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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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Medium · Level 1View options
Freezing point is depressed
Freezing point becomes infinite
Freezing point always increases
Freezing point loses meaning
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Mole fraction of solute
Colour of container
Smell of liquid
Only time
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In a solution with more non-volatile solute particles
In pure solvent
Only in coloured solution
Only in a cold container
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It increases
It decreases
It becomes zero
It will not change
Medium · Level 1View options
Colligative property
Colour-based property
Odour-based property
Shape-based property
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The one with more non-volatile solute particles
The one with only pure solvent
The one with no solute
The one with a dark container colour
Medium · Level 1View options
Mole fraction of solute
Colour of solute
Height of container
Odour of liquid
Medium · Level 1View options
Boiling point increases
Boiling point decreases
Boiling point becomes zero
Boiling point loses meaning
Medium · Level 1View options
Freezing point decreases
Freezing point only increases
Freezing point becomes infinite
Freezing point becomes free from pressure
Medium · Level 1View options
Vapour pressure decreases, so higher temperature is needed for boiling
Vapour pressure increases, so lower temperature is needed for boiling
Mole fraction increases, so solvent disappears
Solute becomes vapour, so boiling stops
Medium · Level 1View options
Nearly same
Always double in the first
Always zero in the second
It will depend on colour
Medium · Level 1View options
Lowering increases
Lowering decreases
Lowering becomes zero
Vapour pressure becomes higher than pure solvent
Medium · Level 1View options
Lowering decreases
Lowering increases
Lowering becomes infinite
Vapour pressure of solvent becomes zero
Medium · Level 1View options
nearly same
completely different
depends only on colour
always zero
Medium · Level 1View options
because solute dissociates into ions and increases particle number
because ions form vapour
because water temperature becomes zero
because vessel shrinks
Medium · Level 1View options
because several particles combine to form fewer particles
because solute completely vaporises
because solvent disappears
because temperature becomes infinite
Medium · Level 1View options
It escapes as vapour
It dissociates into ions and increases effective particle number
It increases solvent mole fraction
It makes vapour pressure independent of temperature
Medium · Level 1View options
Nearly the same
Completely unrelated
One pressure must be zero
Different only due to colour
Medium · Level 1View options
Lower solvent mole fraction and presence of solute particles
Name of container
Colour of liquid
Thickness of glass
Medium · Level 1View options
It will be the same
It will always be more for the heavier solute
It will always be more for the coloured solute
It will occur in neither solution
Medium · Level 1View options
0.05
0.95
5.00
95.00
Medium · Level 1View options
Convert mass into moles
Convert colour into percentage
Draw the shape of the container
Never assume the solution is ideal
Medium · Level 1View options
0.15
0.85
0.50
1.15
Medium · Level 1View options
Vapour pressure decreases, so higher temperature is needed to reach external pressure
The solute itself gives higher vapour pressure
Molar mass of solvent becomes zero
No molecules remain in solution
Medium · Level 1View options
Nearly the same
Always completely different
Zero in one and infinite in the other
Decided only by colour
Question 1MediumLevel 1
What is the general effect on freezing point when vapour pressure is lowered?
Correct answer: A
For a solution containing a non-volatile solute, the liquid solvent has a lower chemical potential and lower vapour pressure than the pure solvent. The solution must be cooled to a lower temperature before its solid and liquid phases are in equilibrium. Thus the freezing point is depressed. This is another colligative effect, not an infinite or meaningless value.
Relative lowering of vapour pressure is related to what?
Correct answer: A
For a solution containing a non-volatile solute, relative lowering is (p° − p)/p°. By Raoult’s law this equals the mole fraction of solute, xsolute, for an ideal dilute solution. Thus it depends on the relative number of solute particles, not on container colour, smell or elapsed time under fixed conditions.
In which solution will lowering of vapour pressure be greater?
Correct answer: A
For comparable amounts of solvent at the same temperature, a greater number of non-volatile solute particles gives a larger solute mole fraction. Raoult’s law then predicts a larger relative lowering of vapour pressure. Pure solvent has no solute and hence no solute-caused lowering; colour and container temperature alone do not determine this comparison.
In a solution containing a non-volatile solute, what happens to lowering of vapour pressure when mole fraction of solute is increased?
