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In Class 12 Chemistry, Chapter 01: Solutions, this topic introduces colligative properties—properties that depend on the number of dissolved solute particles rather than their chemical identity. Students learn relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure for dilute solutions. The topic also develops relationships involving molality, concentration, molar mass, and the van’t Hoff factor, helping students understand the behaviour of electrolytes and the calculation of abnormal molar masses.
TOPIC PRACTICE
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Easy · Level 8View options
The one whose solvent has the larger Kf value
The one containing the darker solute
The one kept in the larger container
The one with the longer solution name
Easy · Level 8View options
Osmotic pressure increases
Osmotic pressure decreases
Osmotic pressure becomes zero
Osmotic pressure depends on colour
Easy · Level 8View options
Colligative property
Chemical colour
Flame colour
Metallic lustre
Easy · Level 8View options
As a correction factor for particle number
As a colour indicator
As a heat source
To remove solubility
Easy · Level 8View options
The number of solute particles
The colour of the solute
The smell of the solution
The height of the vessel
Easy · Level 8View options
Colligative properties
Magnetic properties
Colour properties
Sound properties
Easy · Level 8View options
They depend on the number of solute particles
They depend only on solute colour
They depend only on container shape
They depend only on smell
Easy · Level 8View options
It will increase
It will decrease
It will always be zero
It will have no effect
Easy · Level 8View options
It will decrease
It will increase
It will become infinite
It will equal the solvent mass
Easy · Level 8View options
2
0.5
1
4
Easy · Level 8View options
0.75
1.25
2.00
0.25
Easy · Level 8View options
van’t Hoff factor
Acidity constant
Solution colour factor
Evaporation factor
Easy · Level 8View options
1.5
0.5
2.5
1.0
Easy · Level 8View options
0.5
2
1.5
1
Easy · Level 8View options
Glucose
NaCl
CaCl₂
AlCl₃
Easy · Level 8View options
The effective number of particles doubled
The colour of the solute changed
The name of the solvent changed
The number of particles became half
Easy · Level 8View options
Completely dimerised solute
Glucose
Completely dissociated NaCl
Completely dissociated AlCl3
Easy · Level 8View options
i = actual effective particle number / normal expected particle number
i = normal mass / solvent mass
i = temperature / volume
i = colour / pressure
Easy · Level 8View options
0.02 mol
0.04 mol
0.06 mol
0.08 mol
Easy · Level 8View options
1.6 times
0.6 times
Equal
2.6 times
Easy · Level 8View options
0.8 times
1.25 times
2 times
Equal
Easy · Level 8View options
0.9 times
1.1 times
1.9 times
Equal
Easy · Level 8View options
1.25 times
1.50 times
1.75 times
2.00 times
Easy · Level 8View options
25% higher
50% higher
75% higher
125% higher
Question 1EasyLevel 8
For two non-electrolyte solutions having the same molality, which solution will show the greater depression in freezing point?
Correct answer: A
For a dilute non-electrolyte solution, the depression in freezing point is given by ΔTf = Kf m. In the comparison, the molality m is the same for both solutions, so the magnitude of ΔTf depends on the cryoscopic constant Kf of the solvent. The solvent with the larger Kf therefore produces the greater freezing-point depression; colour, container size, and name are irrelevant.
What happens to osmotic pressure when the number of particles in solution increases?
Correct answer: A
For a dilute solution, osmotic pressure is given by π = iCRT, or equivalently π = cRT when c is the actual particle concentration. At fixed temperature and volume, increasing the number of dissolved particles increases c and therefore increases π. Dissociation is one way to produce more particles. Osmotic pressure does not become zero or depend on colour, so A is correct.
In abnormal molecular mass calculations, which property is directly related to number of solute particles?
Correct answer: A
A colligative property depends on how many solute particles are present, rather than on the chemical identity of those particles. Vapour-pressure lowering, boiling-point elevation, freezing-point depression, and osmotic pressure are examples used for molecular-mass determination. Association and dissociation alter the effective particle number and therefore cause abnormal results. Colour, flame colour, and lustre do not provide this particle-count relation.
How is the van’t Hoff factor used in colligative properties?
