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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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Easy · Level 5View options
In an isotonic solution
In a hypotonic solution
In a hypertonic solution
In pure water
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When the solute is neither dissociated nor associated
When the solute is completely ionised
When the solute forms dimers
When the solute produces three ions
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Elevation of boiling point
Depression of freezing point
Osmotic pressure
Lowering of vapour pressure
Easy · Level 5View options
0.10
0.50
0.90
1.10
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The concentration became half
The concentration became double
The concentration became four times
The concentration became zero
Easy · Level 5View options
The outside solution has higher osmotic pressure
The outside solution has zero osmotic pressure
The cell contains no water
The membrane is completely impermeable
Easy · Level 5View options
Water enters because the outside solution has lower osmotic pressure
The outside solution has zero vapour pressure
The cell temperature suddenly increases
Solute completely disappears from the cell
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They depend on the nature of solute particles, not their number
They depend only on the colour of the solvent
They depend on the number of effective solute particles
They occur only for solid solutes
Easy · Level 5View options
It will increase
It will decrease further
It will remain the same
It will first decrease and then necessarily become zero
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0.005 M
0.010 M
0.020 M
2.00 M
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Viscosity
Surface tension
Freezing-point depression
Refractive index
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0.78 K
100.78 K
200.78 K
−0.78 K
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0.97
0.03
3.00
0.003
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Both depend only on the nature of the solute
Both depend on the nature of the solvent
Both are always identical for every solvent
Neither is related to colligative properties
Easy · Level 5View options
Lowering of vapour pressure
Depression in freezing point
Elevation in boiling point
Osmotic pressure
Easy · Level 5View options
0.01 (शून्य दशमलव शून्य एक)
0.02 (शून्य दशमलव शून्य दो)
0.20 (शून्य दशमलव दो शून्य)
0.98 (शून्य दशमलव नौ आठ)
Easy · Level 5View options
The freezing-point depression will be smaller
The freezing-point depression will be larger
The boiling point will decrease
The vapour pressure will increase
Easy · Level 5View options
Osmotic pressure
Depression in freezing point
Elevation in boiling point
Viscosity
Easy · Level 5View options
It decreases
It increases
It becomes zero
It remains the same
Easy · Level 5View options
1.1 K
271.9 K
273.0 K
544.9 K
Easy · Level 5View options
The number of solvent molecules at the surface decreases
The solute molecules become more volatile
The molar mass of the solvent decreases
The external pressure always increases
Easy · Level 5View options
They depend only on the colour of the solute
They depend only on the number of dissolved particles
They are always independent of the mass of solvent
They occur only with metallic solutes
Easy · Level 5View options
Vapour pressure
Osmotic pressure
Partial pressure
Critical pressure
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1.23 atm
2.46 atm
4.92 atm
9.84 atm
Easy · Level 5View options
The solute is neither associated nor dissociated
The solute is completely dissociated
The solute is completely dimerised
The solute gives three ions
Question 1EasyLevel 5
In which condition is a red blood cell most likely to shrink?
Correct answer: C
A hypertonic solution has a higher effective solute concentration outside the red blood cell than inside it. Water therefore moves out of the cell through its selectively permeable membrane by osmosis, causing the cell to lose volume and shrink; this is called crenation. In an isotonic solution there is no net water movement, while a hypotonic solution or pure water drives water into the cell and may cause swelling or haemolysis.
Under which condition will the van’t Hoff factor be very close to 1?
Correct answer: A
The van’t Hoff factor compares the actual number of solute particles in solution with the number expected from the undissociated formula units. If the solute neither dissociates into smaller particles nor associates into larger particles, the particle number remains unchanged. Consequently, i remains approximately 1. Complete ionisation or formation of dimers changes the particle number and therefore changes i.
Which colligative property is not directly based on vapour-pressure measurement but requires a semipermeable membrane?
Correct answer: C
Osmotic pressure is measured by considering the pressure required to stop the movement of solvent through a semipermeable membrane. It therefore specifically requires a membrane that permits solvent molecules to pass but restricts solute particles. Elevation of boiling point, depression of freezing point, and lowering of vapour pressure are studied through temperature or vapour-pressure relationships and do not require a membrane for their basic definition.
If the relative lowering of vapour pressure is 0.10, what is the mole fraction of the solvent in an ideal solution containing a non-volatile solute?
Correct answer: C
For an ideal solution containing a non-volatile solute, Raoult’s law gives the relative lowering of vapour pressure as (p° − p)/p° = Xsolute. Therefore, Xsolute = 0.10. Since the mole fractions of solute and solvent add to one, Xsolvent = 1 − 0.10 = 0.90. Hence option C is correct; a mole fraction cannot exceed one.