Correct answer: A
For a non-volatile solute, the relative lowering of vapour pressure is related to the solute mole fraction. Increasing x_solute lowers x_solvent and therefore lowers the solvent vapour pressure further. The difference p°−p consequently increases, assuming the same temperature and the usual solution model. It does not become zero or remain unchanged.
Lowering of vapour pressure is considered what type of property?
Correct answer: A
Lowering of vapour pressure is a colligative property because, for dilute ideal solutions, its magnitude depends primarily on the number or concentration of dissolved solute particles rather than their chemical identity. A mole-fraction relation expresses this effect. Colour, odour and particle shape do not define the property.
Which solution will show greater lowering of vapour pressure?
Correct answer: A
For comparable amounts of solvent and at the same temperature, a greater number of non-volatile solute particles produces a larger solute mole fraction and a smaller solvent mole fraction. Raoult's law then predicts a larger fall from the pure-solvent vapour pressure. Pure solvent has no solute-induced lowering, and container colour is irrelevant.
Relative lowering of vapour pressure is related to what when the solute is non-volatile?
Correct answer: A
For a solution containing a non-volatile solute, Raoult’s law gives P = X_solvent P°. The lowering is P° − P = P°(1 − X_solvent) = P°X_solute, because X_solvent + X_solute = 1. Dividing by P° gives (P° − P)/P° = X_solute. Thus relative lowering depends on solute mole fraction, not colour, container height, or odour.
What is the effect of lowered vapour pressure on boiling point of a solution?
Correct answer: A
Boiling occurs when vapour pressure equals external pressure. If a solute lowers the vapour pressure at every given temperature, the solution must be heated further before it reaches that equality. Therefore its boiling point is higher than that of the pure solvent under the same pressure. This increase is called elevation of boiling point.
What is the general effect of lowered vapour pressure on freezing point?
Correct answer: A
Adding a non-volatile solute lowers the chemical potential and escaping tendency of the liquid solvent. Equilibrium between the solid solvent and the solution is therefore reached at a lower temperature than for the pure solvent. This is freezing-point depression. The effect is not an infinite change and does not make the freezing point independent of pressure.
What is the chain of reasoning for boiling point elevation after adding a non-volatile solute?
Correct answer: A
Adding a non-volatile solute lowers the solvent mole fraction and therefore lowers the solution vapour pressure at a given temperature. Boiling requires vapour pressure to equal external pressure, so the solution must be heated further to reach that value. This produces boiling-point elevation. The other choices give the wrong pressure change or mechanism.
If equal mole amounts of two non-volatile solutes are dissolved separately in the same solvent and neither dissociates, how will lowering of vapour pressure compare?
Correct answer: A
Lowering of vapour pressure is colligative: under comparable conditions it depends on the number of solute particles. Equal mole amounts of non-dissociating solutes produce equal numbers of particles, so their lowerings are nearly the same in the same solvent. Chemical identity or colour does not change this ideal colligative comparison.
If a solute dissociates into more particles in solution, what happens to lowering of vapour pressure?
Correct answer: A
Dissociation increases the effective number of solute particles in a given amount of solution. Since lowering of vapour pressure is colligative, the larger particle count gives a larger lowering, often represented by a van't Hoff factor greater than one. The solvent pressure remains below its pure value; it does not become higher merely because dissociation occurs.
If solute particles associate to form fewer particles, what happens to lowering of vapour pressure?
Correct answer: A
Association combines separate solute particles into fewer effective particles. Because the colligative lowering depends on particle number, the effective concentration is smaller than expected and the vapour-pressure lowering decreases. In van't Hoff terms, the factor is less than one. Association does not make the lowering infinite or force the solvent pressure to zero.
If two solutions have the same number of solute particles at the same temperature in the same solvent, how will the lowering of vapour pressure compare?
Correct answer: A
Lowering of vapour pressure is colligative: for the same solvent and temperature it depends on the effective number of dissolved particles. Therefore equal effective particle amounts give nearly equal lowering, even if the solutes are chemically different. The comparison assumes comparable solution conditions; dissociation or association must be included in the actual particle count.
Why can lowering of vapour pressure be greater than expected in a solution of an electrolyte?
Correct answer: A
An electrolyte such as an ionic salt can dissociate into two or more ions in a polar solvent. The number of solute particles then exceeds the number of formula units added, increasing the colligative effect and hence the lowering of solvent vapour pressure. The ions generally remain in solution; their vapour formation is not the reason.
