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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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Up to 25 questions from this page. Select your focus, then start.
25 questions
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Easy · Level 8View options
0.7
0.3
1.3
0.03
Easy · Level 8View options
Molecular kinetic energy increases with temperature
Molecules disappear with temperature
Container colour changes when temperature decreases
Temperature has no relation to molecules
Easy · Level 8View options
The one with strong intermolecular attraction
The one whose molecules escape easily
The one with low boiling point
The one that is more volatile
Easy · Level 8View options
Dynamic equilibrium
Permanent boiling
Complete freezing
Chemical decomposition
Easy · Level 8View options
When its vapour pressure equals one atmosphere pressure
When its vapour pressure is zero
When its colour changes
When it has no solute
Easy · Level 8View options
0.10
0.50
5.0
55
Easy · Level 8View options
Low
High
Infinite
Less than zero
Easy · Level 8View options
65 kilopascal
40 kilopascal
25 kilopascal
15 kilopascal
Easy · Level 8View options
The amount of solute is small
The amount of solute is very large
The solute is volatile and very large in amount
The solvent mole fraction is zero
Easy · Level 8View options
p°
0
2p°
p°/2
Easy · Level 8View options
Negative deviation
Positive deviation
No deviation
Complete vaporization
Easy · Level 8View options
170 kilopascal
200 kilopascal
30 kilopascal
235 kilopascal
Easy · Level 8View options
Because lowering is obtained from the difference between pure-solvent and solution pressures
Because the colour of pure solvent is needed
Because solution mass is zero
Because solute is always volatile
Easy · Level 8View options
fewer solvent molecules are present at surface
solute forms its own vapour
temperature automatically falls
volume of vessel increases
Easy · Level 8View options
mole fraction of solvent
mass of solute
total mass
vessel volume
Easy · Level 8View options
solute-solvent attraction is weaker
solute-solvent attraction is stronger
both components completely separate
solute is highly volatile
Easy · Level 8View options
unlike attraction is stronger than like attraction
unlike attraction is weaker than like attraction
no vapour forms in solution
solution freezes
Easy · Level 8View options
sum of both partial vapour pressures
only from more volatile component
only from less volatile component
difference between both pressures
Easy · Level 8View options
40 kPa
60 kPa
75 kPa
80 kPa
Easy · Level 8View options
obeys Raoult's law at all concentrations
forms only at very high pressure
forms only in water
always releases heat
Easy · Level 8View options
positive deviation
negative deviation
zero deviation
no partial deviation
Easy · Level 8View options
acetone has comparatively weaker intermolecular attraction
acetone molecules never vaporise
water has no attraction
acetone always remains solid
Easy · Level 8View options
kinetic energy of molecules increases
chemical formula changes
vessel becomes heavier
vapour molecules disappear
Easy · Level 8View options
total vapour pressure increases
total vapour pressure decreases
total vapour pressure becomes zero
total vapour pressure remains independent
Easy · Level 8View options
its partial pressure and total pressure
only colour of liquid
only mass of vessel
only smell of liquid
Question 1EasyLevel 8
In an ideal binary solution, the mole fraction of component A is 0.3. What will be the mole fraction of component B?
Correct answer: A
In a binary solution containing only A and B, the mole fractions obey xA + xB = 1. Therefore, xB = 1 − xA = 1 − 0.3 = 0.7. The ideal nature of the solution is relevant to Raoult’s law but does not alter this basic mole-fraction relation. A mole fraction cannot exceed one, and 0.03 is not the complement of 0.3.
What is the main reason vapour pressure depends on temperature?
Correct answer: A
Temperature measures the average kinetic energy of particles. When temperature rises, a larger fraction of liquid molecules has enough energy to overcome intermolecular attractions and enter the vapour phase. The rate of evaporation increases, and in a closed vessel the equilibrium vapour pressure becomes higher. Molecules do not disappear, container colour is irrelevant, and temperature is directly related to molecular motion.
At the same temperature, which liquid will have lower vapour pressure?
Correct answer: A
Strong intermolecular attractions hold liquid molecules more firmly and make it harder for them to escape into the vapour phase. Consequently, fewer molecules are present in the vapour at equilibrium, giving a lower vapour pressure at the same temperature. Easy escape, high volatility, and a low boiling point are generally associated with higher vapour pressure, so options B, C, and D describe the opposite trend.
When the rates of evaporation and condensation become equal, what is established in a closed vessel?
