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In this Class 12 Chemistry topic from Chapter 01: Solutions, students learn how the physical properties of a solution depend on the number of dissolved solute particles rather than their chemical identity. The topic explains lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure. Students also apply colligative-property equations to calculate molar mass, understand dilute solutions, and use the van’t Hoff factor to interpret association or dissociation of solute particles.
TOPIC PRACTICE
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25 questions
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Molar concentration of solute
Mass of solvent
Colour of solution
Melting point of solute
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It becomes half
It becomes double
It becomes zero
It remains unchanged
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0.5 mol kg⁻¹
1.0 mol kg⁻¹
2.0 mol kg⁻¹
5.58 mol kg⁻¹
Easy · Level 1View options
0.26 mol kg⁻¹
1.5 mol kg⁻¹
2.0 mol kg⁻¹
0.67 mol kg⁻¹
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1
2
3
4
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Solute particles hinder the formation of the ordered solid structure
The mass of the solvent always becomes zero
No particles remain in the solution
The solute always changes into a gas
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It will be zero
It will be only from left to right
It will be only from right to left
It will be very rapid
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Hypotonic
Isotonic
Hypertonic
Non-volatile
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0.01 mol kg⁻¹
0.1 mol kg⁻¹
1.0 mol kg⁻¹
10 mol kg⁻¹
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Half
Double
Same
Zero
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It is neither dissociating nor associating
It is giving three ions
It is completely associated
It has become the solvent
Easy · Level 1View options
372.4
373.0
373.6
0.6
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Viscosity
Surface tension
Colligative property
Refractive index
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90
100
10
0.9
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Moles of solute, n
Colour of the solvent
Only the thickness of the membrane
Shape of the container
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The 0.1 mol L⁻¹ solution
The 0.2 mol L⁻¹ solution
Both have the same osmotic pressure
Neither solution has osmotic pressure
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The number of effective solute particles
The colour of the solution
The smell of the solvent
The shine of the glass container
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(0)
(0.5)
(1)
(2)
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Molality
Mole fraction
Mass percentage
Volume percentage
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2 units
3 units
5 units
6 units
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Boiling-point elevation requires Kb
Freezing-point depression requires Kb
No constant is required
Kf and Kb are always equal
Easy · Level 1View options
They depend on the effective number of solute particles
They depend only on the colour of the solute
They depend only on the name of the solvent
They have no relation with concentration
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0.5
1
2
3
Easy · Level 1View options
2
1
0.5
0
Easy · Level 1View options
Elevation of boiling point
Depression of freezing point
Osmotic pressure
Viscosity
Question 1EasyLevel 1
In the osmotic pressure relation π = CRT, what does C represent?
Correct answer: A
For a dilute solution, the osmotic-pressure equation is π = CRT, where π is osmotic pressure, C is the molar concentration of the solute, R is the gas constant and T is the absolute temperature. Thus C is measured in moles per litre, or molarity. It does not represent solvent mass, colour or melting point, so option A is correct.
If the molar concentration of a solution is doubled while the temperature remains constant, what happens to its osmotic pressure?
Correct answer: B
For a dilute solution, osmotic pressure is given by the van’t Hoff equation π = CRT, where C is molar concentration, R is the gas constant, and T is absolute temperature. When temperature and R remain constant, π is directly proportional to C. Therefore, doubling the molar concentration doubles the osmotic pressure, so option B is correct.
For a solution, ΔTf = 3.72 K and Kf = 1.86 K kg mol⁻¹. What is the molality of the non-dissociating solute?
Correct answer: C
For a non-dissociating solute, the van’t Hoff factor is i = 1, so the freezing-point depression equation is ΔTf = Kf m. Rearranging gives m = ΔTf/Kf. Substitution gives m = 3.72/1.86 = 2.0 mol kg⁻¹. Therefore, option C is correct. The units of Kf and the calculated molality are consistent, and no dissociation or association correction is required.
In a solution, ΔTb = 0.78 K and Kb = 0.52 K kg mol⁻¹. What is the molality of the non-dissociating solute?
Correct answer: B
For a non-dissociating solute, i = 1 and the elevation in boiling point is given by ΔTb = Kb m. Hence, m = ΔTb/Kb. Using the given values, m = 0.78/0.52 = 1.5 mol kg⁻¹. Thus option B is correct. The boiling-point constant Kb must be used here because the question gives boiling-point elevation, not freezing-point depression.
