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In this Class 12 Chemistry topic from Chapter 01: Solutions, students learn how vapour pressure arises from the dynamic equilibrium between evaporation and condensation in a liquid. The topic explains the effect of temperature and the presence of a non-volatile solute, including lowering of vapour pressure. Students also connect vapour pressure with mole fraction through Raoult’s law and examine how ideal and non-ideal solutions differ, using equations and basic numerical applications.
TOPIC PRACTICE
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Mole fraction of the solute
Colour of the solute
Height of the container
Odour of the liquid
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It increases because solution particles increase.
It decreases because solvent mole fraction decreases.
It remains unchanged because solvent identity is unchanged.
It becomes zero at every concentration.
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80 units
100 units
125 units
20 units
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Contributions of both solvent and solute
Only colour of container
Only mass of solvent
No particles at all
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The sum of the two partial vapour pressures
Only the lower vapour pressure
Only the larger mass
The sum of the two colours
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Nearly similar in nature
Completely zero
Always extremely different
Only ionic
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When new attractions are weaker
When new attractions are very strong
When no vapour forms
When temperature is always zero
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When new attractions are stronger
When new attractions are weaker
When the solution has no particles
When the liquid has no temperature
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Mole fraction of the solvent
Colour of the solution
Thickness of the container
Only size of solute
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It decreases
It increases
It doubles
It will not change
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Decrease in number of solvent molecules at the surface
Change in colour of container
Change in name of liquid
Increase in brightness of solution
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80 kilopascal
100 kilopascal
20 kilopascal
180 kilopascal
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30 kilopascal
50 kilopascal
60 kilopascal
110 kilopascal
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Sum of partial vapour pressures of all volatile components
Only weight of container
Only colour of solution
Only name of solute
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0.75 times the vapour pressure of pure solvent
Equal to pure solvent
2 times pure solvent
Zero
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P = X_solvent P°
P = X_solute + P°
P = P° / X_solvent
P = P° − X_solvent
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Benzene and toluene
Acetone and chloroform
Ethanol and water
Hydrochloric acid and water
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Because salt lowers the vapour pressure of water
Because salt immediately converts water into vapour
Because salt makes external pressure zero
Because salt destroys the mass of water
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When rates of evaporation and condensation become equal
When liquid completely disappears
When container is opened
When temperature has no effect
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Sum of both partial vapour pressures
Pressure of only the more volatile liquid
Pressure of only the less volatile liquid
Sum of colours of both liquids
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It applies over the entire concentration range
It never applies at a single temperature
It applies only when colour changes
It applies only to solids
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Nearly similar
Completely zero
Always very different
Only ionic
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Greater than expected value
Less than expected value
Always zero
Unrelated to mole fraction
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Less than expected value
Greater than expected value
Always infinite
Always equal to external pressure
Medium · Level 1View options
Positive deviation
Negative deviation
No vapour
Always ideal behaviour
Question 1MediumLevel 1
When the solute is non-volatile, on what does relative lowering of vapour pressure depend?
Correct answer: A
For a solution containing a non-volatile solute, Raoult's law leads to (P° − P)/P° = X_solute. Thus the relative lowering of vapour pressure equals the solute mole fraction in the simple ideal case. It depends on the number of solute particles relative to total particles, not on colour, container height, or smell.
At constant temperature, which statement is correct about vapour pressure when a non-volatile solute is added to a solvent?
Correct answer: B
A non-volatile solute does not add appreciable vapour, but it increases total moles and therefore lowers the solvent mole fraction. At constant temperature, Raoult's law gives p = x_solvent p°, so the solvent vapour pressure decreases. More solution particles do not mean more solvent vapour, and the pressure is not zero at every concentration.
If the mole fraction of the solvent is 0.8 and the vapour pressure of the pure solvent is 100 units, what is the vapour pressure of the solution?
Correct answer: A
For a solution with a non-volatile solute, Raoult’s law gives P_solution = X_solvent P°_solvent. Substituting the given values, P_solution = 0.8 × 100 = 80 units. The value cannot remain 100 because the solvent mole fraction is less than one. Dividing by 0.8 gives 125, which is not the Raoult-law expression, while 20 is only the lowering in pressure.
