dissociation increases particle number and association decreases it, so the measured effect no longer matches the simple molecular formula. The resulting calculated molar mass may therefore appear abnormally low or high.
If solute particles associate with each other, how will the colligative property value compare with the expected value?
Correct answer: A
Association combines two or more solute particles into a larger species. The number of independently moving particles therefore becomes smaller than the number calculated from the original formula units. Since colligative effects depend on particle number, the observed effect is lower than the ideal expected value, and the van’t Hoff factor is generally less than one.
Why can an abnormal molar mass be obtained from colligative properties?
Correct answer: A
Colligative measurements are interpreted through the number of dissolved particles. Ionisation produces more particles than the formula units suggest, whereas association produces fewer. If these changes are ignored and the ideal formula is used directly, the calculated molar mass differs from the true value. The van’t Hoff factor corrects this abnormality.
If a solute dissociates into ions, how may its apparent molar mass generally appear?
Correct answer: A
Dissociation increases the number of particles and therefore makes the observed colligative effect larger than the ideal molecular calculation predicts. If that enhanced effect is interpreted without applying the van’t Hoff factor, the calculation appears to involve more moles of solute than are actually present. Since molar mass equals mass divided by moles, the apparent molar mass becomes lower than the actual value.
If solute particles associate, how may the apparent molar mass generally appear?
Correct answer: A
Association joins several solute molecules into fewer larger particles, reducing the observed colligative effect. If the reduced effect is analysed using the non-associating formula, the calculated number of moles appears smaller than the true number. For a fixed sample mass, mass divided by this underestimated mole value gives an apparent molar mass greater than the actual molar mass.
When i is included in colligative properties, what is the modified relation for boiling point elevation?
Correct answer: A
For a non-electrolyte, the boiling point elevation is ΔT_b = K_bm. If a solute dissociates or associates, the effective number of particles differs from the simple formula-unit count, so the van't Hoff factor i is included: ΔT_b = iK_bm. The expression K_b/m has the wrong dependence, iRT is not the boiling-point relation, and K_fm describes freezing point depression.
When i is included in colligative properties, what is the modified relation for freezing point depression?
Correct answer: A
For a non-electrolyte, freezing point depression is ΔT_f = K_fm. Dissociation increases the number of particles and association decreases it, so the actual effect is corrected by the van't Hoff factor i. The complete relation is therefore ΔT_f = iK_fm. K_bm belongs to boiling point elevation, i/K_f has the wrong form, and CRT is associated with the dilute osmotic-pressure equation.
When can apparent molar mass be obtained less than the actual molar mass?
Correct answer: A
Dissociation changes one formula unit into two or more particles, increasing the effective particle count. Since colligative effects become larger, a calculation that assumes no dissociation interprets the effect as coming from more solute moles. The calculated apparent molar mass, which is inversely related to the inferred amount, therefore becomes lower than the true molar mass.
When can apparent molar mass be obtained greater than the actual molar mass?
Correct answer: A
Association combines several solute particles into fewer units, so the effective number of particles becomes smaller than expected. The observed colligative effect is consequently reduced. If the calculation ignores association, it infers fewer solute moles than are actually present; dividing the solute mass by this underestimated amount gives an apparent molar mass greater than the true value.
If ΔTf is observed greater than expected, which possibility is more suitable?
Correct answer: B
The freezing-point depression is ΔTf = iKf m. If the experimentally observed value is larger than the value calculated with i = 1, the effective particle factor i is greater than one. Ionisation or dissociation produces additional particles and raises i, thereby increasing the depression. Association would lower the number of particles and reduce ΔTf instead.
If ΔTb is observed lower than expected, which possibility is more suitable?
Correct answer: B
Boiling-point elevation follows ΔTb = iKb m. A value lower than the ideal non-electrolyte prediction indicates i < 1, meaning fewer effective particles than expected. Association combines solute particles and causes this decrease in particle number. Ionisation and an increase in particles would make i greater than one and would produce a larger, not smaller, elevation.
In which case can the observed molar mass be less than the actual molar mass?
Correct answer: A
Dissociation breaks solute molecules into two or more particles, increasing the number of effective particles in solution. Consequently, the observed colligative effect is larger than expected for the undissociated solute. If the ordinary formula is used without correcting for this increase, the calculated or observed molar mass becomes smaller than the actual value. Association generally causes the opposite error.
If the normal molar mass of a solute is M and its observed molar mass in solution is M/2, what is its van't Hoff factor?
Correct answer: A
For abnormal molar-mass behaviour, the van't Hoff factor is related to molar masses by i = normal molar mass/observed molar mass. Substituting the given values gives i = M/(M/2) = 2. An observed molar mass smaller than the normal value means that the solute has produced more effective particles, usually through dissociation. Therefore, the correct answer is 2.
If the observed molar mass of a substance is greater than its normal molar mass, which process is possible in solution?
Correct answer: A
A greater observed molar mass corresponds to a van't Hoff factor less than one. This happens when two or more solute molecules join to form larger associated species, thereby reducing the number of independently behaving particles. The reduced particle count produces a smaller colligative effect and makes the calculated molar mass appear larger. Dissociation and ionisation would increase particle number and generally lower the observed molar mass.
A solution is found to have i < 1. What does this indicate about the solute?
Correct answer: A
The van’t Hoff factor i compares the actual number of solute particles in solution with the number expected if no association or dissociation occurred. When i is less than one, individual solute molecules combine to form larger associated units, such as dimers. The total number of particles consequently decreases. Because colligative effects depend on particle number, the observed effect is smaller than the normal calculated value, and the experimentally determined molar mass appears larger.
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