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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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Easy · Level 12View options
6 kPa
69 kPa
75 kPa
8 kPa
Easy · Level 12View options
straight line
closed circle
discontinuous line
completely irregular
Easy · Level 12View options
more molecules get enough energy to leave the surface
mass of molecules continuously increases
solvent mole fraction always decreases
solute forms automatically
Easy · Level 12View options
the liquid is at boiling condition
the liquid is necessarily freezing
there is no vapour in the liquid
the liquid temperature is zero
Easy · Level 12View options
0.40
0.36
0.54
2.50
Easy · Level 12View options
nearly ideal solution
positive deviation solution
negative deviation solution
non-volatile solid mixture
Easy · Level 12View options
it may become lower than expected
it may become higher than expected
it will always become zero
total pressure will not be less than partial pressure
Easy · Level 12View options
because it is ratio of two similar pressures
because it is mass
because it is temperature
because it is volume
Easy · Level 12View options
0.80
0.20
1.20
20
Easy · Level 12View options
because surface area may affect rate but not equilibrium pressure
because surface area is always zero
because vaporisation stops
because condensation becomes impossible
Easy · Level 12View options
higher
lower
always zero
equal to mole fraction
Easy · Level 12View options
100 kPa
55 kPa
45 kPa
10 kPa
Easy · Level 12View options
0.25
0.40
0.75
24
Easy · Level 12View options
0.24
0.76
1.76
0.52
Easy · Level 12View options
at fixed temperature equilibrium vapour pressure depends on nature of liquid
equilibrium vapour pressure is always equal to amount of liquid
vapour pressure is decided only by height of vessel
vapour pressure has no relation with boiling point
Easy · Level 12View options
60 kPa
20 kPa
80 kPa
75 kPa
Easy · Level 12View options
0.15
0.20
0.25
0.75
Easy · Level 12View options
The solute increases the mass of solvent molecules
The escaping tendency of solvent molecules from the surface decreases
The solute removes external pressure
The chemical formula of solvent changes
Easy · Level 12View options
Vapour molecules become heavier
More molecules gain enough energy to enter the vapour phase
The name of the liquid changes
The solvent mole fraction necessarily becomes one
Easy · Level 12View options
10 kPa
40 kPa
50 kPa
60 kPa
Easy · Level 12View options
It obeys Raoult's law only in very dilute condition
It obeys Raoult's law at all concentrations
It always shows positive deviation
It always releases heat
Easy · Level 12View options
Total vapour pressure decreases
Total vapour pressure increases
Total vapour pressure becomes zero
Total vapour pressure becomes independent of composition
Easy · Level 12View options
0.4
0.6
0.1
1.4
Easy · Level 12View options
When the solute mole fraction is very high
When the solute mole fraction is nearly zero
When the solvent mole fraction is zero
When the solute is necessarily ionic
Easy · Level 12View options
Component A
Component B
Both will remain equal
Neither component will enter the vapour
Question 1EasyLevel 12
A solution has solvent mole fraction 0.92. The vapour pressure of pure solvent is 75 kPa. What is the lowering of vapour pressure?
Correct answer: A
The solute mole fraction is x solute = 1 − 0.92 = 0.08. For a non-volatile solute, the lowering is Δp = x solute p° solvent. Hence Δp = 0.08 × 75 = 6 kPa. Equivalently, the solution pressure is 0.92 × 75 = 69 kPa and the lowering is 75 − 69 = 6 kPa. Thus 69 kPa is the solution pressure, not the lowering.
For an ideal solution, what is the usual nature of the graph between total vapour pressure and liquid composition?
Correct answer: A
For an ideal binary solution, p A = x A p° A and p B = (1 − x A)p° B. Each partial pressure therefore changes linearly with liquid composition. Their sum, P = p A + p B, is also a linear function of x A, joining the pure-B and pure-A vapour-pressure values. Non-ideal solutions may show curved positive or negative deviations instead.
The vapour pressure of a liquid increases rapidly with temperature. What is the most suitable reason?
Correct answer: A
When temperature rises, the average kinetic energy of liquid molecules increases and the energy distribution broadens. A larger fraction of molecules can overcome intermolecular attractions and escape from the surface. In a closed vessel, more escaping molecules produce a higher equilibrium vapour pressure. Molecular mass does not continuously change, and solvent mole fraction is irrelevant for a pure liquid unless a solution is being considered.
If vapour pressure has become equal to external pressure, what can be said about the liquid?
Correct answer: A
The defining condition for boiling is that the liquid’s equilibrium vapour pressure equals the pressure acting on its surface. At this point vapour bubbles can form throughout the liquid and grow instead of collapsing. Equality alone does not imply freezing, zero temperature or absence of vapour; the actual boiling temperature depends on the liquid and the external pressure.
