For two liquids A and B to form an ideal solution

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Correct — A, Enthalpy of mixing should be zero. An ideal solution is one that obeys Raoult's law over the entire composition range, which means the partial vapour pressure of each component equals its mole fraction times the vapour pressure of that component in the pure state: p(A) = x(A)·p°(A) and p(B) = x(B)·p°(B). Physically, that can only hold if the A–B attraction is the same in strength as the A–A and B–B attractions the molecules already felt in the pure liquids. If nothing changes in the energy of interaction when the molecules are shuffled together, then no heat has to be supplied or given out, and no contraction or expansion occurs — so the two measurable signatures of ideality are ΔH(mixing) = 0 and ΔV(mixing) = 0. That is exactly what option (a) states. The other two thermodynamic quantities in the options cannot be zero, and knowing why is what makes this question quick. Mixing always increases the number of ways the molecules can be arranged, so the entropy of mixing is strictly positive: ΔS(mixing) = −nR[x(A) ln x(A) + x(B) ln x(B)], which for an equimolar binary mixture works out to R ln 2 = 5.76 joules per kelvin per mole of solution. And because ΔG = ΔH − TΔS, an ideal solution has ΔG(mixing) = 0 − TΔS(mixing) = −TΔS(mixing), a negative number — about −1.7 kilojoules per mole of solution at 298 K for the equimolar case. A negative free energy is precisely why two such liquids mix spontaneously when you pour them together. So only the enthalpy vanishes, and (a) is the answer.

Why the other options are wrong — (b) The exact opposite of the truth, and the most attractive wrong answer because it sits next to the right one in the same 'should be zero' pattern. Entropy of mixing is positive for every real mixture, ideal or not — that is the whole statistical meaning of mixing, since a mixed state can be realised in vastly more molecular arrangements than a separated one. If ΔS(mixing) were zero there would be no thermodynamic driving force to mix at all.  (c) A zero free-energy change would mean the mixed and unmixed states are equally stable — the liquids would sit in their separate layers with no tendency to combine. Since ΔG(mixing) = −TΔS(mixing) for an ideal solution and ΔS(mixing) is positive, ΔG(mixing) is always negative. Mixing is spontaneous, and its spontaneity here is entirely entropy-driven.  (d) Fails because option (a) is a correct and complete statement of the defining condition. An escape option only wins when each of the other three can be shown to fail, and here the first of them is the standard textbook criterion for ideality.

Concept: Raoult's law, published by François-Marie Raoult in 1887, says that in a solution of volatile liquids the partial vapour pressure of each component is proportional to its mole fraction, the constant of proportionality being the vapour pressure of the pure component. A solution that obeys it at every composition and temperature is called ideal. Ideality is a statement about forces: the molecules must not be able to tell whether their neighbours are of their own kind or the other kind. Real pairs come close when the two liquids are structurally and chemically similar — benzene with toluene, n-hexane with n-heptane, chlorobenzene with bromobenzene, ethyl bromide with ethyl iodide. Where the forces differ, the solution deviates. If A–B attraction is weaker than the average of A–A and B–B, escaping tendency rises, vapour pressure exceeds the Raoult prediction and ΔH(mixing) is positive — a positive deviation, as in ethanol and water. If A–B attraction is stronger, as when chloroform hydrogen-bonds to acetone, vapour pressure falls below the prediction, ΔH(mixing) is negative and the mixture warms up.

State PCS 2025 Q2 — Science & Technology PYQ with Answer

State PCS 2025 Q2 — Science & Technology PYQ with Answer