4: Vapour Pressure

For component A in an ideal solution, Raoult’s law is PA = XA PA°. Substituting the given values gives PA = 0.4 × 200 = 80 units. Thus option A is correct. The value 200 is the pressure of pure A, while 500 comes from an incorrect division and 0.2 is not the pressure obtained from the stated data.

The solvent's vapour pressure represents, in a simple kinetic sense, its tendency to escape from the liquid surface. Adding a non-volatile solute lowers the solvent mole fraction and reduces the fraction of surface or liquid particles able to escape as solvent vapour. Thus the escaping tendency decreases; the solvent does not transform into solute or disappear.

For a non-volatile solute, Raoult’s law gives Psolution = Xsolvent P°. If the solvent mole fraction is high and close to one, multiplying P° by this fraction gives a pressure close to P°, though normally slightly lower. Thus option A is correct. The pressure is not necessarily near zero or higher than the pure-solvent value.

Stronger intermolecular attractions hold liquid molecules more firmly, so fewer can escape into the vapour phase at a given temperature. Therefore vapour pressure and volatility are lower. Weaker attraction generally has the opposite effect. Also, lowering temperature normally lowers vapour pressure, so the word “always” makes option D incorrect.

Heating increases the average kinetic energy of liquid molecules. Consequently, a larger fraction of molecules has enough energy to overcome intermolecular attractions and escape from the surface into the vapour phase. More escaping molecules produce a higher equilibrium vapour pressure. Heating does not make particles massless, stop their motion, or reduce their size to zero.

A liquid boils when its vapour pressure becomes equal to the external pressure. Lowering the external pressure means that this equality is reached at a lower temperature, because the liquid needs to develop a smaller vapour pressure. Therefore, the boiling temperature decreases and option A is correct. This is why liquids boil at lower temperatures at high altitudes.

Boiling occurs when vapour pressure becomes equal to external pressure. Increasing external pressure raises the target value, so the liquid must be heated to a higher temperature to produce sufficient vapour pressure. Therefore the boiling point rises. Boiling never requires zero vapour pressure, and mole fraction alone does not state the required temperature.

For an ideal solution, Raoult’s law gives PA = XA PA°. At constant temperature, the pure-component vapour pressure PA° is constant. Thus a decrease in XA produces a proportional decrease in PA. Option A is correct. PA equals PA° only when XA is one, and it cannot become infinite or increase when the multiplying factor becomes smaller.

A volatile component has an appreciable tendency to enter the vapour phase. Therefore, when both components are volatile, each contributes a partial vapour pressure, and the total pressure is their sum under Dalton's law. Calling one component only a solvent does not remove its volatility, while the container does not supply the liquid vapour.

For a solution containing a non-volatile solute, Raoult’s law is P = Xsolvent P°. Substituting the data gives P = 0.75 × 200 = 150 units. Therefore, option A is correct. The value 200 is the pressure of the pure solvent before adding solute, 75 omits the factor of 200, and 275 incorrectly adds the quantities.

Assuming the horizontal axis represents increasing temperature and the vertical axis represents vapour pressure, an upward-sloping line means the pressure value increases as temperature increases. Heating raises molecular kinetic energy and increases evaporation, which supports this trend. The axes must always be checked before interpreting any graph.

Vapour pressure depends strongly on temperature, so a meaningful comparison between a solution and its pure solvent must be made at the same temperature. Otherwise a pressure difference could arise simply from heating or cooling rather than from adding solute. Colour, container name, and experimenter identity have no role in this thermodynamic comparison.

First determine whether the solute can enter the vapour phase. A non-volatile solute contributes negligible vapour and lowers the solvent pressure through Raoult's law. A volatile component contributes its own partial pressure, so the total pressure requires a sum of partial pressures. Colour, container appearance, and word length provide no chemical information for solving the problem.