Why can lowering of vapour pressure be less than expected for an associating solute?
Correct answer: A
An associating solute forms larger species, such as dimers, from smaller dissolved particles. The effective number of particles becomes lower than the number calculated from the amount added, so the colligative lowering of vapour pressure is smaller than the ideal prediction. This is a particle-count effect, not solvent disappearance or solute vaporisation.
Why can an electrolyte solute show greater lowering of vapour pressure than expected?
Correct answer: B
An electrolyte such as an ionic compound can dissociate into two or more ions in solution. The effective number of solute particles then becomes larger than the number of formula units added, increasing the colligative effect and lowering solvent vapour pressure more. The magnitude is represented by the van’t Hoff factor when dissociation is considered.
If two solutions have the same solvent and same number of effective solute particles at the same temperature, how will their lowering of vapour pressure compare?
Correct answer: A
Lowering of vapour pressure is a colligative property. Under the same temperature and with the same solvent, its magnitude depends on the effective number of dissolved particles relative to solvent particles. If those effective particle conditions are the same, the lowering will be nearly the same, even if the chemical identities differ. Association or dissociation must be included in counting particles.
Observed lowering of vapour pressure in a solution can be directly linked with what?
Correct answer: A
For a non-volatile solute, adding particles reduces the solvent mole fraction from one and lowers the solvent’s escaping tendency. Raoult’s law expresses this as p_solution = x_solvent p°_solvent, or Δp/p° = x_solute. Thus the pressure lowering is linked to solvent composition and effective solute-particle number, not container name, colour, or glass thickness.
If two different non-volatile solutes are separately dissolved in equal moles in equal amounts of the same solvent and both remain non-dissociated, how will the lowering of vapour pressure compare?
Correct answer: A
For a non-volatile, non-dissociating solute, the lowering depends on the solute mole fraction. Equal moles placed in equal moles of the same solvent give equal mole fractions and equal numbers of dissolved particles. Thus the lowering is the same, regardless of molar mass, colour, or chemical name. The conclusion assumes comparable ideal dilute-solution behaviour.
If the vapour pressure of a pure solvent is 100 units and that of the solution is 95 units, what is the relative lowering of vapour pressure?
Correct answer: A
Relative lowering of vapour pressure is calculated as (p⁰ − p)/p⁰, where p⁰ is the vapour pressure of the pure solvent and p is that of the solution. Here the absolute decrease is 100 − 95 = 5 units. Dividing by the pure-solvent pressure gives 5/100 = 0.05. Thus 0.05, or 5%, is correct; 5 is only the absolute decrease, not the relative value.
What should be the first precaution in a numerical question on vapour pressure?
Correct answer: A
Raoult’s law and the colligative-property equations use mole fractions, not masses directly. Therefore, when masses are supplied, first divide each mass by the appropriate molar mass to obtain moles, then form the required mole fraction. The solution’s ideality must still be checked from the wording, but converting mass to moles is the essential first calculation step.
In an ideal solution, pure solvent vapour pressure is 100 kPa and solution vapour pressure is 85 kPa. What is the mole fraction of the non-volatile solute?
Correct answer: A
The relative lowering is (p° − p)/p° = (100 − 85)/100 = 15/100 = 0.15. For a solution containing a non-volatile solute, this relative lowering equals the solute mole fraction, xsolute. Hence xsolute = 0.15. The value 0.85 is the solvent mole fraction, and 1.15 cannot be a mole fraction because mole fractions range from zero to one.
Why does adding a non-volatile solute increase the boiling point of a solution?
Correct answer: A
A non-volatile solute contributes essentially no vapour, but it lowers the solvent mole fraction and hence the solution vapour pressure. Boiling requires the solution vapour pressure to equal the external pressure. Since the pressure starts lower at any given temperature, the solution must be heated further to reach equality, producing boiling-point elevation.
If two solutions have the same solvent and the same number of effective solute particles at the same temperature, how will lowering of vapour pressure compare?
Correct answer: A
Vapour-pressure lowering is a colligative property. Under the same solvent and temperature, its value depends primarily on the number of independent dissolved particles, not on their chemical identity. Thus equal effective particle numbers give nearly equal lowering in the ideal dilute limit. Association, dissociation, concentration, or non-ideal interactions must be included if they change the effective count or assumptions.
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