In a binary solution, total vapour pressure is 90 kPa and partial pressure of component A is 36 kPa. What is mole fraction of A in vapour?
Correct answer: A
For a gas mixture, Dalton’s law relates vapour-phase mole fraction to partial pressure: y A = p A/P total. Substitution gives y A = 36/90 = 0.40. The number 0.36 is the partial pressure expressed without its unit, not the mole fraction, and 2.50 is impossible because a mole fraction must lie between zero and one. The liquid mole fraction cannot be inferred without more data.
When like and unlike molecular attractions in a mixture of two liquids are nearly equal, what type of solution is formed?
Correct answer: A
An ideal solution is one in which the interactions between unlike molecules are approximately equal to those between like molecules. Replacing an A–A or B–B contact with an A–B contact then causes little change in the escaping tendency, enthalpy or volume. The vapour pressures consequently follow Raoult’s law closely. Stronger unlike attraction would give negative deviation, while weaker attraction would give positive deviation.
If unlike molecular attraction becomes very strong while mixing two liquids, what happens to vapour pressure?
Correct answer: A
Very strong unlike-molecule attraction stabilises the mixed liquid phase and makes escape into vapour more difficult. The partial pressures therefore become lower than the values predicted by Raoult’s law, giving negative deviation. The pressure does not have to become zero, because some molecules still enter the vapour. Also, total pressure is the sum of non-negative partial pressures and cannot be less than an individual partial pressure.
Why does relative lowering of vapour pressure have no unit?
Correct answer: A
Relative lowering is defined as (p° − p)/p°, where both the numerator and denominator are pressures measured in the same unit. Dividing one pressure by another cancels units such as pascal or kilopascal, leaving a pure number. It may be written as a decimal or percentage. It is not itself a mass, temperature or volume, so those options do not explain its unitless nature.
The vapour pressure of a solution is 20 percent lower than that of pure solvent. Assuming non-volatile solute, what is the mole fraction of solvent?
Correct answer: A
A 20% lowering means (p° − p)/p° = 0.20. For a non-volatile solute, this relative lowering equals x solute, so x solute = 0.20. Since the solvent and solute mole fractions sum to one, x solvent = 1 − 0.20 = 0.80. A mole fraction cannot be 1.20 or 20, and 0.20 refers to the solute, not the solvent.
Why does increasing the surface area of a liquid not change its equilibrium vapour pressure?
Correct answer: A
Increasing surface area can increase the initial rate at which molecules leave the liquid, so equilibrium may be reached more quickly. In a closed vessel at fixed temperature, however, the final equilibrium condition requires the rates of evaporation and condensation to be equal. The pressure at that condition is determined by temperature and liquid nature, not surface area. Evaporation and condensation do not become impossible.
What is the pure vapour pressure of a more volatile component like?
Correct answer: A
Volatility is the tendency of a substance to pass from liquid to vapour. At the same temperature, a more volatile liquid has molecules that escape more readily, so its equilibrium vapour pressure in the pure state is higher. Pure vapour pressure is a pressure, whereas mole fraction is a dimensionless composition quantity; they cannot generally be equal. A zero vapour pressure would indicate no appreciable vapour formation.
If in an ideal solution the partial pressure of A is 55 kPa and that of B is 45 kPa, what is total pressure?
Correct answer: A
For a mixture of vapours, Dalton’s law states that the total pressure equals the sum of the partial pressures of all components. Therefore P = p A + p B = 55 + 45 = 100 kPa. Neither individual partial pressure is the total pressure, and 10 kPa is their difference rather than their sum. The ideal-solution condition is consistent with using the calculated partial pressures.
In a solution, pure solvent vapour pressure is 96 kPa and lowering of vapour pressure is 24 kPa. What is the relative lowering?
Correct answer: A
Relative lowering is defined as the lowering divided by the vapour pressure of the pure solvent: Δp/p° = 24/96. Simplifying gives 0.25, a dimensionless quantity, or 25% if expressed as a percentage. The value 0.75 is the corresponding solvent mole fraction for a non-volatile solute, while 24 is the absolute lowering and not the relative value.
If solvent mole fraction in a solution is 0.76, what is the relative lowering of vapour pressure?
Correct answer: A
For a binary solution containing a non-volatile solute, x solute = 1 − x solvent = 1 − 0.76 = 0.24. Raoult’s law then gives relative lowering (p° − p)/p° = x solute = 0.24. The value 0.76 is the solvent mole fraction itself, while 1.76 is impossible for a mole fraction and 0.52 has no basis in the given data.
Which option gives the most exam-useful correct statement related to vapour pressure?
Correct answer: A
At a fixed temperature, the equilibrium vapour pressure of a liquid depends on the nature of the liquid, especially the strength of its intermolecular attractions. It is not equal to the amount of liquid and is not determined only by vessel height. Vapour pressure is directly related to boiling point: boiling occurs when it equals external pressure, so a lower vapour pressure generally corresponds to a higher boiling point at the same external pressure.
