Solutions and Phase Equilibria Questions and Answers

Answer: D

The correct relation is the one with A's enthalpy of fusion, A's melting point and a negative exponent: \(1-x_{B}=\)\(\exp[\frac{-\Delta H^{o}_{fus(A)}}{R}(\frac{1}{T}-\frac{1}{T_{A}})]\).

The solution freezes when its component A is in equilibrium with pure solid A, so the condition involves A's enthalpy of fusion and A's melting point. For an ideal solution:

\[ \begin{aligned} \ln x_{A} &= -\frac{\Delta H_{fus(A)}}{R}\left(\frac{1}{T} - \frac{1}{T_{A}}\right) \\ x_{A} &= 1 - x_{B} \end{aligned} \]

Exponentiating gives the last option. A check: T lies below T_A, so 1/T - 1/T_A is positive, the exponent is negative and x_A is less than 1, as it must be.

The two options with B's enthalpy and T_B describe solid B separating instead, and the option with a positive exponent would make x_A greater than 1.

Answer: C

At Q, solid B, solid AB₂ and liquid are in equilibrium.

The vertical line for AB₂ does not reach the liquid region: it stops at a horizontal line, so AB₂ melts incongruently. At that temperature AB₂ decomposes into solid B and a liquid of composition Q. Q is the peritectic point, where the liquidus coming down from the B side meets the horizontal peritectic line.

On the peritectic line three phases coexist: solid B, solid AB₂ and liquid of composition Q. With three phases in a two-component system at fixed pressure, there are no degrees of freedom left, which is why the line is horizontal.

Solid A is not involved at Q: A appears only at the eutectic on the left. A three-solid equilibrium, or a four-phase one, is impossible at an invariant point of a binary system.

Answer: D

The slopes are \(m_{1} \lt 0\) and \(m_{2} \gt 0\).

The Clapeyron equation gives the slope of the solid-liquid line:

\[ \frac{dP}{dT} = \frac{\Delta H_{fus}}{T\,\Delta V_{fus}} \]

\(\Delta H_{fus}\) is always positive, so the sign of the slope is the sign of the volume change on melting. Ice is less dense than liquid water, so water contracts on melting (\(\Delta V_{fus} \lt 0\)) and its melting line slopes backwards: \(m_{1} \lt 0\). Solid CO₂ is denser than the liquid, like most substances, so \(\Delta V_{fus} \gt 0\) and \(m_{2} \gt 0\).

Water is the unusual case, so any option giving both slopes the same sign, or reversing the two, is wrong.

Answer: A

The vapour mole fractions are 0.688 hexane and 0.312 heptane.

Treat 4 and 6 as the relative amounts, so the liquid mole fractions are 0.4 and 0.6. By Raoult's law the partial pressures are

\[ \begin{aligned} p_{\mathrm{hex}} &= 0.4 \times 0.198 = 0.0792\ \mathrm{atm} \\ p_{\mathrm{hep}} &= 0.6 \times 0.06 = 0.036\ \mathrm{atm} \end{aligned} \]

\[ y_{\mathrm{hex}} = \frac{0.0792}{0.1152} = 0.688 \]

So hexane is 0.688 and heptane 0.312 of the vapour. The vapour is richer in the more volatile hexane than the liquid is. 0.400 and 0.600 are the liquid composition, and 0.312 and 0.688 swaps the two components.

Answer: A

The degrees of freedom are 0, 1, 2.

For one component the Gibbs phase rule gives

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

• Point i is where three boundary lines meet, the triple point, with three phases in equilibrium: F = 0, so both p and T are fixed.
• Point j lies on a boundary line, where two phases coexist: F = 1, so fixing T fixes p.
• Point k lies inside a single-phase region: F = 2, so p and T can both vary.

The option 3, 2, 1 forgets to subtract the phases, and 2, 0, 1 and 0, 2, 1 assign the counts to the wrong points.

Answer: A

The correct option is P = 3, C = 2, F = 1.

• Phases: Pb₃O₄(s), PbO(s) and O₂(g). Each solid is a separate phase and the gas is one more, so P = 3.
• Components: three species linked by one equilibrium, so C = 3 - 1 = 2 (for example PbO and O₂).

\[ F = C - P + 2 = 2 - 3 + 2 = 1 \]

The system is univariant, like CaCO₃ decomposing: fixing the temperature fixes the equilibrium O₂ pressure. P = 3, C = 3 ignores the reaction relation. The options with P = 2 count the two solids as one phase, but two distinct solids are always two phases.

Answer: C

The molality is 2m.

Take 1 L of solution. Its mass is \(1000 \times 1.86 = 1860\) g, and it holds 3 mol of X, which weigh \(3 \times 120 = 360\) g. The solvent (water) is the rest:

\[ \begin{aligned} m_{\text{water}} &= 1860 - 360 = 1500\ \text{g} = 1.5\ \text{kg} \\ \text{molality} &= \frac{3}{1.5} = 2\ \text{mol kg}^{-1} \end{aligned} \]

3m confuses molality with molarity; 1m and the others come from forgetting to subtract the solute's mass or using the solution's volume as the solvent's mass.

Answer: C

The highest \(K_{b}\) is that of Carbon tetrachloride.

Approximate ebullioscopic constants (K kg mol^-1) are: carbon tetrachloride 5.03, chloroform 3.63, acetic acid 3.07, cyclohexane 2.79 and carbon disulphide 2.34.

\(K_{b} = \frac{RT_{b}^{2}M}{1000\,\Delta_{vap}H}\) is large for a heavy solvent with a modest enthalpy of vaporisation, which is why \(CCl_{4}\) tops the list.

Answer: A

The most soluble gas is formaldehyde.

Formaldehyde is a small polar molecule that forms hydrogen bonds with water and reacts with it to form the hydrate (methanediol); its 37-40% aqueous solution is sold as formalin. It is far more soluble than the other gases.

CO₂ is only moderately soluble, vinyl chloride is slightly soluble, and the non-polar gases methane and argon are poorly soluble, their solubility governed by a large Henry's law constant.

Answer: C

The mole fraction of nitrogen is \(10^{-4}\).

Henry's law states that the partial pressure of a gas over a solution is proportional to its mole fraction in the solution, \(p = K_{H} x\). Writing both pressures in bar (76.4 kbar = 76400 bar):

\[ \begin{aligned} x &= \frac{p}{K_{H}} = \frac{7.648}{76400} \\ &= 1.0 \times 10^{-4} \end{aligned} \]

The key step is converting kilobar to bar; leaving the constant as 76.4 would give 0.1, which is not even close to a dissolved gas fraction. \(10^{-3}\), \(10^{-5}\) and the other powers are each at least a factor of ten away from the computed value. A large \(K_{H}\) means low solubility, so a small mole fraction is expected for nitrogen.