Correct answer: A
Colligative properties depend on the number of dissolved particles, not merely on the chemical formula of the solute. Dissociation increases particle number, while association decreases it. The van’t Hoff factor i accounts for this change, so it is used as a particle-number correction factor in equations for osmotic pressure, boiling-point elevation and freezing-point depression.
Colligative properties depend primarily on which factor?
Correct answer: A
Colligative properties depend on how many dissolved particles are present, not on the chemical identity of those particles in the ideal dilute limit. Vapour-pressure lowering, boiling-point elevation, freezing-point depression and osmotic pressure all respond to particle number. Association and dissociation matter because they change this number.
The concept of abnormal molar mass is most closely related to which type of properties?
Correct answer: A
Abnormal molar mass is usually inferred from measurements such as osmotic pressure, boiling-point elevation or freezing-point depression. These are colligative properties and depend on the number of dissolved particles. Association and dissociation change that number, so the inferred molar mass becomes abnormal.
Which statement is correct for colligative properties and abnormal molecular mass?
Correct answer: A
Colligative properties depend on the number of dissolved particles, not on their individual colour, smell, or chemical identity in the ideal approximation. Association and dissociation change that number, so they alter the measured property and can produce an abnormal calculated molar mass. Container shape also does not determine a colligative property when concentration and temperature are fixed.
What is the effect of dissociation on depression in freezing point?
Correct answer: A
Depression in freezing point is a colligative property, so it depends on the total number of dissolved particles. Dissociation changes one solute unit into two or more ions and therefore increases the particle count. Since ΔTf = iKf m, a value of i greater than one makes the observed depression larger than the normal calculated value. It does not make the depression zero.
What happens to elevation in boiling point due to association?
Correct answer: A
Elevation in boiling point is a colligative effect and is proportional to the number of dissolved particles. Association combines separate solute molecules into larger species, thereby reducing the effective particle count. In ΔTb = iKb m, association gives i less than one, so the observed elevation is smaller than the value calculated without association. It cannot become infinite or equal a mass.
If the measured depression in freezing point is double the expected value, what is i?
Correct answer: A
The wording of the question states that the measured freezing-point depression is twice the value expected for unchanged particles. Since ΔTf(observed) = i × ΔTf(normal), the ratio observed/normal is directly i. Therefore i = 2. Such a value indicates an increased number of particles, usually due to dissociation rather than association.
If the measured elevation in boiling point is 0.75 times the expected value, what is i?
Correct answer: A
The abnormal boiling-point elevation is related to the normal value by ΔTb(observed) = i × ΔTb(normal). The question gives the observed-to-expected ratio as 0.75, so i = 0.75. Since this is less than one, the effective particle number is reduced, which is consistent with association. It is not the complement 0.25.
Which factor is included in colligative-property formulas for an abnormal solution?
Correct answer: A
The van’t Hoff factor i corrects the ideal colligative-property equation for the actual number of solute particles. It is multiplied into expressions such as ΔTb = iKbm, ΔTf = iKfm, and π = iCRT. Association gives i below one, while dissociation gives i above one. Acidity, colour, and evaporation factors do not provide this particle-number correction.
If the measured depression in freezing point is 1.5 times the expected value, what is the van’t Hoff factor (i)?
Correct answer: A
For a fixed solvent, solute concentration, and temperature conditions, depression in freezing point is proportional to the number of dissolved particles: ΔTf = iKf m. The expected value corresponds to i = 1. If the measured depression is 1.5 times that value, then the multiplying factor is i = 1.5. This indicates more particles than predicted for a non-associated, non-dissociated solute.
If measured osmotic pressure is half the expected value, what is the value of i?
Correct answer: A
Osmotic pressure is given by π = iCRT. When concentration, gas constant, and temperature are unchanged, the ratio of actual pressure to the normal expected pressure equals i. A pressure equal to one-half of the expected value therefore gives i = 0.5. Such a value indicates fewer effective particles, commonly because of association rather than dissociation.
Among solutions of equal molality, which gives the maximum depression in freezing point if dissociation is complete?
