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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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Medium · Level 4View options
Heat may be released
Heat is never released
Heat is always infinite
Heat has no relation to molecular interactions
Medium · Level 4View options
A mixture showing positive deviation
A mixture showing negative deviation
A perfectly ideal mixture
Only a pure solvent
Medium · Level 4View options
A mixture showing negative deviation
A mixture showing positive deviation
Any ideal gas
Only a non-volatile solid
Medium · Level 4View options
Specific attraction between unlike molecules
Complete dissociation of both
Zero vapour pressure of both
Both becoming solid
Medium · Level 4View options
The solute dissociates into ions and increases particle number
The solvent mass becomes zero
Ions necessarily vaporise
External pressure vanishes
Medium · Level 4View options
The lowering may be less than expected
The lowering always becomes infinite
Its temperature relation disappears
Vapour pressure always becomes zero
Medium · Level 4View options
Component A
Component B
Both will always be equal
Neither will enter the vapour phase
Medium · Level 4View options
Vapour pressure is an equilibrium property, while rate is also affected by external conditions
Both are exactly the same
Vapour pressure is only colour
Rate is only mole fraction
Medium · Level 4View options
Attractions between like and unlike molecules are nearly equal
No molecules are present
Vapour pressure becomes zero
The liquid becomes solid
Medium · Level 4View options
it dissociates into ions and increases number of particles
it does not affect solvent vaporisation
it is always volatile
it converts liquid into gas
Medium · Level 4View options
less than expected value
always more than expected value
infinite
independent of pure solvent pressure
Medium · Level 4View options
heat may be released on mixing
heat is always absorbed strongly
no molecules attract
solution formation is impossible
Medium · Level 4View options
volume increase is possible
volume will always remain exactly zero
volume must decrease
volume has no relation with molecules
Medium · Level 4View options
150 kPa
200 kPa
250 kPa
300 kPa
Medium · Level 4View options
because it is related to solute mole fraction and particle number
because it only tells colour
because it tells height of vessel
because it changes name of solvent
Medium · Level 4View options
70 kPa
50 kPa
80 kPa
100 kPa
Medium · Level 4View options
80 kPa
110 kPa
130 kPa
200 kPa
Medium · Level 4View options
0.30
0.40
0.70
1.00
Medium · Level 4View options
Attraction between unlike molecules is very strong
Attraction between unlike molecules is comparatively weak
There is no volatile component in the liquid
All molecules are completely fixed
Medium · Level 4View options
Positive deviation
Negative deviation
Perfect ideality
Zero vaporisation
Medium · Level 4View options
A mixture showing ideal behaviour
A mixture showing positive deviation
A mixture showing negative deviation
A mixture of non-volatile solids
Medium · Level 4View options
Above the ideal vapour-pressure line
Below the ideal vapour-pressure line
Always zero vapour pressure
Only the vapour pressure of a pure liquid
Medium · Level 4View options
Because all molecules become vapour
Because attractions between like and unlike molecules are nearly equal
Because solute is necessarily ionic
Because there is no solvent in solution
Medium · Level 4View options
Benzene and toluene
Acetone and chloroform
Water and nitric acid
Ethanol and water
Medium · Level 4View options
Strong specific attraction between unlike molecules
Complete repulsion between unlike molecules
Both components become non-volatile
Both components become ions and disappear
Question 1MediumLevel 4
In a solution showing negative deviation, which statement about heat of mixing is generally correct?
Correct answer: A
Negative deviation generally indicates that unlike molecules attract each other more strongly than the original like molecules. Formation of these stronger interactions can lower the enthalpy of the mixture and release heat, giving an exothermic heat of mixing. The word “may” is appropriate because the exact heat depends on the system; it is never infinite.
Which type of mixture can form a minimum-boiling azeotrope?
Correct answer: A
A positive-deviation mixture has a total vapour pressure higher than the Raoult-law value. At a given external pressure, higher vapour pressure corresponds to boiling at a lower temperature. If the deviation is sufficiently large and the vapour and liquid compositions coincide at a composition, a minimum-boiling azeotrope results. Negative deviation gives the opposite type.
Which type of mixture can form a maximum-boiling azeotrope?
Correct answer: A
Negative deviation means the actual vapour pressure is lower than the ideal value, usually because unlike molecules attract strongly. A lower vapour pressure requires a higher temperature to equal the external pressure, so the boiling temperature can show a maximum. When liquid and vapour compositions coincide at that point, a maximum-boiling azeotrope is formed.
What is considered the reason for negative deviation in a mixture of acetone and chloroform?
