4: Vapour Pressure

Correct answer: A. Heat may be released

Explanation: 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.

Correct answer: A. A mixture showing positive deviation

Explanation: 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.

Correct answer: A. A mixture showing negative deviation

Explanation: 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.

Correct answer: A. Specific attraction between unlike molecules

Explanation: 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.

Correct answer: A. The solute dissociates into ions and increases particle number

Explanation: 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.

Correct answer: A. The lowering may be less than expected

Explanation: 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.

Correct answer: A. Component A

Explanation: 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.

Correct answer: A. Vapour pressure is an equilibrium property, while rate is also affected by external conditions

Explanation: 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.

Correct answer: A. Attractions between like and unlike molecules are nearly equal

Explanation: 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.

Correct answer: A. it dissociates into ions and increases number of particles

Explanation: 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.

Correct answer: A. less than expected value

Explanation: 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.

Correct answer: A. heat may be released on mixing

Explanation: 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.

Correct answer: A. volume increase is possible

Explanation: 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.

Correct answer: A. 150 kPa

Explanation: 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.

Correct answer: A. because it is related to solute mole fraction and particle number

Explanation: 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.

Correct answer: A. 70 kPa

Explanation: 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.

Correct answer: B. 110 kPa

Explanation: 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.

Correct answer: C. 0.70

Explanation: 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.

Correct answer: B. Attraction between unlike molecules is comparatively weak

Explanation: 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.

Correct answer: B. Negative deviation

Explanation: 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.

Correct answer: B. A mixture showing positive deviation

Explanation: 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.

Correct answer: B. Below the ideal vapour-pressure line

Explanation: 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.

Correct answer: B. Because attractions between like and unlike molecules are nearly equal

Explanation: 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.

Correct answer: A. Benzene and toluene

Explanation: 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.

Correct answer: A. Strong specific attraction between unlike molecules

Explanation: 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.