In an ideal solution, the vapour pressure of pure solvent is 80 kPa and the mole fraction of solvent is 0.75. What is the vapour pressure of the solution?
Correct answer: A
For an ideal solution with a non-volatile solute, Raoult’s law gives p solution = x solvent p° solvent. Substituting the values, p solution = 0.75 × 80 = 60 kPa. The value 20 kPa is the lowering, 80 kPa is the pressure of pure solvent, and 75 is the mole fraction written as a number rather than a pressure. The given mole fraction belongs to the solvent, so it is used directly.
If the vapour pressure of pure solvent is 120 kPa and the vapour pressure of the solution is 90 kPa, what is the mole fraction of the solute?
Correct answer: C
For a solution containing a non-volatile solute, the relative lowering of vapour pressure equals the mole fraction of the solute: x_solute = (P° − P)/P°. Here, the lowering is 120 − 90 = 30 kPa. Therefore, x_solute = 30/120 = 0.25. Thus, option C is correct. The value 0.75 is the solvent mole fraction, not the solute mole fraction.
What is the deeper reason for decrease in vapour pressure when a non-volatile solute is added to a liquid?
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 solvent molecules, so fewer solvent molecules enter the vapour at equilibrium. This is a physical effect described by Raoult’s law, not a change in solvent mass, formula, or external pressure.
Liquid and vapour are in equilibrium in a closed vessel. Why does equilibrium vapour pressure increase when temperature rises?
Correct answer: B
Raising temperature increases the average kinetic energy of liquid molecules. A larger fraction can overcome intermolecular attractions and enter the vapour phase. The new equilibrium is established at a higher vapour pressure. Molecular mass, the name of the liquid, and solvent mole fraction do not provide this temperature effect.
If the pure solvent vapour pressure is 50 kPa and the solute mole fraction is 0.2, what is the vapour pressure of a solution containing a non-volatile solute?
Correct answer: B
Because the solute is non-volatile, only the solvent contributes to the vapour pressure. The solvent mole fraction is x_solvent = 1 − x_solute = 1 − 0.2 = 0.8. By Raoult’s law, P_solution = x_solvent P°_solvent = 0.8 × 50 = 40 kPa. Hence, option B is correct. The pressure is lower than 50 kPa because adding a non-volatile solute reduces the fraction of solvent molecules at the surface.
Which statement is most accurate for an ideal solution?
Correct answer: B
An ideal solution obeys Raoult’s law over the entire composition range, not merely at infinite dilution. Its A–B interactions are approximately comparable to A–A and B–B interactions, so mixing has nearly zero enthalpy and volume change. Positive deviation and heat release are non-ideal behaviours, so B is the only correct statement.
In a volatile binary ideal solution, what happens to total vapour pressure when the liquid mole fraction of the more volatile component is increased?
Correct answer: B
The more volatile component has the larger pure-component vapour pressure, say p_A° > p_B°. For an ideal solution, P = x_A p_A° + (1 − x_A)p_B°. Increasing x_A replaces some contribution from the lower-pressure component B with the higher-pressure contribution from A, so total pressure rises. It does not become zero or composition-independent; those choices contradict Raoult’s law.
In a binary solution, the partial pressure of component A is 40 kPa and the total pressure is 100 kPa. What is the mole fraction of component B in the vapour phase?
Correct answer: B
For a binary vapour mixture, the partial pressure of B is p_B = P − p_A = 100 − 40 = 60 kPa. Dalton’s law relates vapour mole fraction to partial pressure: y_B = p_B/P. Hence y_B = 60/100 = 0.60. Option A is the mole fraction of A in the vapour, while 1.4 is impossible because a mole fraction must lie between zero and one.
In which situation will the vapour pressure of a solution remain nearly equal to that of the pure solvent?
Correct answer: B
For a non-volatile solute, Raoult’s law gives p_solution = x_solvent p°_solvent. If the solute mole fraction is nearly zero, x_solvent is nearly one, so p_solution is nearly p°_solvent. A high solute fraction substantially lowers pressure, and zero solvent fraction leaves no solvent vapour pressure. Whether the solute is ionic alone does not determine this conclusion without concentration and dissociation information.
In an ideal binary solution, the pure vapour pressure of component A is lower than that of component B. At equal liquid mole fractions, which component will be less abundant in the vapour?
Correct answer: A
At equal liquid mole fractions, Raoult’s law gives p_A = x_Ap_A° and p_B = x_Bp_B°. Since x_A = x_B but p_A° < p_B°, the partial pressure of A is lower. In the vapour phase, y_i = p_i/P, so A also has the smaller vapour-phase mole fraction. B is more volatile and therefore becomes enriched in the vapour; equal liquid fractions do not imply equal vapour fractions.
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