Correct answer: D
For equal molality, ΔTf = iKf m, so the solution with the largest van’t Hoff factor gives the greatest depression. Glucose is a non-electrolyte and gives i = 1. NaCl gives two ions, CaCl₂ gives three, and AlCl₃ gives four ions on complete dissociation. Therefore AlCl₃ has the largest i and produces the maximum freezing-point depression, assuming ideal complete dissociation.
A colligative property doubles while the masses of solute and solvent remain the same. What is the most likely reason?
Correct answer: A
Colligative properties depend on the number of dissolved particles, not on their colour or chemical name alone. With the same masses and concentration conditions, a doubled effect corresponds to approximately i = 2, meaning twice as many effective particles as in the normal molecular case. Dissociation of an electrolyte is a common way to produce this increase.
At the same molality, which solution gives the lowest osmotic pressure near the ideal limiting case?
Correct answer: A
At fixed temperature and molality, osmotic pressure is proportional to i. Complete dimerisation changes two solute molecules into one particle, so i = 1/2. Glucose has i about 1, NaCl gives i about 2 when fully dissociated, and AlCl3 gives i about 4. Therefore the completely dimerised solute has the fewest particles and the lowest osmotic pressure.
Which statement gives the correct relation between i and particle number?
Correct answer: A
The van’t Hoff factor compares the number of particles actually present in solution with the number expected if the solute neither dissociated nor associated. Thus i is the ratio of actual effective particle number to normal expected particle number. Dissociation gives i greater than one, association gives i less than one, and no abnormality gives i equal to one.
If 0.02 mol of AB₂ dissociates completely, how many moles of effective particles are formed?
Correct answer: C
The governing idea is that complete dissociation changes one formula unit into all of its constituent ions. Each AB₂ unit produces one A ion and two B ions, so the total number of particles is three per formula unit. Therefore, effective particle moles = 0.02 × 3 = 0.06 mol. Option A ignores dissociation, while options B and D use incorrect particle multipliers. Thus option C is correct.
If a solute has i = 1.6, how will its osmotic pressure compare with that of a normal solute of the same molality?
Correct answer: A
For dilute solutions at the same temperature and molality, osmotic pressure is proportional to the van’t Hoff factor i. A normal nonelectrolyte has i = 1, while the given solute has i = 1.6. Therefore its osmotic pressure is 1.6 times that of the normal solution. The increase indicates more effective particles, usually because of dissociation.
If i = 0.8 and the substance undergoes only dimerisation, how will its boiling-point elevation compare with that of a normal solute at the same molality?
Correct answer: A
For dilute solutions, boiling-point elevation is ΔTb = iKb m. At the same molality and with the same solvent, Kb and m are unchanged, so the ratio of elevations is simply the ratio of i values. A normal nonelectrolyte has i = 1, whereas this solute has i = 0.8; its elevation is therefore 0.8 times as large.
If a solute has i = 0.9, how will its boiling-point elevation compare with that of a normal solute of the same molality?
Correct answer: A
The boiling-point elevation is Delta Tb = i Kb m. For a normal non-electrolyte, i = 1. At the same molality and in the same solvent, the ratio of elevations is therefore 0.9/1 = 0.9. Thus the solution shows 0.9 times the normal elevation, reflecting a slight reduction in particles due to association.
If KCl is 75% dissociated, how many times is its osmotic pressure compared with glucose at the same concentration?
Correct answer: C
KCl dissociates into two ions, so for degree of dissociation alpha its van’t Hoff factor is i = 1 + alpha. With alpha = 0.75, iKCl = 1.75. Glucose is a non-electrolyte and does not dissociate, so iGlucose = 1. At the same concentration and temperature, osmotic pressure is proportional to i; therefore the ratio is 1.75. Option C is correct.
If a solute has i = 1.25, how much higher will its colligative effect be than the normal value?
Correct answer: A
For fixed concentration, solvent, and temperature, a colligative effect is proportional to the van’t Hoff factor i. A normal non-electrolyte has i = 1, whereas this solute has i = 1.25. Its effect is therefore 1.25 times the normal value. The increase is (1.25 − 1) × 100 = 25%. Hence option A is correct; 125% describes the total relative value, not the increase.
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