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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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Medium · Level 2View options
Dissociation of solute particles
Association of solute molecules
Complete vaporisation of the solute
Ideal behaviour of the solution
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0.67
1.0
1.5
2.0
Medium · Level 2View options
0.5
1
2
3
Medium · Level 2View options
Association of solute molecules
Dissociation of the solute
Solute molecules remaining as separate monomers
Volatility of the solute
Medium · Level 2View options
The observed molar mass will be greater than the normal molar mass
The observed molar mass will be less than the normal molar mass
The observed molar mass will always be zero
The molar mass cannot be measured
Medium · Level 2View options
Greater than the normal molar mass
Equal to the normal molar mass
Less than the normal molar mass
Always infinite
Medium · Level 2View options
Glucose in water
Sodium chloride in water
Benzoic acid in benzene
Urea in water
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The substance is completely dissociated
The substance forms dimers
The substance gives three ions
The substance is non-volatile
Medium · Level 2View options
0.5
1
2
4
Medium · Level 2View options
0.5
1
2
4
Medium · Level 2View options
When the observed value differs from the normal value
When the value is expressed in grams
When water is used as the solvent
When the temperature is room temperature
Medium · Level 2View options
40 g mol^-1
60 g mol^-1
120 g mol^-1
240 g mol^-1
Medium · Level 2View options
0.5
1
2
3
Medium · Level 2View options
29.25 g mol^-1
58.5 g mol^-1
87.75 g mol^-1
117 g mol^-1
Medium · Level 2View options
Greater than the true molar mass
Less than the true molar mass
Equal to the true molar mass
Will be zero
Medium · Level 2View options
2/3
1.0
3/2
2.0
Medium · Level 2View options
0.80
1.00
1.25
2.50
Medium · Level 2View options
Lower than the true value
Higher than the true value
Always equal to the true value
Always zero
Medium · Level 2View options
Lower than the true value
Higher than the true value
Zero
It cannot differ from the true value
Medium · Level 2View options
1/2
1/3
1
2
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Half
Equal
Double
Three times
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0.5
1
2
3
Medium · Level 2View options
Association of solute particles
Complete dissociation of the solute
Increased ionisation of the solute
Vapour formation of the solvent
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Increase in the number of particles due to dissociation
Decrease in the number of particles due to association
Change in the colour of the solvent
Complete disappearance of the solute
Medium · Level 2View options
Because the number of effective solute particles decreases
Because the number of effective solute particles increases
Because the solvent becomes ionised
Because the temperature always decreases
Question 1MediumLevel 2
A solute’s apparent molar mass determined from osmotic pressure is greater than its actual molar mass. What does this indicate?
Correct answer: B
For a dilute solution, osmotic pressure is related to the number of solute particles by π = iCRT. Association joins two or more solute molecules into larger units, decreasing the number of particles and lowering the observed osmotic pressure. If the ordinary formula is used, this smaller pressure gives an apparent molar mass larger than the true value. Therefore, the observation indicates association.
If the observed molar mass of KCl is \(\frac{2}{3}\) of its true molar mass, what is the value of the van’t Hoff factor \(i\)?
Correct answer: C
For abnormal molar-mass calculations, the van’t Hoff factor is defined as \(i=\frac{M_{\text{true}}}{M_{\text{observed}}}\), or equivalently \(M_{\text{observed}}=\frac{M_{\text{true}}}{i}\). Since the observed molar mass is \(\frac{2}{3}M_{\text{true}}\), \(i=\frac{M_{\text{true}}}{(2/3)M_{\text{true}}}=\frac{3}{2}=1.5\). Thus, option C is correct. A value greater than one indicates effective dissociation of KCl particles in solution.
The normal molar mass of a solute is 120 g mol⁻¹, while its observed molar mass from the osmotic-pressure method is 60 g mol⁻¹. What is the van’t Hoff factor?
Correct answer: C
For abnormal molar mass, the van’t Hoff factor is related to the normal and observed molar masses by i = Mnormal/Mobserved. Substitution gives i = 120/60 = 2. A value greater than one means that the number of solute particles is greater than expected for undissociated molecules. This result is consistent with dissociation, although the numerical question only asks for the factor.
The experimental molar mass of a solute determined from osmotic pressure is greater than its actual molar mass. What behaviour does this mainly indicate?
Correct answer: A
Osmotic pressure depends on the number of dissolved particles. When solute molecules associate, several original molecules combine into fewer particles, reducing the observed osmotic pressure. If the usual formula is then used without correcting for association, the calculated or apparent molar mass becomes larger than the true molar mass. Dissociation would increase the particle count and produce a lower apparent molar mass, so association is indicated.
A solute associates to form dimers in a solution. How will this affect the observed molar mass calculated from a colligative property?
