NCERT Solutions for Class 12 Chemistry Chapter 1 - Download PDF

Structural angle. The cleanest mental picture is a 3 × 3 table indexed by the state of the solute (rows) and the state of the solvent (columns), which gives the nine canonical cases.

A binary solution mixes exactly two components and is the simplest case to study; the same ideas extend to ternary and higher mixtures. The key requirement is that the mixture be homogeneous, meaning no observable boundaries between regions of different composition, even under a microscope.

1. Gaseous solvent. Any gas is miscible with any other gas in all proportions (no intermolecular forces strong enough to resist mixing). Liquids and solids enter a gaseous solvent only as their vapours: humid air contains water vapour; the smell of camphor in a room is camphor vapour dispersed in air.
2. Liquid solvent. The most common laboratory case. Gases (O2, CO2) dissolve to varying extents in water. Two miscible liquids form a single phase (ethanol + water); immiscible liquids do not form a solution. Solids dissolve up to their solubility limit (glucose in water, NaCl in water).
3. Solid solvent. Less obvious but very important. Hydrogen gas occupies the interstitial spaces of solid palladium. Liquid mercury soaks into solid sodium to form an amalgam. Two solids together form alloys: brass (Cu + Zn), bronze (Cu + Sn), gold jewellery (gold + copper). These are genuine solid solutions, not mere mixtures.

Alternative classification: by number of components. A binary solution has two components, a ternary has three, and so on. NCERT restricts attention to binaries because the ideas (mole fraction, Raoult's law, _mixH) extend naturally with no new physics, only more book-keeping. A second axis of classification is by particle size: true solutions (<1 nm), colloids (1 – 1000 nm), and suspensions (>1000 nm). The nine-cell table refers strictly to true solutions.

Concept linkage. The very definition of ``solution'' is deceptively rich. Homogeneity requires that thermodynamic intensive variables (density, refractive index, chemical potential of each species) be the same in every macroscopic sub-volume. That is exactly the condition for a single thermodynamic phase in the Gibbs phase rule, F = C - P + 2. So when we say ``a solution is a homogeneous mixture'', we are silently invoking the phase rule with P=1.

Numerical cross-check. The textbook's nine examples (3 × 3) are not theoretical; they are the only combinations that exist in nature. Try to think of a tenth and you will find it collapses into one of the nine.

Why this matters. Recognising the type of solution sets expectations: gases dissolved in liquids obey Henry's law, two miscible liquids obey Raoult's law, and the colligative properties later in the chapter assume a solid (or non-volatile liquid) solute dissolved in a liquid solvent. JEE Main routinely asks one-mark ``identify the type of solution'' questions (e.g. JEE Main 2023 Jan 25 shift 2 listed ``soda water'' and asked the type of solution).

Nine types of solutions, grouped under three families by the state of the solvent (gas, liquid, solid).

Quick reading. The textbook table of solution types lists nine entries; one of them is exactly the case asked here, ``gas dissolved in solid'', with the standard example ``hydrogen in palladium''.

1. Definition restated: in a solid solution the solvent is solid and the solute can be solid, liquid or gas. We need the last case.
2. Pick the textbook's stock example: solid Pd as solvent, H2 gas as solute. Palladium's face-centred-cubic lattice has octahedral voids large enough to host hydrogen atoms after the H2 molecule dissociates on the metal surface.
3. State the result.

Alternative examples worth knowing. (i) H2 in titanium or zirconium (used as ``getter'' materials in vacuum systems); (ii) O2 or N2 dissolved in solid metals during electrolytic refining; (iii) CO2 trapped in solid H2O as gas hydrate (strictly a clathrate but textbook treats it as a solid solution). Any of these qualifies as a gas-in-solid solution.

Concept linkage. Two structural mechanisms make a gas dissolve in a solid: interstitial (small atoms like H, C, N, O occupying gaps between metal atoms) and substitutional (when the dissolved atom is comparable in size to the host, replacing host atoms). Hydrogen in Pd is purely interstitial because the H atom (r ∼ 0.5 ) is much smaller than the Pd atom (r ∼ 1.4 ).

Why this matters. Such ``interstitial'' solid solutions underlie hydrogen-storage alloys (LaNi5H6, Mg2NiH4) used in nickel-metal-hydride batteries. They also explain hydrogen embrittlement of steel (atomic H seeping into the iron lattice) and the reversible loading of H2 in fuel-cell electrodes. NEET 2022 and JEE Main 2024 (Apr 6) both had MCQs of the form ``which of the following is an example of solid solution where solute is a gas?'' with H/Pd as the correct answer.

Hydrogen in palladium (interstitial solid solution).

Strategic angle. Group the four units by ``what is in the numerator'' and ``what is in the denominator''. Numerator can be moles or mass; denominator can be moles, mass or volume. That gives a table from which the four useful combinations drop out.