The osmotic pressure of a solution becomes double at the same temperature. If i remains unchanged, what change must have occurred in concentration?
Correct answer: B
For a dilute solution, osmotic pressure is expressed as π = iCRT. When temperature T and van’t Hoff factor i are unchanged, π is directly proportional to concentration C. Consequently, if the osmotic pressure changes from π to 2π, the concentration must change from C to 2C. It cannot become half or four times, because those changes would produce π/2 or 4π under the stated constant conditions.
If a blood cell is placed in a hypertonic solution, why will water move outward?
Correct answer: A
A hypertonic solution has a greater concentration of osmotically active particles than the fluid inside the blood cell. Consequently, its osmotic pressure is higher. Across a selectively permeable membrane, water moves from the region of lower effective solute concentration toward the region of higher effective solute concentration. Therefore water leaves the cell, and option A is correct.
A cell placed in a hypotonic solution may swell. What is the correct colligative reason?
Correct answer: A
A hypotonic solution has a lower concentration of effective solute particles and therefore a lower osmotic pressure than the cell interior. The cell membrane permits water movement more readily than solute movement. Water consequently enters the cell from the outside, where effective particle concentration is lower, toward the cell fluid, where it is higher. This influx can make the cell swell.
Which statement is most accurate for colligative properties?
Correct answer: C
Colligative properties depend primarily on the number of solute particles present in a given amount of solvent, rather than on the chemical identity of those particles. Examples include relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure. If dissociation or association occurs, the effective particle number changes, so the van’t Hoff factor is used.
If the amount of a non-volatile solute is increased in a solution, how will the vapour pressure of the solution change compared with that of the pure solvent?
Correct answer: B
A non-volatile solute does not itself contribute appreciably to the vapour phase. Its presence lowers the mole fraction and escaping tendency of the solvent, so the solution has a lower vapour pressure than the pure solvent. Increasing the amount of solute lowers the solvent mole fraction further and produces a further decrease in vapour pressure. It does not necessarily become exactly zero.
A solution contains 0.010 mol of solute in 1.00 L of solution. If its van’t Hoff factor is i = 2.0, what effective concentration should be used to calculate its osmotic pressure at 300 K?
Correct answer: C
The actual molarity is C = n/V = 0.010 mol/1.00 L = 0.010 M. For an electrolyte or any solute represented by a van’t Hoff factor, osmotic pressure is π = iCRT. Thus the concentration of effective particles is iC = 2.0 × 0.010 M = 0.020 M. This effective concentration, rather than the analytical molarity alone, must be used in the osmotic-pressure expression.
Which of the following properties ideally depends only on the number of solute particles present in a solution and not on their chemical nature?
Correct answer: C
Freezing-point depression is a colligative property. In a sufficiently dilute solution, its value depends on the number of dissolved particles, represented by molality and the van’t Hoff factor, rather than on the identity of the solute. Viscosity, surface tension, and refractive index depend strongly on molecular interactions and chemical nature.
If the boiling point of a solution is \(100.78^\circ C\) and that of pure water is \(100.00^\circ C\), what is the elevation in boiling point?
Correct answer: A
The elevation in boiling point is the difference between the boiling point of the solution and that of the pure solvent: \(\Delta T_b=T_b(\text{solution})-T_b^0(\text{solvent})\). Hence \(\Delta T_b=100.78-100.00=0.78^\circ C\). A temperature difference has the same numerical value in kelvin, so the answer is 0.78 K.
If the mole fraction of a non-volatile solute in an ideal dilute solution is 0.03, what is the relative lowering of vapour pressure?
Correct answer: B
For a solution containing a non-volatile solute, Raoult’s law gives the relative lowering of vapour pressure as (p° − p)/p° = X_solute, where p° is the vapour pressure of the pure solvent and p is that of the solution. Since X_solute = 0.03, the relative lowering is 0.03, or 3%. The solvent mole fraction, 0.97, is not the requested quantity.
Which statement about the ebullioscopic constant \(K_b\) and cryoscopic constant \(K_f\) is correct?
Correct answer: B
The constants \(K_b\) and \(K_f\) are characteristic properties of the solvent. They depend on quantities such as the solvent’s enthalpy of vaporisation or fusion, molar mass, and boiling or freezing temperature. They do not depend on the chemical identity of a dilute, non-electrolyte solute, although the observed change also depends on molality.
Which colligative property is directly used to explain the bursting or shrinking of blood cells?
Correct answer: D
Blood-cell membranes are selectively permeable, so water can move across them while many solutes cannot. In a hypotonic medium, water enters the cell and it may swell or burst; in a hypertonic medium, water leaves and the cell shrinks. These effects are explained by osmosis and differences in osmotic pressure, so option D is correct.
