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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 2View options
Because it is molar mass
Because it is a ratio of two pressures
Because it is temperature
Because it is volume
Easy · Level 2View options
The colour of the solvent
The mole fraction of the solute
The mass of the container
The smell of the solution
Easy · Level 2View options
It becomes approximately double
It becomes half
It becomes zero
It remains unchanged
Easy · Level 2View options
Number of non-volatile solute particles has increased
Colour of container has changed
Identity of solvent has disappeared
External pressure has always become zero
Easy · Level 2View options
Mole fraction of the solute
Mass of the solvent
Colour of the solution
Volume of the container
Easy · Level 2View options
Boiling point increases
Boiling point decreases
Boiling point becomes zero
Boiling point has no relation to vapour pressure
Easy · Level 2View options
Non-volatile solute
Volatile solute
Gaseous solvent
Only coloured solute
Easy · Level 2View options
Colligative property
Property based only on colour
Property based only on shape
Nuclear property
Easy · Level 2View options
In the solution having more solute particles
In the solution having fewer solute particles
In the pure solvent
Only in a larger container
Easy · Level 2View options
Difference between vapour pressure of pure solvent and solution divided by vapour pressure of pure solvent
Multiplying solution vapour pressure by solute mass
Dividing solvent mass by solution colour
Subtracting temperature from solution volume
Easy · Level 2View options
0.20
0.80
1.20
2.00
Easy · Level 2View options
Because it is a ratio of two pressures of the same kind
Because it is a temperature difference
Because it is a measure of mass
Because it is molar mass
Easy · Level 2View options
Lowering increases
Lowering decreases
Lowering becomes zero
Lowering has no relation to mole fraction
Easy · Level 2View options
14 kPa
26 kPa
40 kPa
35 kPa
Easy · Level 2View options
It enters vapour phase and increases total pressure
It lowers vapour pressure by decreasing solvent mole fraction
It makes external pressure zero
It chemically changes the solvent
Easy · Level 2View options
Colligative property
Chemical property
Colour property
Odour property
Easy · Level 2View options
Colligative property
Electrical property
Surface property
Nuclear property
Easy · Level 2View options
Freezing point of solution is lower than pure solvent
Freezing point of solution is higher than pure solvent
Boiling point of solution decreases
Colour of solvent changes
Easy · Level 2View options
Flow of solvent through a semipermeable membrane
Melting of a metal
Burning of a gas
Change in solute colour
Easy · Level 2View options
\(\pi = CRT\)
\(\Delta T_b = K_bm\)
\(\Delta T_f = K_fm\)
\(p = x p^0\)
Easy · Level 2View options
0
1
2
Depends only on the molar mass of glucose
Easy · Level 2View options
\(\Delta T_f = iK_fm\)
\(\Delta T_b = iK_bm\)
\(\pi = iCRT\)
\(pV = nRT\)
Easy · Level 2View options
Because it does not depend on volume change with temperature
Because it always shows colour
Because it applies only to gases
Because solvent mass is not considered
Easy · Level 2View options
Elevation increases
Elevation decreases
Elevation becomes zero
There is no relation
Easy · Level 2View options
Colour of solution
Lowering of vapour pressure
Elevation in boiling point
Osmotic pressure
Question 1EasyLevel 2
Why is relative lowering of vapour pressure unitless?
Correct answer: B
Relative lowering is defined as (p° − p)/p°, where both p° and p are pressures. The numerator and denominator therefore carry the same pressure unit, such as pascal or kilopascal, and the units cancel during division. The resulting number has no unit. It should not be confused with vapour pressure itself, which does have pressure units.
In a dilute solution containing a non-volatile solute, the lowering of vapour pressure is approximately proportional to what?
Correct answer: B
For a non-volatile solute, Raoult’s law gives Δp/p° = x_solute, or Δp = p°x_solute. At a fixed temperature and fixed pure-solvent pressure, the lowering is therefore directly proportional to the solute mole fraction. The relation depends on the number of solute particles, not on colour, smell, or container mass. This particle-based dependence is why the property is colligative.
The mole fraction of solute in a solution is doubled, while the solution remains dilute. What happens to the relative lowering of vapour pressure?
Correct answer: A
For a non-volatile solute, the relative lowering of vapour pressure is Δp/p° = x_solute. This is a direct proportionality in the dilute-solution treatment. Therefore, if the solute mole fraction changes from x to 2x while temperature and the solvent remain otherwise comparable, the relative lowering changes from x to approximately 2x. It does not become half or stay unchanged; the statement that the solution remains dilute supports this approximation.