If the solute in a solution is volatile, what can be included in total vapour pressure?
Correct answer: A
A volatile solute can enter the vapour phase along with the solvent. Consequently, the total pressure above the solution is the sum of the partial pressures of both components, provided the vapour behaves approximately ideally. This differs from a non-volatile-solute case, where the solute's own vapour contribution is neglected.
In an ideal solution of two volatile liquids, what is the total vapour pressure equal to?
Correct answer: A
In an ideal solution of two volatile liquids, each component contributes a partial vapour pressure according to Raoult’s law: p_A = X_A P°_A and p_B = X_B P°_B. Dalton’s law then gives the total pressure as P_total = p_A + p_B. Thus both volatile components contribute to the total; it is not determined simply by the lower pressure, mass, or colour of the liquids.
In an ideal solution, how are attractions between the components considered?
Correct answer: A
An ideal solution is one in which interactions between unlike molecules are approximately comparable to interactions between like molecules: A–B is similar to A–A and B–B. Consequently, mixing causes little enthalpy change and little volume change, and Raoult's law is obeyed. The attractions are not zero and need not be purely ionic.
When can positive deviation from Raoult's law be observed?
Correct answer: A
In positive deviation, the observed total vapour pressure is higher than Raoult's-law prediction. This occurs when unlike A–B attractions are weaker than the original A–A and B–B attractions. Molecules are then held less strongly in the liquid and escape more readily. Stronger unlike attraction would instead favour negative deviation.
When can negative deviation from Raoult's law be observed?
Correct answer: A
Negative deviation means that the actual vapour pressure is lower than the ideal Raoult-law value. It can occur when unlike A–B attractions are stronger than the original attractions, holding molecules more firmly in the liquid. Fewer molecules escape, so pressure decreases. Weaker unlike attractions produce positive deviation instead.
According to Raoult's law, the partial vapour pressure of solvent in an ideal solution is proportional to what?
Correct answer: A
For an ideal solution, Raoult’s law states that the partial vapour pressure of a volatile component equals its mole fraction multiplied by its vapour pressure in the pure state: pi = xi pi°. Therefore, the solvent’s partial pressure is directly proportional to its mole fraction. Colour, container thickness and particle size are not the governing variables.
If the mole fraction of solvent decreases, what happens to its vapour pressure according to Raoult's law?
Correct answer: A
Raoult's law gives p_solvent = x_solvent p°_solvent at fixed temperature. Since p° is constant for the chosen solvent and temperature, reducing x_solvent directly reduces p_solvent in the same proportion. No particular doubling can be inferred, and the pressure does not remain unchanged unless the mole fraction remains unchanged.
What is the main reason for lowering of vapour pressure?
Correct answer: A
When a non-volatile solute is added, the proportion of solvent molecules among all particles decreases. Consequently, fewer solvent molecules are available at the surface and fewer escape into the vapour phase at a given temperature. This lowers the solvent's equilibrium vapour pressure. Colour, name, and brightness have no role in this explanation.
If the mole fraction of solvent in a solution is 0.8 and the vapour pressure of pure solvent is 100 kilopascal, what is the vapour pressure of solvent in the solution?
Correct answer: A
For an ideal solution, Raoult’s law gives psolvent = xsolvent p°solvent. Substituting the given values, psolvent = 0.8 × 100 kPa = 80 kPa. The answer is lower than 100 kPa because the solvent mole fraction is less than one. The value 20 kPa is the lowering, not the partial vapour pressure itself.
If vapour pressure of pure solvent is 50 kilopascal and mole fraction of solvent in solution is 0.6, what will be the partial vapour pressure?
Correct answer: A
Use Raoult’s law for the solvent: p = x p°. Here x = 0.6 and p° = 50 kPa, so p = 0.6 × 50 = 30 kPa. Since the solvent mole fraction is below one, its partial vapour pressure must be below the pure-solvent value of 50 kPa. The other choices do not follow the required multiplication.
In an ideal solution, total vapour pressure is obtained from what?
Correct answer: A
Dalton’s law states that the total pressure of a mixture of gases or vapours is the sum of the partial pressures contributed by its components. Thus, for an ideal solution containing volatile components, Ptotal = pA + pB + …, with each partial pressure obtained from Raoult’s law. Container weight, colour and the solute’s name do not determine the total vapour pressure.