Correct answer: A
Acetone and chloroform can form a strong specific interaction, commonly described through hydrogen bonding involving the carbonyl oxygen of acetone and the acidic hydrogen of chloroform. This unlike attraction reduces the escaping tendency of both components, so the observed vapour pressure becomes lower than the Raoult-law prediction. It is not caused by complete dissociation or solidification.
Why can lowering of vapour pressure due to an electrolyte be greater than the simple expected value?
Correct answer: A
An electrolyte such as an ionic compound can dissociate into two or more ions in solution. The number of dissolved particles is then greater than the number of formula units added, increasing the colligative effect and often making vapour-pressure lowering larger than the non-electrolyte estimate. The ions generally remain solvated; their compulsory vaporisation is not the reason.
If solute particles associate with each other, what happens to lowering of vapour pressure?
Correct answer: A
Association combines several solute particles into fewer larger species. Since colligative effects depend on the number of dissolved particles, the actual particle count becomes smaller than the count assumed for separate particles. Consequently the lowering of vapour pressure is less than the ideal non-association prediction. It does not become infinite or necessarily reduce the pressure to zero.
In an ideal solution, pure vapour pressure of A is greater than that of B. Their liquid-phase mole fractions are equal. Which component will have the higher proportion in the vapour phase?
Correct answer: A
For an ideal solution, the partial pressure is Pi = xiP°i. Since the liquid mole fractions of A and B are equal, the component with the larger pure vapour pressure has the larger partial pressure. Thus PA is greater than PB. Vapour-phase mole fraction is proportional to partial pressure, so A is enriched in the vapour. Equal liquid fractions do not force equal vapour fractions.
What is the most important difference between vapour pressure and rate of evaporation?
Correct answer: A
Equilibrium vapour pressure is the pressure established by vapour and liquid at equilibrium, mainly at a specified temperature and for a specified liquid composition. Evaporation rate is a kinetic quantity: it can change with temperature, surface area, air movement and humidity. A larger surface may increase rate but does not change the final equilibrium pressure at the same temperature.
Why is the heat of mixing nearly zero when an ideal solution forms?
Correct answer: A
During mixing, like–like interactions are partly replaced by unlike interactions. In an ideal solution, these interactions have nearly the same strength, so the energy required to separate old neighbours is nearly balanced by the energy released when new neighbours form. Hence enthalpy or heat of mixing is approximately zero. This is separate from vapour pressure becoming zero or a phase change.
Why can an electrolyte solute show comparatively greater lowering of vapour pressure?
Correct answer: A
Colligative effects depend on the number of dissolved particles. If an electrolyte dissociates, one formula unit can produce two or more ions, increasing the effective particle count. The solvent mole fraction then decreases more than it would for an undissociated solute of the same amount, so vapour-pressure lowering is larger. The actual increase depends on the degree of dissociation.
If solute particles associate to form larger particles, how can the observed lowering of vapour pressure be affected?
Correct answer: A
Association combines several solute particles into fewer larger units. Since vapour-pressure lowering is a colligative effect, it depends on the number of effective particles, not merely on the mass of solute added. Association therefore gives fewer particles and a smaller lowering than the value calculated by assuming no association. The exact deviation depends on the extent of association.
If actual vapour pressure of a solution is lower than ideal value, what indication about energy change on mixing may be obtained?
Correct answer: A
A vapour pressure below the Raoult-law value indicates negative deviation. This commonly arises when unlike molecules attract more strongly than the original like molecules. Formation of these stronger interactions can release energy, so the enthalpy of mixing may be negative. The word “may” is important because vapour-pressure deviation alone does not establish the exact heat quantitatively; it gives a useful indication.
If actual vapour pressure of a solution is greater than ideal value, what indication about volume change on mixing may be obtained?
Correct answer: A
A vapour pressure above the Raoult-law value indicates positive deviation, usually associated with weaker unlike-molecule attractions. The molecules may pack less efficiently after mixing, so an increase in volume can occur. However, positive deviation does not make volume expansion compulsory in every system; the question appropriately says it is possible. Ideal solutions have nearly zero volume change on mixing.
In an ideal binary solution, pure vapour pressure of A is 300 kPa and that of B is 100 kPa. Mole fraction of A is 0.25. What is total vapour pressure?
Correct answer: A
The liquid mole fraction of B is x B = 1 − 0.25 = 0.75. By Raoult’s law, p A = 0.25 × 300 = 75 kPa and p B = 0.75 × 100 = 75 kPa. The total pressure is the sum of these partial pressures: P = 75 + 75 = 150 kPa. The pure pressures are not added directly because each is weighted by its liquid mole fraction.
Why is relative lowering of vapour pressure useful in finding molar mass of a solute?