If the relative lowering of vapour pressure of a solution is 0.02, what is the approximate mole fraction of the solute in a dilute solution?
Correct answer: B
For a solution containing a non-volatile solute, Raoult’s law states that the relative lowering of vapour pressure is equal to the mole fraction of the solute: (P° − P)/P° = Xsolute. Therefore, when the relative lowering is 0.02, Xsolute is approximately 0.02. This relation is especially applicable to dilute solutions, where the solvent mole fraction is close to one and the ideal-solution approximation is valid. Hence, option B is correct.
If a solvent has a higher Kf, which conclusion is correct for the same molality of a non-volatile solute?
Correct answer: B
For a non-electrolyte at a fixed molality, the depression in freezing point is ΔT_f = K_fm. If the solvent’s cryoscopic constant K_f is larger while m remains unchanged, the product K_fm becomes larger. Thus the solution shows a greater freezing-point depression. This conclusion concerns freezing behavior, not a decrease in boiling point or an increase in vapour pressure.
Which of the following is not a colligative property?
Correct answer: D
Colligative properties depend primarily on the number of dissolved particles, rather than on their chemical identity. Osmotic pressure, depression in freezing point, elevation in boiling point, and relative lowering of vapour pressure are colligative properties. Viscosity is not colligative because it depends on the nature of the solute and solvent, molecular interactions, and temperature. Hence, viscosity is the correct answer.
If the temperature is increased while measuring the osmotic pressure and the concentration remains constant, what happens to the osmotic pressure?
Correct answer: B
For a dilute solution, osmotic pressure is expressed as π = iCRT. If concentration C and the van’t Hoff factor i remain constant, osmotic pressure is directly proportional to the absolute temperature T. Thus, increasing temperature increases the osmotic pressure. It does not become zero or remain unchanged, so option B is correct.
If the freezing point of a solution is 271.9 K and that of the pure solvent is 273.0 K, what is the depression in freezing point?
Correct answer: A
The depression in freezing point is the decrease in the freezing point caused by dissolving a solute. It is calculated as the freezing point of the pure solvent minus the freezing point of the solution: ΔTf = Tf° − Tf = 273.0 K − 271.9 K = 1.1 K. Therefore, option A is correct. The temperature difference in kelvin has the same numerical size as the difference in degrees Celsius.
If the vapour pressure of a solution is lower than that of the pure solvent, what is the main reason?
Correct answer: A
When a non-volatile solute is added to a solvent, solute particles occupy some positions at the liquid surface and reduce the mole fraction of the solvent. Consequently, fewer solvent molecules can escape from the surface into the vapour phase. According to Raoult’s law, the partial vapour pressure of the solvent becomes lower than its vapour pressure in the pure state. Therefore, option A is correct.
Which statement correctly describes the characteristic of colligative properties?
Correct answer: B
Colligative properties depend primarily on the number of solute particles present in a given amount of solvent, not on the chemical identity of those particles. The important factor is the effective particle count, which can change when an electrolyte dissociates or when molecules associate. Vapour-pressure lowering, boiling-point elevation, freezing-point depression, and osmotic pressure are examples. Therefore, option B gives the correct characteristic.
What is the minimum external pressure required to stop the entry of solvent through a semipermeable membrane into a solution called?
Correct answer: B
Osmosis is the movement of solvent through a semipermeable membrane toward the solution with the greater effective solute concentration. The minimum external pressure that must be applied to the solution to just stop this solvent movement is called osmotic pressure. Vapour pressure and partial pressure describe gas-phase or vapour behaviour, while critical pressure is unrelated to osmosis. Therefore, option B is correct.
A nonelectrolyte solution has an osmotic pressure of 2.46 atm at a concentration of 0.1 M. What will its osmotic pressure be at 0.2 M and the same temperature?
Correct answer: C
For a dilute nonelectrolyte solution, the osmotic pressure follows π = CRT because i = 1. At constant temperature, osmotic pressure is directly proportional to molar concentration. The concentration changes from 0.1 M to 0.2 M, so it doubles. Thus the new pressure is 2 × 2.46 = 4.92 atm. Therefore, option C is correct.
A solution has i = 1. Which situation is most suitable for it?
Correct answer: A
The van’t Hoff factor compares the actual number of solute particles in solution with the number expected from the formula units initially dissolved. If i = 1, there is no net change in particle number. This is expected when the solute neither associates into larger units nor dissociates into ions. Complete dissociation or dimerisation would make i greater or less than 1, respectively. Therefore, option A is correct.
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