If CaCl₂ dissociates completely, what is its ideal van’t Hoff factor (i)?
Correct answer: C
One formula unit of calcium chloride, CaCl₂, separates completely in water according to CaCl₂ → Ca²⁺ + 2Cl⁻. This produces one calcium ion and two chloride ions, giving a total of three solute particles. For complete dissociation under ideal conditions, the van’t Hoff factor equals the number of particles produced from one formula unit. Therefore, i = 3 and option C is correct.
Why is the freezing point of a solution lower than that of pure water?
Correct answer: A
Freezing requires solvent molecules to arrange themselves into an ordered solid structure. When a non-volatile solute is dissolved, its particles become distributed among the solvent molecules and disturb this orderly arrangement. Consequently, the solution must be cooled to a temperature lower than the normal freezing point of pure water before the solvent can crystallize. This lowering is called depression of freezing point and is a colligative property.
If two solutions are isotonic, what will be the net flow of solvent across a semipermeable membrane?
Correct answer: A
Isotonic solutions have equal osmotic pressures at the same temperature. Solvent molecules may still cross the semipermeable membrane in both directions, but the rates of movement are equal in opposite directions. Consequently, the two-way movements cancel and the net solvent flow is zero. Therefore, isotonicity means absence of net osmosis, not that molecular movement completely stops.
A solution has a higher osmotic pressure than another solution. What is it called relative to the other solution?
Correct answer: C
At the same temperature, osmotic pressure is related to the effective concentration of solute particles. A solution with greater osmotic pressure has a greater effective solute-particle concentration than the reference solution. It is therefore described as hypertonic relative to that solution. Hypotonic means lower osmotic pressure, while isotonic means equal osmotic pressure. Thus the correct answer is hypertonic.
If 0.1 mol of a non-dissociating solute is dissolved in 1 kg of solvent, what is the molality?
Correct answer: B
Molality is defined as the number of moles of solute divided by the mass of solvent in kilograms: m = moles of solute / kilograms of solvent. Here, the solute amount is 0.1 mol and the solvent mass is 1 kg. Therefore, m = 0.1/1 = 0.1 mol kg⁻¹. The fact that the solute does not dissociate means no particle-factor correction is needed.
If a solute has i = 2, how will ΔT_f = iK_fm compare with the non-dissociated case?
Correct answer: B
For a non-dissociated solute, the van’t Hoff factor is i = 1, so the freezing-point depression is ΔT_f = K_fm. If the solute produces twice as many effective particles and i becomes 2, the equation becomes ΔT_f = 2K_fm. Thus, at the same K_f and molality, the depression is twice the non-dissociated value.
A solution has i = 1. In this simple context, what can be inferred about the solute?
Correct answer: A
A van’t Hoff factor of i = 1 means that the effective number of particles is equal to the number expected from the solute formula. In the simple idealized interpretation, there is no significant dissociation that would make i greater than 1 and no association that would make i less than 1. Thus, the solute is treated as neither dissociating nor associating.
The boiling-point elevation of a solution is 0.6 and the boiling point of the pure solvent is 373. What is the boiling point of the solution?
Correct answer: C
Boiling-point elevation is defined as the increase in boiling point caused by adding a non-volatile solute. Therefore, the boiling point of the solution is T_b = T_b° + ΔT_b. Substituting the values gives T_b = 373 + 0.6 = 373.6. The value must be added, not subtracted, so option C is correct.
Which of the following properties depends only on the number of solute particles present in a solution, not on their chemical nature?
Correct answer: C
A colligative property depends primarily on the number of dissolved solute particles, rather than on the chemical identity of those particles. Important examples are relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure. Viscosity, surface tension, and refractive index generally depend on the nature of the substances as well.
If the mole fraction of the solvent is 0.9 and the vapour pressure of the pure solvent is 100, what will be the vapour pressure of the solvent in the solution?
Correct answer: A
For a non-volatile solute, Raoult’s law gives the solvent vapour pressure as p = x_solvent p°. Here, the solvent mole fraction is 0.9 and the pure-solvent vapour pressure p° is 100. Therefore, p = 0.9 × 100 = 90. The vapour pressure is lowered from 100 because some of the liquid particles are solute particles.
If V and T are constant in the equation πV = nRT, osmotic pressure is proportional to which quantity?