Correct answer: A
Dimerisation combines two original solute molecules into one effective particle. Consequently, the number of particles and the colligative effect become smaller than expected for the same mass of solute. The calculation therefore interprets the sample as containing fewer moles, so the observed molar mass is higher than the true molar mass.
Due to dissociation of an electrolyte, how is the observed molar mass generally obtained from a colligative property?
Correct answer: C
Dissociation changes one solute unit into two or more particles, so the number of effective particles increases. Colligative effects therefore become larger than expected for the undissociated substance. When the larger effect is used in the usual formula, the calculated number of moles is too high and the observed molar mass becomes lower than the true value.
In which solution will the apparent molar mass determined from colligative properties be greater than the true molar mass because of association of solute particles?
Correct answer: C
Benzoic acid associates through hydrogen bonding and commonly forms dimers in non-polar benzene. This lowers the number of effective solute particles, so the van’t Hoff factor is less than one. Because the colligative effect is smaller, the calculated apparent molar mass becomes greater than the true molar mass. Therefore option C is correct.
The observed molar mass of a substance in benzene is found to be twice its normal molar mass. Which conclusion is correct?
Correct answer: B
For abnormal molar mass caused by association, the van’t Hoff factor is related to molar masses by i = normal molar mass/observed molar mass. If the observed molar mass is twice the normal value, i = 1/2. Complete dimerisation changes two original molecules into one associated particle, giving i = 0.5. Thus the substance forms dimers, so option B is correct.
If the observed molar mass of a solute is half its true molar mass, what is the value of the van’t Hoff factor i?
Correct answer: C
For abnormal molar masses, the relation between observed molar mass and true molar mass is Mobserved = Mtrue/i, where i is the van’t Hoff factor. The question states Mobserved = Mtrue/2. Comparing the two expressions gives Mtrue/i = Mtrue/2, so i = 2. A value greater than one is consistent with dissociation, because dissociation increases the number of particles in solution and makes the observed molar mass smaller.
If the observed molar mass of a solute is twice its true molar mass, what is the value of the van’t Hoff factor i?
Correct answer: A
The relation for abnormal molar mass is Mobserved = Mtrue/i. Here, Mobserved = 2Mtrue. Substitution gives 2Mtrue = Mtrue/i. Cancelling the non-zero true molar mass gives 2 = 1/i, so i = 1/2 = 0.5. A value below one indicates association, in which solute particles combine and the effective number of particles becomes smaller than expected, producing an observed molar mass greater than the true value.
In which situation is the molar-mass value called abnormal?
Correct answer: A
The normal molar mass is the value expected from the molecular formula of a substance. In colligative-property measurements, association or dissociation can change the effective number of particles in solution. As a result, the molar mass calculated from the observed colligative effect may differ from the theoretical value. Such a deviation is called abnormal molar mass; it is not defined by the unit, solvent, or room temperature.
If the true molar mass is 120 g mol^-1 and the van't Hoff factor i = 2 for a solute, what observed molar mass will be obtained from a colligative-property measurement?
Correct answer: B
For abnormal molar mass caused by dissociation or association, the observed molar mass is related to the true molar mass by M_observed = M_true/i. Since i = 2, the solute produces twice as many particles as expected for an undissociated solute. Therefore, M_observed = 120/2 = 60 g mol^-1. The observed value is lower because dissociation increases the number of particles. Hence, option B is correct.
The true molar mass of a substance is 60 g mol^-1, but the freezing-point method gives an observed molar mass of 30 g mol^-1. What is the value of the van't Hoff factor i?
Correct answer: C
The relation between true and observed molar mass is M_observed = M_true/i. Substituting the given values, 30 = 60/i. Multiplying by i gives 30i = 60, so i = 2. A value greater than one indicates that the solute produces more particles than expected, usually because of dissociation. Therefore, option C is correct.
If NaCl is completely dissociated and its true molar mass is 58.5 g mol^-1, what will be its ideal observed molar mass approximately?
Correct answer: A
Complete dissociation of NaCl produces two particles, Na+ and Cl-, from one formula unit. Therefore, the ideal van't Hoff factor is i = 2. The observed molar mass obtained from a colligative property is M_observed = M_true/i. Thus, M_observed = 58.5/2 = 29.25 g mol^-1. The lower apparent molar mass reflects the increased number of particles. Hence, option A is correct.
While determining molar mass by the lowering of vapour pressure method, if a solute undergoes complete dimerisation in benzene, how will its observed molar mass compare with its true molar mass?
