4: Vapour Pressure

Adding more non-volatile solute, while keeping temperature and the solvent basis comparable, increases the solute mole fraction and decreases the solvent mole fraction. Raoult's law then predicts a further decrease in solvent vapour pressure. The solute does not itself supply appreciable vapour, and pressure cannot become infinite or acquire a colour value.

When the comparison is made at the same temperature and pressure conditions, a lower vapour pressure than that of the pure solvent is the characteristic effect of dissolving a non-volatile solute. The solute lowers the solvent mole fraction, and Raoult’s law gives P_solution = X_solvent P°_solvent, which is less than P° when X_solvent is below one. The other conclusions are physically meaningless or irrelevant.

Vapour pressure is produced by molecules that escape from the liquid and establish an equilibrium vapour phase. A higher value at the same temperature indicates a greater tendency of the liquid to form vapour, usually because attractions are weaker or particles have greater escaping tendency. It does not mean that no liquid remains or that evaporation is unrelated.

A non-volatile solute contributes negligible vapour pressure of its own and lowers the solvent mole fraction. Consequently, fewer solvent molecules escape at a given temperature, so the solution vapour pressure is lower than that of the pure solvent. This is the first qualitative conclusion; numerical questions can then use Raoult's law to calculate the exact value.

For a solution containing a non-volatile solute, Raoult's law is P = Xsolvent P°. Pure solvent has Xsolvent = 1, so its vapour pressure is P°. After solute is added, Xsolvent becomes less than 1; therefore P is smaller than P°. Colour, container volume, and the solvent's name do not determine this lowering.

For a solution containing a non-volatile solute, Raoult’s law gives the vapour pressure as P = Xsolvent × P°. Substituting the given values, P = 0.7 × 100 = 70 units. Therefore, option A is correct. The value cannot remain 100 because the solute lowers the pressure, and 30 is the lowering rather than the solution pressure.

For a solution containing a non-volatile solute, Raoult’s law gives P = X_solvent P°, where P° is the vapour pressure of the pure solvent. Dividing by P° gives P/P° = X_solvent. Since the solution pressure is 80% of P°, P/P° = 80/100 = 0.80. Therefore, the solvent mole fraction is 0.8; 0.2 would represent the solute mole fraction in a binary solution.

For a dilute ideal solution with a non-volatile solute, the relative lowering of vapour pressure is (P° − P)/P° = Xsolute. This follows because only the solvent contributes to the vapour, and its escaping tendency is reduced according to its mole fraction. Thus option A is correct. Container shape, colour, and odour do not determine this colligative relation.

For a non-volatile solute, the relative lowering of vapour pressure equals the solute mole fraction: (P° − P)/P° = Xsolute. The given value is Xsolute = 0.2, so the relative lowering is 0.2, or 20% if expressed as a percentage. Option B is the solvent mole fraction in a binary solution, not the requested relative lowering.

Raoult's law provides the ideal reference vapour pressure. If the observed pressure is higher than this reference, the solution shows positive deviation. Usually, unlike-component attractions are weaker than the original like-component attractions, so molecules escape more easily. Negative deviation would instead mean an observed pressure lower than the ideal value.

Raoult’s law for a volatile component A is PA = XA × PA°. Therefore, its partial vapour pressure depends on its mole fraction in the liquid phase and on the vapour pressure of pure A at the same temperature. Option A correctly includes both factors. Container height and colour are irrelevant, while mass alone cannot determine the pressure without composition information.

For a volatile component A in an ideal solution, Raoult’s law gives PA = XA PA°. At constant temperature, PA° remains fixed. Consequently, increasing XA increases PA in direct proportion. Therefore, option A is correct. A decrease would occur only if the component’s mole fraction decreased; colour has no role, and the pressure does not become zero merely because composition changes.

The vapour above a mixture contains contributions from each volatile component. According to Dalton’s law of partial pressures, the total pressure is the sum of these contributions: Ptotal = PA + PB. Thus option A is correct. The more volatile component may contribute more, but the less volatile component still contributes if it is volatile; neither component can be ignored automatically.

The total pressure of a mixture of vapours is obtained by adding the partial pressures of all components, according to Dalton’s law. Thus, P_total = P_A + P_B = 40 + 60 = 100 units. Multiplication, giving 2400, is not used for pressure addition, while 20 is the difference and 60 is only the pressure of component B, not the total pressure.

A non-volatile solute has negligible vapour pressure at the stated temperature, so it contributes essentially no vapour. The vapour above the solution therefore comes mainly from the solvent, whose pressure is lowered according to P = Xsolvent P°. The container and outside air are not sources of the equilibrium vapour pressure of the solution.

For a non-volatile solute, lowering of vapour pressure is a colligative effect and increases with the number of dissolved particles, provided temperature and solvent amount are comparable. More particles make the solvent mole fraction smaller and cause greater lowering. No solute gives pure-solvent pressure, while a volatile solute requires a different analysis.

At a fixed temperature, higher vapour pressure means that a larger fraction of molecules can escape from the liquid surface; hence the liquid is more volatile. Therefore A is more volatile than B. It generally also has a lower normal boiling point, although equal boiling points do not follow from the given statement, and every liquid can boil under suitable pressure.

Boiling begins when the liquid's vapour pressure equals the external pressure. A liquid that has a higher vapour pressure at every comparable temperature reaches this equality at a lower temperature. Thus it has a lower normal boiling point. The alternatives reverse or deny the meaning of volatility and boiling.

Positive deviation means that the observed vapour pressure is higher than the Raoult-law prediction. This commonly occurs when unlike-component attractions are weaker than the attractions in the pure liquids. Molecules are then held less strongly and escape more readily. The interaction need not be completely ionic, and mole fraction is unrelated to that description.

In negative deviation, unlike molecules attract each other more strongly than expected from the pure components. This stabilizes the liquid phase and reduces the number of molecules escaping into vapour. Consequently, the actual vapour pressure lies below the Raoult-law value. Weak attraction would produce the opposite, positive deviation.

The governing relation for relative lowering of vapour pressure is (P° − P)/P°, where P° is the pure-solvent pressure and P is the solution pressure. The absolute lowering is 80 − 72 = 8 units. Dividing by the original pressure gives 8/80 = 0.10. Therefore option A, 0.1, is correct. The value 8 represents absolute lowering, not relative lowering.

For a non-volatile solute, the relative lowering of vapour pressure is equal to the solute mole fraction: (P° − P)/P° = Xsolute. Therefore, when the solute mole fraction increases, the relative lowering also increases. This is a direct relationship, so option A is correct. It is not inverse, and neither colour nor the absence of a relationship explains the colligative behaviour.

Increasing temperature raises the average kinetic energy of liquid molecules. More molecules can overcome intermolecular attractions and escape into the vapour phase, so evaporation initially increases. In the closed vessel a new equilibrium is eventually established at a higher vapour pressure. The other options contradict the effects of heating and phase equilibrium.

Dynamic vapour-pressure equilibrium requires evaporation and condensation to occur in a closed space, so vapour remains available to return to the liquid. In an open container, escaping vapour is continuously removed and its concentration cannot build up to a stable equilibrium value above the liquid. The liquid surface and temperature still exist and remain important.

In PA = XA PA°, PA is the partial vapour pressure of component A in the solution, XA is its liquid-phase mole fraction, and PA° is the vapour pressure of pure A at the same temperature. Since PA° is fixed at fixed temperature, PA changes directly with XA. The equation does not state that pressure is zero or that XA is always one.