Solutions and Phase Equilibria Questions and Answers

Answer: B

The air used by scuba divers contains 11.7 % helium.

At the high pressures underwater, nitrogen dissolves in the blood (Henry's law) and forms bubbles when the diver surfaces, causing the bends, while at depth it can also act as a narcotic. To reduce this, divers' tanks are filled with air diluted with helium, which is far less soluble in blood. The textbook composition is 11.7 % helium, 56.2 % nitrogen and 32.1 % oxygen.

56.2 % and 32.1 % are the nitrogen and oxygen figures from the same mixture, and the other values are not part of it.

Answer: D

The highest freezing point belongs to H₂O.

Water freezes at 273 K, because its molecules are held by strong, extensive hydrogen bonding into an ordered solid.

The others freeze far below this: chloroform at about 210 K, carbon disulphide at about 162 K, ethanol at about 159 K and diethyl ether at about 157 K. Ethanol can hydrogen-bond, but only through one OH group per molecule, and its ethyl group disrupts packing.

Answer: A

The lowest solubility is that of Argon.

By Henry's law, \(p = K_{H}x\): the larger the Henry's law constant, the smaller the mole fraction dissolved at a given pressure. At 298 K, approximate \(K_{H}\) values (kbar) are argon 40.3, methane 0.413, vinyl chloride 0.611, carbon dioxide 1.67 and formaldehyde \(1.83 \times 10^{-5}\).

Argon has by far the largest \(K_{H}\), so it is the least soluble. Formaldehyde is extremely soluble because it hydrogen-bonds with and reacts with water.

Answer: A

K_f for benzene is 4.3°C kg mol^-1.

Use ΔT_f = K_f m, where m is the molality of naphthalene:

\[ \begin{aligned} \Delta T_{f} &= 5.5 - 2.8 = 2.7\,^{\circ}\text{C} \\ n &= 1.60/128 = 0.0125\ \text{mol} \\ m &= 0.0125/0.020 = 0.625\ \text{mol kg}^{-1} \end{aligned} \]

\[ \begin{aligned} K_{f} &= \frac{2.7}{0.625} \\ &= 4.32\,^{\circ}\text{C kg mol}^{-1} \end{aligned} \]

The unit follows from dividing kelvin (or °C) by mol kg^-1, which gives °C kg mol^-1. The options with g mol^-1 or mol kg^-1 have the right number but the wrong unit, and 5.1 would come from using benzene's molar mass where it does not belong.

Answer: D

The activity of water is 0.9864.

The activity of the solvent is the ratio of its vapour pressure over the solution to that of the pure solvent at the same temperature. At 100°C pure water boils, so its vapour pressure is 760 torr:

\[ a_{w} = \frac{p}{p^{*}} = \frac{749.7}{760} = 0.9864 \]

The activity of water in a solution must be less than 1, because the dissolved KNO₃ lowers the vapour pressure. That rules out 1.0023 and 1.0230, which are above 1. 0.9218 is far too low for a 0.5 M solution, whose vapour pressure falls by only about 10 torr.

Answer: A

The values are 3 components, 3 phases and 2 degrees of freedom.

There are four species, Fe, H₂O, FeO and H₂, linked by one equilibrium, so the number of components is C = 4 − 1 = 3. The phases are solid Fe, solid FeO and one gas phase containing H₂O and H₂, so P = 3. The phase rule gives

\[ F = C - P + 2 = 3 - 3 + 2 = 2 \]

Counting the two gases as separate phases (option B) is wrong, because gases always mix into one phase. Options C and D follow from miscounting P or F.

Answer: D

The value for the equimolar solution is more negative than −13 kJ mol⁻¹.

For an ideal solution the Gibbs energy of mixing is entirely entropic:

\[ \Delta G_{mix} = nRT\,(x_{1}\ln x_{1} + x_{2}\ln x_{2}) \]

The bracket is most negative at x₁ = x₂ = 0.5, where it equals ln 0.5 = −0.693, compared with −0.671 at the given composition (0.606 and 0.394). For the same amount of liquid, the equimolar mixture therefore has the more negative ΔG_mix.

ΔG_mix depends on composition, ruling out option B, it is not a pure number like 'unity', and 'less negative' (C) has the dependence the wrong way round.

Answer: A

ΔH_mix is zero.

An ideal solution is one in which the interactions between unlike molecules (benzene-toluene) are the same as those between like molecules. Mixing then involves no net breaking or making of interactions, so no heat is absorbed or released, and ΔH_mix = 0 at every composition. The whole of the negative ΔG_mix in such a solution comes from the entropy of mixing, ΔG_mix = −TΔS_mix.

A positive ΔH_mix describes a solution with positive deviations from Raoult's law, and a negative one a solution with negative deviations, so neither applies to an ideal solution, and the figure 1.0 kJ mol⁻¹ has no basis.

Answer: B

The point representing the supercritical condition is B.

On a pressure-temperature phase diagram, the liquid-vapour boundary rises from the triple point and ends at the critical point. Beyond that point, at temperatures and pressures above the critical values, there is no distinction between liquid and gas: the substance is a supercritical fluid. Point B lies just past the upper end of the curved boundary, in that region.

Point A lies in the region above the steep (solid-liquid) boundary, point C sits close to the triple point between the phase boundaries, and point D lies below the vapour-pressure curve in the gas region. None of these is beyond the critical point.

Answer: B

Statements (I) and (III) are correct, so the answer is (I), (III).

Henry's law, which states that the solubility of a gas is proportional to its partial pressure, holds for gases that dissolve without reacting with the solvent, such as O₂ in water (I), and it implies that solubility increases with pressure (III).

HCl does not obey Henry's law in water because it ionises completely to H₃O⁺ and Cl^-, so (II) is false. The unit in (IV) is also wrong for the usual form p = K_Hx, where K_H has units of pressure (atm or bar), not atm^-1; the constant's unit depends on how the law is written, so atm^-1 is not 'the' unit.