Correct answer: A
Complete dimerisation means that two solute molecules combine to form one particle in solution. Therefore, the number of solute particles becomes half of the expected number, and the colligative effect, including lowering of vapour pressure, becomes smaller. For complete association into dimers, the van’t Hoff factor is i = 1/2. Since M observed = M true/i, the observed molar mass is 2M true, so it is greater than the true molar mass. This is why association produces an abnormally high observed molar mass.
If the observed molar mass of a solute is 2/3 of its true molar mass, what is the van’t Hoff factor (i)?
Correct answer: C
The relation between observed molar mass and true molar mass is M observed = M true/i, where i is the van’t Hoff factor. The question gives M observed = (2/3)M true. Substitution gives (2/3)M true = M true/i. After cancelling M true, 2/3 = 1/i, so i = 3/2 = 1.5. Thus, option C is correct. The ratio 2/3 is the observed-to-true molar-mass ratio, not the van’t Hoff factor itself.
If the observed molar mass is 1.25 times the true molar mass, what is the van’t Hoff factor (i)?
Correct answer: A
For abnormal molar masses, the observed molar mass is related to the true molar mass by M observed = M true/i. Here M observed = 1.25M true. Therefore, 1.25M true = M true/i, and after cancelling M true, i = 1/1.25 = 0.80. Thus, option A is correct. A value of i below one indicates association, because association decreases the number of solute particles and makes the observed molar mass greater than the true value.
If a solute undergoes dissociation, how will the molar mass calculated without applying the van’t Hoff factor compare with the true molar mass?
Correct answer: A
Dissociation breaks solute molecules or formula units into two or more particles, so the number of particles becomes larger than expected. Consequently, the observed colligative effect is greater. If the van’t Hoff factor is ignored, the calculation attributes this larger effect to more solute moles, and molar mass, calculated as mass divided by moles, appears lower than the true value. Therefore, option A is correct.
If a solute undergoes association, how will the molar mass obtained by a colligative-property method compare with its true molar mass?
Correct answer: B
Association combines two or more solute particles into a larger associated species, so the number of independent particles decreases. The measured colligative effect therefore becomes smaller than the effect expected for non-associated particles. If this reduction is ignored, the calculated number of solute moles is too small. Since molar mass equals mass divided by moles, the apparent molar mass becomes higher than the true value. Thus, option B is correct.
If solute AB completely dissociates into A⁺ and B⁻, the observed molar mass from the colligative method will be what fraction of the true value?
Correct answer: A
Complete dissociation of one AB formula unit produces two particles, A⁺ and B⁻. Therefore, the van’t Hoff factor is i = 2. For a colligative-property measurement, the observed molar mass is related to the true molar mass by M_observed = M_true/i. Hence M_observed = M_true/2, or one-half of the true value. Option A is correct.
If a solute forms complete dimers, the observed molar mass from the colligative method will be how many times the true value?
Correct answer: C
In complete dimerisation, two solute molecules associate to form one dimer particle. Thus the number of particles becomes half the original number, giving a van’t Hoff factor i = 1/2. Since the observed molar mass obtained from colligative properties is M_observed = M_true/i, it becomes M_true/(1/2) = 2M_true. Therefore, it is double the true value, so option C is correct.
If the normal molar mass is 120 g mol⁻¹ and the observed molar mass is 60 g mol⁻¹, what is the van’t Hoff factor?
Correct answer: C
The relation between normal and observed molar masses is i = normal molar mass/observed molar mass. Therefore, i = 120/60 = 2. A van’t Hoff factor greater than one indicates that the number of solute particles has increased, usually because the solute dissociates or ionises in solution. Thus, option C is correct.
If the observed molar mass is greater than the normal molar mass, what is the simplest possible reason?
Correct answer: A
For abnormal molar masses, i = normal molar mass/observed molar mass. If association occurs, several solute molecules combine to form larger units, so the number of particles decreases. The colligative effect becomes smaller than expected, giving i less than one and an observed molar mass greater than the normal value. Hence association is correct.
If an electrolyte gives an observed molar mass lower than its normal molar mass during molar-mass determination, what is the reason?
Correct answer: A
Dissociation of an electrolyte produces two or more ions from each formula unit, thereby increasing the effective number of solute particles. Since colligative properties depend on particle number, the observed effect becomes larger than expected for the undissociated substance. When the ordinary formula is used to calculate molar mass, this larger effect gives a lower observed molar mass. Hence, option A is correct.
If a solute associates to form dimers, why does the molar mass obtained from colligative properties generally increase?
Correct answer: A
When two solute molecules associate to form one dimer, the number of independently acting solute particles becomes smaller. Colligative properties depend on the number of particles, so the observed effect becomes less than expected for the unassociated solute. If this smaller effect is interpreted using the ordinary formula, the calculated number of moles appears smaller and the molar mass, mass divided by moles, appears larger. This is called an abnormal molar mass due to association.
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