For an ideal solution containing a non-volatile solute, Raoult's law states Psolvent = Xsolvent P°solvent. Adding solute increases the total number of moles while the solvent moles remain fixed, so Xsolvent decreases. At the same temperature P°solvent is unchanged; consequently the solution vapour pressure decreases. No automatic temperature change is required.

A non-volatile solute does not enter the vapour phase. When its amount increases, the mole fraction of the solvent decreases, so fewer solvent molecules can escape from the liquid surface. By Raoult’s law, P = Xsolvent P°, a lower solvent mole fraction gives a lower vapour pressure. Therefore, the vapour-pressure lowering becomes greater.

For a solution containing a non-volatile solute, Raoult’s law gives P = Xsolvent P°. Substituting the given values, P = 0.8 × 120 = 96. The answer has the same pressure unit as the given pure-solvent pressure. The values 120 and 24 result from using the wrong relation or subtracting incorrectly.

For a solution containing a non-volatile solute, Raoult’s law gives P = X_solvent P°. Dividing both sides by P° gives P/P° = X_solvent. The given pressure ratio is 75%, which must be written as 75/100 = 0.75. Therefore, the solvent mole fraction is 0.75. Option A represents the solute mole fraction in this ideal case, while 1.25 and 75 are not valid mole fractions.

For a non-volatile solute, the solute contributes no vapour pressure, so Raoult’s law gives P = X_solvent P°. The relative lowering is (P° − P)/P°. Substituting P = X_solvent P° gives (P° − P)/P° = 1 − X_solvent = X_solute, because the mole fractions of solute and solvent add to one. Hence option A is correct; the other quantities do not determine this colligative relation.

For a dilute ideal solution with a non-volatile solute, the relative lowering of vapour pressure is equal to the mole fraction of the solute: (P° − P)/P° = X_solute. Since X_solute is directly given as 0.15, the relative lowering is 0.15. The value 0.85 is the solvent mole fraction, 1.15 is impossible for a mole fraction, and 15 is the unconverted percentage-like number.

Relative lowering of vapour pressure is calculated as (P° − P)/P°, where P° is the vapour pressure of the pure solvent and P is that of the solution. Substitution gives (100 − 90)/100 = 10/100 = 0.10. Thus option B is correct. The value 0.9 is P/P°, while 10 and 90 are pressure differences or pressure values without the required normalization.

Both liquids are volatile, so each contributes vapour above the solution. Raoult’s law gives PA = XA P°A and PB = XB P°B. Dalton’s law then states that the total pressure is Ptotal = PA + PB. Therefore, the partial pressures must be added; multiplying mole fractions or considering only one component would omit part of the vapour pressure.

Raoult’s law states that the partial vapour pressure of a volatile component in an ideal solution equals its liquid-phase mole fraction multiplied by the vapour pressure of the pure component at the same temperature. Therefore, for component A, P_A = X_A P_A°. The other expressions either add unrelated quantities, divide in the wrong direction, or subtract quantities with incompatible meanings and units.

For an ideal solution, Raoult’s law gives the partial vapour pressure of A as P_A = X_A P_A°. Substituting the data, P_A = 0.3 × 160 = 48 units. Hence option B is correct. The value 160 is the vapour pressure of pure A, not its partial pressure in the solution; 0.3 is only a mole fraction, and 190 does not result from the required multiplication.

In a binary vapour phase, Dalton’s law states that the total pressure is the sum of the partial pressures of the components. Therefore, P_total = P_A + P_B = 45 + 55 = 100 units. Option C is correct. Subtraction gives the difference, multiplication gives an unrelated product, and 55 is only the partial pressure of component B rather than the total pressure.

A binary solution contains only two components, A and B. Therefore, their mole fractions must add up to 1: X_A + X_B = 1. Substituting X_A = 0.35 gives X_B = 1 − 0.35 = 0.65. Thus, option B is correct. Option A merely repeats X_A, while 1.35 and 3.5 cannot be valid mole fractions here because a mole fraction lies between 0 and 1.