If lowering of vapour pressure increases in a solution, what is the most likely reason at same solvent and same temperature?
Correct answer: A
At fixed temperature and with the same solvent, vapour-pressure lowering is a colligative effect. Increasing the number of effective non-volatile solute particles lowers the solvent mole fraction and its escaping tendency, so the lowering becomes larger. Container colour, solvent identity disappearing, or external pressure being zero does not explain the stated colligative change.
For a solution containing a non-volatile solute, the relative lowering of vapour pressure is equal to what?
Correct answer: A
Let p⁰ be the vapour pressure of the pure solvent and p its pressure in the solution. For a non-volatile solute, Raoult’s law gives p = x_solvent p⁰. Therefore, the relative lowering is (p⁰ − p)/p⁰ = 1 − x_solvent. Since x_solvent + x_solute = 1, this becomes x_solute. It depends on the number of solute particles, not on colour, container volume, or solvent mass alone.
What is the effect of lowering the vapour pressure of a solution on its boiling point?
Correct answer: A
A liquid boils when its vapour pressure reaches the external pressure. If adding a non-volatile solute lowers the vapour pressure at every given temperature, the solution must be heated to a higher temperature before its pressure can reach that same external pressure. Consequently, the boiling point rises; this is called elevation of boiling point. It does not fall to zero, and boiling point is directly related to vapour pressure.
Which type of solute most directly lowers the vapour pressure of a solvent?
Correct answer: A
A non-volatile solute contributes negligibly to the vapour phase, while its presence lowers the mole fraction of the solvent in the liquid. Raoult’s law therefore gives a lower solvent partial pressure than for the pure solvent. A volatile solute contributes its own vapour and requires a different total-pressure treatment. Colour and the physical state of the solvent are irrelevant to this principle.
Lowering of vapour pressure is what type of property?
Correct answer: A
Lowering of vapour pressure is a colligative property because, for dilute solutions, its magnitude depends primarily on the number of dissolved solute particles relative to solvent particles, not on the chemical identity of those particles. Association and dissociation can change the effective particle number, which explains abnormal colligative behaviour. Colour, shape, and nuclear properties are unrelated.
In which situation will lowering of vapour pressure be greater?
Correct answer: A
For comparable solutions with a non-volatile solute, more solute particles produce a larger solute mole fraction and a smaller solvent mole fraction. From Δp = p°xsolute, the lowering therefore increases with the effective number of particles. A pure solvent has no solute and hence no solute-caused lowering. Container size alone does not determine the equilibrium vapour pressure.
Which is the correct mathematical form of relative lowering of vapour pressure?
Correct answer: A
If p° is the vapour pressure of the pure solvent and p is the pressure over the solution, the lowering is p° − p. Relative lowering compares this decrease with the original reference value p°, so it is (p° − p)/p°. The denominator must be the pure-solvent pressure; multiplying by mass or using colour, volume, or temperature is not the definition.
In an ideal solution containing a non-volatile solute, the mole fraction of the solute is 0.20. What is the relative lowering of the vapour pressure of the solvent?
Correct answer: A
For an ideal solution with 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 its vapour pressure in the solution. Since x_solute = 0.20, the relative lowering is 0.20, or 20%. The value 0.80 is the solvent’s mole fraction, not the relative lowering.
Why does relative lowering of vapour pressure have no unit?
Correct answer: A
Relative lowering of vapour pressure is defined as (p° − p)/p°, where p° is the vapour pressure of the pure solvent and p is that of the solution. Both the numerator and denominator are pressures, measured in the same unit. Their units cancel during division, leaving a pure number. It may be written as a decimal or percentage, but it is not measured in pascals, grams, or any other physical unit. Hence option A is correct.
If solute mole fraction is increased and the solute is non-volatile, what happens to lowering of vapour pressure?
Correct answer: A
For a non-volatile solute, Raoult’s law gives relative lowering Δp/p° = x_solute in the dilute ideal limit. Increasing x_solute therefore increases the fractional lowering and lowers the solution pressure further. The statement assumes the same solvent and temperature and no unusual association or dissociation; in such cases the effective particle concentration must be considered.
A solution has solvent mole fraction 0.65. The vapour pressure of the pure solvent is 40 kPa. What is the lowering of vapour pressure?