In a solution, mole fraction of solvent is 0.75. Compared with pure solvent, how will its vapour pressure be?
Correct answer: A
For an ideal solution, Raoult's law gives p_solvent = x_solvent p°_solvent. With x_solvent = 0.75, p_solvent = 0.75 p°_solvent. Thus the solution vapour pressure is three-fourths of the pure-solvent value, not equal to it, twice it, or zero. This comparison assumes the same temperature.
Which is the correct form of Raoult's law when the solute is non-volatile?
Correct answer: A
When the solute is non-volatile, essentially only the solvent contributes to the vapour pressure. Raoult’s law states that the partial vapour pressure of the solvent is P = X_solvent P°, where X_solvent is its mole fraction and P° is the vapour pressure of the pure solvent at the same temperature. The other options use an incorrect sum, inverse relation, or subtraction without consistent units.
Which of the following liquid mixtures is an example of an approximately ideal solution and approximately obeys Raoult’s law?
Correct answer: A
Benzene and toluene are chemically similar, and the attractions between unlike molecules are close to those in the two pure liquids. Their mixture therefore shows nearly ideal behaviour and approximately follows Raoult's law. Acetone–chloroform has strong specific attraction, while ethanol–water and HCl–water show significant non-ideal effects.
For a sufficiently dilute solution, dissolved salt behaves as a non-volatile solute and lowers the mole fraction and vapour pressure of water. Boiling requires vapour pressure to equal external pressure, so the salt solution must be heated to a higher temperature. This is boiling-point elevation, not conversion of water to gas or removal of external pressure.
When does vapour pressure become constant in a closed container?
Correct answer: A
In a closed container, evaporation transfers molecules from liquid to vapour while condensation returns molecules to the liquid. Initially evaporation may dominate, but vapour accumulation increases condensation until the two rates become equal. This dynamic equilibrium gives a constant vapour pressure at fixed temperature, even though both processes continue microscopically.
In an ideal solution of two volatile liquids, total vapour pressure is equal to what?
Correct answer: A
In an ideal solution, each volatile component obeys Raoult's law: p_A = x_A P°_A and p_B = x_B P°_B. Dalton's law then gives P_total = p_A + p_B. Both components contribute, even if one is more volatile than the other; the more volatile component does not replace the contribution of the less volatile one.
An ideal solution has similar intermolecular interactions between unlike molecules and between like molecules. Consequently, each component's partial vapour pressure remains proportional to its mole fraction throughout the composition range, not merely at infinite dilution. Thus Raoult's law applies over the entire concentration range for an ideal solution at a fixed temperature.
In an ideal solution, how are attractions between components considered?
Correct answer: A
An ideal solution is one in which interactions between unlike molecules, A–B, are approximately equal to interactions between like molecules, A–A and B–B. Because mixing does not cause a major change in intermolecular forces, each component follows Raoult’s law over the relevant composition range. Attractions are not zero and need not be ionic; they are simply nearly similar in strength.
In positive deviation from Raoult's law, how is the actual vapour pressure?
Correct answer: A
Positive deviation means that the observed partial or total vapour pressure is greater than the value predicted by Raoult's law. This commonly occurs when unlike-molecule attractions are weaker than the average like-molecule attractions, allowing particles to escape more easily. It does not mean pressure is independent of composition or necessarily infinite.
In negative deviation from Raoult's law, how is the actual vapour pressure?
Correct answer: A
Negative deviation means that the measured vapour pressure is lower than the ideal value calculated using Raoult's law. Stronger unlike-molecule attractions hold particles more firmly in the liquid, reducing their escape tendency. The pressure is not necessarily zero or equal to external pressure; it is simply below the corresponding ideal prediction at that composition and temperature.
Which deviation from Raoult's law can occur when new attractions are weaker?
Correct answer: A
If the new unlike-molecule attractions are weaker than the original like-molecule attractions, particles are held less strongly after mixing. Their escaping tendency and vapour pressure become higher than Raoult's-law predictions, giving positive deviation. The mixture still has vapour; weaker attraction does not imply no vapour, and it does not guarantee ideal behaviour.
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