Correct answer: A
Starting from πV = nRT, divide both sides by V to obtain π = nRT/V. If V and T are fixed, and R is the universal gas constant, then every factor except n remains constant. Consequently, π increases directly with the number of solute moles: doubling n doubles π under the stated conditions. Therefore, option A is correct.
Two solutions of equal volume have concentrations 0.1 mol L⁻¹ and 0.2 mol L⁻¹. At the same temperature, which solution has the higher osmotic pressure?
Correct answer: B
For a dilute solution, osmotic pressure is given by π = CRT, where C is the molar concentration and T is absolute temperature. At the same temperature, π is directly proportional to C. Since 0.2 mol L⁻¹ is twice 0.1 mol L⁻¹, the 0.2 mol L⁻¹ solution has the greater osmotic pressure, approximately twice as large if the solutes have the same van’t Hoff factor. Hence B is correct.
What is the common basis of vapour-pressure lowering, freezing-point depression, boiling-point elevation, and osmotic pressure in a solution?
Correct answer: A
These four effects are colligative properties. For an ideal dilute solution, their magnitude depends primarily on the number of independently moving solute particles relative to the amount of solvent, not on the chemical identity of those particles. Dissociation increases the effective particle count, whereas association decreases it. Therefore, the common basis is the number of effective solute particles, making A correct.
What is the sum of the mole fraction of the solute and the mole fraction of the solvent in a binary solution?
Correct answer: C
A binary solution contains two components: the solute and the solvent. If their mole fractions are represented by x_solute and x_solvent, the total number of moles is the sum of the moles of both components. Therefore, x_solute + x_solvent = 1. Mole fraction is dimensionless, and the sum of the mole fractions of all components in any solution is always unity. Hence, option C, (1), is correct.
Which concentration unit is used directly in calculating freezing-point depression and boiling-point elevation?
Correct answer: A
For a dilute solution containing a non-volatile solute, freezing-point depression is given by ΔTf = iKf m and boiling-point elevation is given by ΔTb = iKb m. In both equations, m denotes molality, which is the number of moles of solute per kilogram of solvent. Molality is preferred because it does not change with temperature-related volume changes.
The osmotic pressure of a solution is 2 units. If its concentration is tripled at the same temperature, what will be the new osmotic pressure?
Correct answer: D
For a dilute solution, osmotic pressure is expressed as π = iCRT. When temperature and the nature of the solute remain unchanged, π is directly proportional to concentration C. Therefore, tripling the concentration triples the osmotic pressure: πnew = 3 × 2 = 6 units. Hence option D is correct.
A student used Kf while calculating ΔTb. What is the mistake?
Correct answer: A
The symbol ΔTb denotes the elevation in boiling point, so its relation is ΔTb = iKb m. The subscript b refers to boiling, whereas Kf belongs to freezing-point depression, ΔTf = iKf m. Kb and Kf are solvent-specific constants and are not generally equal. Therefore, using Kf for ΔTb is the error.
Which of the following statements about colligative properties is correct?
Correct answer: A
Colligative properties are properties of dilute solutions that depend primarily on the number of dissolved solute particles, not on their chemical identity. The four main examples are relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure. Dissociation and association change the effective particle number through the van’t Hoff factor.
If 0.1 mol of NaCl completely dissociates to produce 0.2 mol of effective particles, what is the van’t Hoff factor i?
Correct answer: C
The van’t Hoff factor is defined as the ratio of the actual number of solute particles in solution to the number expected if no dissociation or association occurred. Here, i = moles of effective particles / initial moles of NaCl = 0.2/0.1 = 2. This agrees with complete dissociation of NaCl into Na⁺ and Cl⁻ ions.
If two solute particles associate to form one particle, what is the ideal value of the van’t Hoff factor i?
Correct answer: C
The van’t Hoff factor compares the number of particles actually present after association or dissociation with the number initially expected. When two particles combine to form one particle, the effective number becomes half of the original number. Therefore, i = actual particles/initial particles = 1/2 = 0.5. This is the ideal value for complete pairwise association.
Which of the following is not a colligative property that depends on the number of solute particles present in a solution?
Correct answer: D
Viscosity is not a colligative property. It depends on the nature of both solute and solvent, intermolecular forces, temperature, and the shape or size of dissolved particles. In contrast, elevation of boiling point, depression of freezing point, osmotic pressure, and relative lowering of vapour pressure are the standard colligative properties because they depend mainly on the number of solute particles.
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