Correct answer: A
In a closed vessel, liquid molecules continuously evaporate and vapour molecules continuously condense. When the forward and reverse rates become equal, the macroscopic amounts of liquid and vapour remain constant, although molecular motion continues. This state is dynamic equilibrium. It is not permanent boiling, freezing or chemical decomposition.
In which condition is the normal boiling point of a liquid defined?
Correct answer: A
A liquid boils when its vapour pressure becomes equal to the external pressure. The normal boiling point is specifically defined under a standard external pressure of one atmosphere. Therefore, the correct condition is vapour pressure equal to one atmosphere. Zero vapour pressure would not permit boiling, and colour or the presence of solute is not the defining condition, although solutes can change a solution’s boiling point.
In a solution, the lowering of vapour pressure is 5 kilopascal and the vapour pressure of the pure solvent is 50 kilopascal. What is the relative lowering?
Correct answer: A
Relative lowering is obtained by dividing the absolute lowering of vapour pressure by the vapour pressure of the pure solvent: relative lowering = Δp/p°. The given values give Δp/p° = 5/50 = 0.10. This is a ratio without a unit, equivalent to 10%. It is not 5.0, because the lowering must be normalized by the pure-solvent pressure, and it is not 0.50 because that would be incorrect division.
If a liquid attains vapour pressure equal to external pressure at a lower temperature, how will its boiling point be?
Correct answer: A
Boiling begins when the vapour pressure of a liquid equals the pressure acting on its surface. If this equality is reached at a lower temperature, the temperature required for boiling is lower; therefore, the liquid has a lower boiling point. The conclusion does not imply that the boiling point must be below zero or infinite. A higher boiling point would require reaching the same external pressure only at a higher temperature.
In an ideal solution, the partial vapour pressure of component A is 40 kilopascal and that of component B is 25 kilopascal. What is the total vapour pressure?
Correct answer: A
For a mixture containing volatile components, Dalton’s law states that the total pressure is the sum of the partial pressures. Therefore, ptotal = pA + pB = 40 + 25 = 65 kPa. The total pressure cannot be found by subtracting the values; subtraction gives only their difference, 15 kPa. It also must be greater than either individual partial pressure when both components contribute positively.
If adding a non-volatile solute causes only a small lowering of vapour pressure, what can be inferred about the amount of solute?
Correct answer: A
For a non-volatile solute, lowering of vapour pressure increases with the effective solute mole fraction or particle concentration. A small lowering therefore indicates that the solute has a small effect, normally because its amount is small relative to the solvent. A very large amount would cause a larger lowering, while zero solvent fraction would make the stated comparison invalid.
According to Raoult’s law, what will be the value of p when x = 1?
Correct answer: A
Raoult’s law for a component is p = x p°, where p is its partial vapour pressure, x is its mole fraction, and p° is the vapour pressure of the pure component. Substituting x = 1 gives p = 1 × p° = p°. This represents the pure-component limit: when the component makes up the entire liquid phase, its solution pressure equals its pure vapour pressure, not zero or a multiple of it.
If the actual vapour pressure of a solution is lower than the pressure predicted by Raoult’s law, what type of deviation is it?
Correct answer: A
Raoult’s law predicts the ideal vapour pressure. When the observed pressure lies below that prediction, the solution shows negative deviation. The usual molecular reason is stronger attraction between unlike molecules, which keeps more molecules in the liquid phase. Positive deviation would mean a pressure above the ideal value, while no deviation means equality.
In a solution containing a non-volatile solute, the mole fraction of solvent is 0.85. If the vapour pressure of the pure solvent is 200 kilopascal, what is the vapour pressure of the solution?
Correct answer: A
Because the solute is non-volatile, the measurable vapour pressure is due to the solvent. Raoult’s law gives psolution = xsolvent p°solvent. Hence psolution = 0.85 × 200 = 170 kPa. The pressure is lower than 200 kPa because the solvent occupies only 0.85 of the total mole fraction. The value 30 kPa is the lowering, not the solution pressure, and 235 kPa is physically inconsistent here.
Why is the vapour pressure of pure solvent needed while measuring lowering of vapour pressure?
Correct answer: A
Lowering of vapour pressure is a comparative quantity. It is defined as Δp = p° − p, so both the pure-solvent pressure p° and the solution pressure p must be known. Without p°, the difference and the relative lowering Δp/p° cannot be calculated. Colour, zero mass and universal solute volatility have no role in this definition.
Why does the vapour pressure of a solution decrease when a non-volatile solute is added to a liquid?
