4: Vapour Pressure

A larger surface provides more sites for evaporation, so it can alter the time required to reach equilibrium and the initial evaporation rate. However, at fixed temperature the equilibrium condition makes evaporation and condensation rates equal at a pressure determined by the liquid’s nature. Once equilibrium is reached, changing surface area does not change that saturated pressure.

An actual pressure above the ideal line represents positive deviation from Raoult’s law. Positive deviation means unusually high vapour pressure, so the mixture reaches the surrounding pressure at a lower temperature than expected. A composition with this minimum boiling temperature can form a minimum-boiling azeotrope. Maximum-boiling behaviour is associated with negative deviation below the ideal line.

A pressure below the ideal line indicates negative deviation, caused by stronger unlike-molecule attractions. The reduced vapour pressure means the mixture must be heated to a higher temperature before boiling, so a maximum in the boiling-temperature curve is possible. This corresponds to a maximum-boiling azeotrope; a minimum-boiling azeotrope is linked with positive deviation.

For a solution containing a non-volatile solute, Raoult’s law gives (p⁰ − p)/p⁰ = xsolute, where p⁰ is the vapour pressure of pure solvent and p is that of the solution. Thus relative lowering equals the solute mole fraction. Mole fraction uses moles of solute divided by total moles of solute plus solvent, not a mass fraction or volume.

For an ideal solution containing a non-volatile solute, Raoult’s law gives p = x1 p°. Therefore (p° − p)/p° = 1 − x1 = x2, where x2 is the mole fraction of the solute. The result depends on composition, not simply on solvent mass, colour, or the square of external pressure. This relation assumes ideal dilute behaviour.

A non-volatile solute does not itself escape into the vapour phase under the stated conditions. Its presence lowers the mole fraction and the effective availability of solvent molecules at the liquid surface, so fewer solvent molecules escape per unit time. Consequently the equilibrium vapour pressure decreases, as expressed by p = xsolvent p°solvent. The solute does not completely vaporise.

For an ideal solution containing a non-volatile solute, Raoult’s law gives the relative lowering of vapour pressure as (P° − P)/P° = Xsolute. The stated relative lowering is 0.02, so the mole fraction of the solute is also 0.02. The solvent mole fraction would be approximately 0.98 in this dilute solution, but it is not the quantity asked for.

For a solution containing a non-volatile solute, Raoult’s law gives Psolution = xsolvent P°solvent. Since the mole fractions add to one, increasing the solute mole fraction decreases the solvent mole fraction. The solvent then contributes less vapour, so the total vapour pressure falls. Decreasing solute mole fraction would produce the opposite effect, while isotonicity and changing a name do not determine vapour pressure.

Raoult’s law gives the relative lowering of vapour pressure as x₂ = n₂/(n₁ + n₂). For a dilute solution, n₂ is much smaller than n₁, so this becomes approximately n₂/n₁. Hence 0.02 = n₂/9.8, giving n₂ = 0.02 × 9.8 = 0.196 mol. Rounded suitably, this is 0.2 mol, so option B is correct.

For a dilute solution, the lowering of vapour pressure is usually small, especially when the solute concentration is low. Accurately measuring the vapour pressure of the pure solvent and the solution, and then determining their small difference, requires careful apparatus and control of temperature. Experimental uncertainty can therefore produce a significant error in the calculated molar mass. Thus, option A is correct.

For an ideal solution containing a non-volatile solute, Raoult’s law gives the relative lowering of vapour pressure as the solute mole fraction: \(\Delta p/p^0=X_{\text{solute}}\). Thus, \(X_{\text{solute}}=0.02\). Since mole fraction equals solute moles divided by total moles, \(n_{\text{solute}}=0.02\times5.0=0.10\,\mathrm{mol}\). Therefore, option C is correct.