Correct answer: A
For a dilute solution with a non-volatile, non-electrolyte solute, relative lowering satisfies Δp/p° = x solute, approximately n solute/n solvent. The solute moles are obtained from its mass divided by molar mass, so measured pressure lowering can be combined with known solvent data to calculate the unknown molar mass. Association or dissociation must be considered if present.
In an ideal solution, liquid mole fraction of A is 0.60. Pure pressure of A is 50 kPa and of B is 100 kPa. What is the total pressure?
Correct answer: A
The mole fraction of B is x B = 1 − 0.60 = 0.40. Raoult’s law gives p A = 0.60 × 50 = 30 kPa and p B = 0.40 × 100 = 40 kPa. By Dalton’s law, total pressure is P = p A + p B = 30 + 40 = 70 kPa. The pure-component pressures cannot be used directly without multiplying by their liquid mole fractions.
In an ideal binary solution, the pure vapour pressure of component A is 200 kPa and that of component B is 50 kPa. If the mole fraction of A is 0.4, what is the total vapour pressure?
Correct answer: B
Raoult’s law gives the partial pressure of each component as its mole fraction multiplied by its pure-component vapour pressure. For A, p_A = 0.4 × 200 = 80 kPa. Since the solution is binary, x_B = 1 − 0.4 = 0.6, so p_B = 0.6 × 50 = 30 kPa. The total pressure is p_A + p_B = 80 + 30 = 110 kPa, so option B is correct.
In an ideal solution, the partial pressures of components A and B are 30 kPa and 70 kPa, respectively. What is the mole fraction of B in the vapour phase?
Correct answer: C
The total pressure of the vapour is the sum of the component partial pressures: P = 30 + 70 = 100 kPa. By Dalton’s law, the vapour-phase mole fraction of B is y_B = p_B/P. Substitution gives y_B = 70/100 = 0.70. The value 0.30 is the vapour mole fraction of A, while the total mole fraction must be 1.00, not the fraction of B alone.
If the vapour pressure of a real solution is much higher than the value predicted by Raoult’s law, what molecular behaviour is likely?
Correct answer: B
A vapour pressure higher than the Raoult-law prediction is called a positive deviation. It occurs when unlike interactions, such as A–B attractions, are weaker than the average A–A and B–B attractions. Molecules then escape from the liquid more easily, increasing vapour pressure. Strong unlike attraction would instead hold molecules in the liquid and produce negative deviation, so option A is opposite.
If the actual vapour pressure of a solution is lower than the ideal value and heat is released on mixing, what deviation is shown?
Correct answer: B
A vapour pressure below the Raoult-law value indicates negative deviation. If mixing releases heat, the process is exothermic, usually showing that unlike A–B attractions are stronger than the original A–A and B–B attractions. These stronger interactions make escape into the vapour phase more difficult, so the observed pressure decreases. Positive deviation is associated with weaker unlike attractions and often heat absorption.
Which mixture is most likely to form a minimum-boiling azeotrope?
Correct answer: B
Positive deviation means the mixture has a vapour pressure higher than the ideal prediction. At a given external pressure, it therefore reaches the boiling condition at a lower temperature. A composition at which liquid and vapour have the same composition can form a minimum-boiling azeotrope. Negative deviation is associated with maximum boiling.
Maximum-boiling azeotrope is associated with which type of vapour-pressure behaviour?
Correct answer: B
A maximum-boiling azeotrope results from negative deviation from Raoult’s law. The unlike molecular attraction is relatively strong, so the actual vapour pressure lies below the ideal line. A lower vapour pressure requires a higher temperature to equal the external pressure, producing a maximum in boiling temperature. Thus B is correct.
Why are heat of mixing and volume change nearly zero in an ideal solution?
Correct answer: B
In an ideal solution, A–B interactions are nearly equal in strength to A–A and B–B interactions. Replacing like contacts by unlike contacts therefore causes little energy change, so ΔH_mix is nearly zero. The molecules also pack with nearly no net expansion or contraction, giving ΔV_mix close to zero. Ionic character is not required.
Which pair is more likely to form a nearly ideal solution?
Correct answer: A
Benzene and toluene are chemically similar, have comparable molecular sizes, and possess broadly similar intermolecular forces. Consequently, their unlike interactions are close to their like interactions and deviations from Raoult’s law are small. Acetone–chloroform, water–nitric acid, and ethanol–water have stronger specific interactions and are more non-ideal.
What is the main reason for negative deviation in a mixture of acetone and chloroform?
Correct answer: A
Acetone and chloroform form a strong specific interaction, commonly described as hydrogen bonding involving the acidic hydrogen of chloroform and the carbonyl oxygen of acetone. This stabilizes the liquid mixture, makes escape more difficult, and lowers vapour pressure below the Raoult-law value. Complete repulsion would instead favour positive deviation.
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