Correct answer: A
A non-volatile solute does not enter the vapour phase appreciably. Its particles reduce the fraction of the liquid surface occupied by solvent molecules and lower the solvent's escaping tendency. Therefore fewer solvent molecules are present in equilibrium vapour, so the vapour pressure decreases. Temperature, vessel volume and solute vapour are not the stated cause.
According to Raoult's law, the partial vapour pressure of the solvent in an ideal solution is proportional to what?
Correct answer: A
For an ideal solution, Raoult's law is pA = xA pA°. Thus the partial pressure of component A is directly proportional to its mole fraction in the liquid, provided temperature is fixed. If the solvent mole fraction decreases, its partial vapour pressure decreases in the same relation; mass and vessel volume are not the governing quantities.
In which situation is negative deviation from Raoult's law observed?
Correct answer: B
Negative deviation means that the observed vapour pressure is lower than the value calculated from Raoult's law. This occurs when unlike, solute-solvent attractions are stronger than the corresponding like-like attractions. Molecules are then held more tightly and escape less readily. Weak attraction generally produces positive, not negative, deviation.
In which situation is positive deviation from Raoult's law more likely?
Correct answer: B
Positive deviation occurs when the actual total vapour pressure is higher than the Raoult-law prediction. If unlike molecules attract each other less strongly than like molecules do, they escape from the liquid more easily. The increased escaping tendency raises vapour pressure. Stronger unlike attraction would instead cause negative deviation.
For an ideal solution of two volatile liquids, how is total vapour pressure obtained?
Correct answer: A
Both liquids are volatile, so each contributes molecules to the vapour phase. Dalton's law states that the total pressure of a gas mixture equals the sum of the partial pressures: ptotal = pA + pB. The more volatile component may contribute more, but the less volatile component still contributes; therefore neither one alone nor their difference is correct.
If the vapour pressure of pure solvent is 80 kPa and the mole fraction of solvent is 0.75, what is the partial vapour pressure of solvent in an ideal solution?
Correct answer: B
Raoult’s law states that the partial vapour pressure of a component in an ideal solution equals its mole fraction multiplied by the vapour pressure of that pure component. Thus, p = x × p° = 0.75 × 80 = 60 kPa. Therefore, option B is correct. The answer is lower than 80 kPa because the solvent is only 75% of the liquid-phase mole fraction.
Which statement correctly identifies an ideal solution?
Correct answer: A
An ideal solution obeys Raoult's law for each component throughout the composition range, not merely at one concentration. Its unlike and like intermolecular interactions are approximately similar, so enthalpy and volume changes on mixing are approximately zero. Ideal solutions are not restricted to water, high pressure, or heat-releasing mixtures.
If the actual vapour pressure of a solution is less than the pressure predicted by Raoult's law, what type of deviation is it?
Correct answer: B
Deviation is identified by comparing observed pressure with the ideal Raoult-law value. If pactual is smaller than pideal, the difference is negative and the solution shows negative deviation. Stronger unlike attractions commonly cause this lower pressure. Positive deviation would require an observed pressure greater than the predicted value.
Why does a liquid like acetone have higher vapour pressure than water?
Correct answer: A
At the same temperature, vapour pressure depends on how readily molecules escape from the liquid surface. Acetone has weaker overall intermolecular attraction than water, whose molecules form strong hydrogen-bond interactions. Consequently, acetone molecules escape more easily and its volatility and vapour pressure are higher. It does not mean acetone never vapourises, and water certainly does have intermolecular attraction.
Why does the equilibrium vapour pressure of a liquid generally increase when temperature is increased?
Correct answer: A
Increasing temperature raises the average kinetic energy of the liquid molecules. A larger fraction of molecules then has enough energy to overcome intermolecular attractions and escape into the vapour phase. At the new equilibrium, the vapour exerts a higher pressure. The liquid’s chemical formula and the vessel’s mass do not cause this trend, and vapour molecules do not disappear; their equilibrium amount changes.
If the mole fraction of the more volatile component increases in an ideal binary solution, what happens to total vapour pressure?
Correct answer: A
The more volatile component has the larger pure-liquid vapour pressure. In an ideal solution, increasing its liquid mole fraction increases its partial-pressure contribution, while decreasing the other component's contribution. The weighted total therefore moves toward the higher pure-component pressure and increases, although it need not double or become infinite.
The mole fraction of a component in the vapour phase is related to what?
Correct answer: A
For an ideal-gas vapour mixture, Dalton's law gives ptotal = Σpi and the vapour-phase mole fraction is yi = pi/ptotal. Thus a component's partial pressure divided by total pressure gives its proportion in vapour. Liquid colour, vessel mass and smell do not provide the quantitative vapour composition.
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