Correct answer: A
The solute mole fraction is x_solute = 1 − x_solvent = 1 − 0.65 = 0.35. For a non-volatile solute, the lowering in vapour pressure is Δp = x_solute p°_solvent. Therefore, Δp = 0.35 × 40 kPa = 14 kPa. Equivalently, the solution pressure is 0.65 × 40 = 26 kPa, and 40 − 26 = 14 kPa. Thus 26 kPa is the solution pressure, not the lowering, so option A is correct.
Which statement explains the effect of a non-volatile solute most correctly?
Correct answer: B
A non-volatile solute has negligible vapour pressure under the stated conditions, so it does not add a significant partial pressure to the vapour. Its presence reduces the mole fraction of the solvent, and Raoult’s law then gives a lower solvent partial pressure and lower total pressure. This is a physical solution effect, not necessarily a chemical reaction with the solvent.
Which property depends only on the number of solute particles and not on their nature?
Correct answer: A
A colligative property depends on the number of dissolved particles relative to the solvent, provided the solution is sufficiently dilute and particle interactions are handled appropriately. Important examples are relative lowering of vapour pressure, boiling-point elevation, freezing-point depression, and osmotic pressure. Dissociation and association matter because they change the effective number of particles.
Elevation in boiling point is what type of property?
Correct answer: A
Elevation in boiling point is a colligative property because, for a dilute solution, its magnitude depends mainly on the number of dissolved particles relative to the solvent, not on the chemical identity of those particles. The other choices describe unrelated categories and do not represent the standard solution properties used in this chapter.
Depression in freezing point means that the solution freezes at a temperature lower than the freezing point of the pure solvent. A non-volatile solute lowers the chemical potential or escaping tendency of the solvent, so solid–liquid equilibrium is reached only at a lower temperature. It is not a fall in boiling point or a colour change.
Osmotic pressure is a colligative property related to which process?
Correct answer: A
Osmosis is the movement of solvent molecules through a semipermeable membrane from the dilute solution, or pure solvent, toward the more concentrated solution. Osmotic pressure is the external pressure required to stop this movement. It depends mainly on the number of dissolved particles, concentration, and temperature, not on colour, burning, or melting.
Which simple relation is used for the osmotic pressure of a dilute ideal solution?
Correct answer: A
For a dilute ideal solution, osmotic pressure follows the van’t Hoff equation \(\pi = CRT\). Here, \(\pi\) is osmotic pressure, \(C\) is the molar concentration of the solute, \(R\) is the gas constant, and \(T\) is absolute temperature in kelvin. Option B describes boiling-point elevation, option C describes freezing-point depression, and option D is Raoult’s law for vapour pressure.
A student dissolves 1.0 mol of glucose in 1.0 kg of water. What will be the value of the van't Hoff factor if glucose does not ionise?
Correct answer: B
Glucose behaves as a non-electrolyte in this idealised situation and remains as intact molecules rather than splitting into ions. Therefore one dissolved molecule gives one effective solute particle, so the ratio of actual to calculated particles is i = 1. The amount of water and glucose affects molality, but not this factor when there is no association or dissociation.
Which relation is correct for depression in the freezing point?
Correct answer: A
The depression in freezing point is given by \(\Delta T_f = iK_fm\). In this expression, \(i\) is the van’t Hoff factor, \(K_f\) is the cryoscopic constant of the solvent, and \(m\) is the molality of the solution. The symbol \(f\) identifies freezing-point depression. Option B is the corresponding formula for elevation in boiling point, while C and D describe osmotic pressure and the ideal-gas equation.
Why is molality used in boiling point and freezing point formulas?
Correct answer: A
Molality is defined as moles of solute per kilogram of solvent. It uses mass, which remains essentially unchanged when temperature changes, whereas solution volume can expand or contract. This makes molality convenient for temperature-dependent colligative-property calculations. It does include solvent mass and is not restricted to gases.
If the molality of a solution is increased, what happens to elevation in boiling point?
Correct answer: A
For a dilute solution, ΔTb = iKb m. If the solvent and solute behaviour remain the same, i and Kb are constant, so ΔTb is directly proportional to molality. Increasing m therefore increases the boiling-point elevation. It does not make the elevation zero or remove the relation; the statement assumes the usual dilute-solution conditions.
Which property is not included among colligative properties?
Correct answer: A
The standard colligative properties are relative lowering of vapour pressure, elevation in boiling point, depression in freezing point and osmotic pressure. They depend primarily on the number of dissolved particles. Colour depends on the nature and electronic structure of the substance, so it is not classified as a colligative property.
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