1. Moles / moles ⇒ mole fraction x_i = n_i / ∑ n_j. Dimensionless and additive (∑ x_i = 1).
2. Moles / mass-of-solvent ⇒ molality m = n_solute / w_{solvent (kg)}. Units mol kg^-1.
3. Moles / volume-of-solution ⇒ molarity M = n_solute / V_{soln (L)}. Units mol L^-1.
4. Mass / total-mass ⇒ mass percentage w% = (w_A / w_total) × 100. Dimensionless.

Alternative bookkeeping units (worth knowing). Beyond the four NCERT-named units there are three more in routine use:

• Parts per million (ppm): mass-ratio × 10⁶. Used for trace pollutants (1 ppm = 10^-4 %).
• Volume percent (% v/v): used for liquid-in-liquid mixtures where volumes are easier to measure (alcoholic drinks: ``40% ABV'').
• Normality (N): equivalents of solute per litre of solution. Standard in acid-base and redox titrations.

Concept linkage. Mole fraction is the right variable for vapour pressure because Raoult's and Henry's laws are thermodynamic statements about the chemical potential _i = _i^∘ + RT ln x_i, which counts particles, not mass. By contrast, molality is the right variable for Δ T_f and Δ T_b because the cryoscopic and ebullioscopic constants K_f and K_b depend on the solvent's mass-based properties (latent heat per kg). Molarity is preferred whenever a volumetric pipette or burette is involved.

Numerical cross-check. For dilute aqueous solutions, m ≈ M because the density is ≈ 1 g mL^-1 and the solute mass is small. The exact link is M = m ρ / (1 + m M_w/1000), which reduces to M ≈ m when m M_w ≪ 1000.

Why this matters. Confusing molarity with molality is a classic exam trap. They differ by a factor of solution density and the mass of solute; for dilute aqueous solutions at room temperature the two are numerically close, but never assume they are equal. NEET 2021 and JEE Main 2022 (July 25 shift 2) both had paired questions where a student had to flip between molarity and molality using density.

Definitions as stated in the steps above.

Strategic angle. Use the compact ``shortcut'' that drops out of the four-step calculation: M = 10 × (mass %) × dM_w, where d is the density of solution in g mL^-1 and M_w is the molar mass of solute in g mol^-1. This avoids picking a sample size each time.

1. Derive the shortcut. Take 100 g of solution. Mass of solute =w g, moles =w/M_w. Volume in litres =(100/d)/1000 = 0.1/d. Substituting, M = w/M_w0.1/d = 10 w dM_w, and w here is numerically the mass percentage.
2. Plug in the values. M = 10 × 68 × 1.50463 = 1022.7263 = 16.23mol/L.
3. Sanity check. Commercial concentrated HNO3 is sold as ∼ 16 M; our answer matches the reference value.

Why this matters. The shortcut M = 10 w d/M_w is one of the most reused formulae in titration chemistry: any time you are handed a (% by mass, density) pair, you can read off molarity in seconds. JEE Main 2021 (Feb 26 shift 1) directly tested this shortcut for HCl.

M(HNO3) = 16.23mol/L.

Strategic angle. The cleanest route is to fix the sample at 1 kg of solvent, because molality is exactly the moles of solute per kilogram of solvent. Then convert that same picture to volume to read off molarity.

1. Take 1000 g of water. Since the solution is 10% w/w, glucose is 10/90 as massive as water, so it weighs 10/90 × 1000 = 111.11 g.
2. Moles of glucose = 111.11/180 = 0.617. Hence m = 0.617 mol per kg of water = 0.617mol/kg.
3. Mole fraction of glucose: moles of water = 1000/18 = 55.55, total moles =56.17, x_glucose = 0.617/56.17 = 0.0110, x_water = 0.989.
4. For molarity, use the shortcut M = 10 w dM_w = 10 × 10 × 1.2180 = 0.667mol/L.

Why this matters. The same 10 % sugar syrup can be quoted as ``0.6 M'', ``0.6 m'' or ``x = 0.011'' depending on the unit. Carrying density along lets you flip between molarity and molality without losing information. JEE Main 2020 (Sept 5 shift 2) and NEET 2019 both featured glucose-in-water concentration conversions of this exact form.

Numerical cross-check. Since d = 1.2 > 1, we expect M > m - confirmed (0.667 > 0.617). For a 10% solution M/m = d/(1 - 0.1 M_w/1000) approximately; substituting gives M/m ≈ 1.2/(1 - 0.018) ≈ 1.22 × d^-1 - close to the observed ratio 0.667/0.617 = 1.08. Density and water-mass correction together explain the small mismatch.

m = 0.617 mol kg^-1, x_glu = 0.0110, x_w = 0.989, M = 0.67 mol L^-1.

Strategic angle. Look at the average ``molar mass per formula unit of mixture'' and the average ``mol of HCl per formula unit of mixture'', so the answer reduces to a single division.

1. For an equimolar mixture, the average molar mass is M̄ = (106 + 84)/2 = 95g/mol. Moles of ``mixture units'' in 1 g: n_mix = 1/95.
2. Each ``mixture unit'' contains 12 mol Na2CO3 and 12 mol NaHCO3, so it consumes 12(2) + 12(1) = 1.5 mol HCl.
3. Total HCl needed: n_HCl = 1.5/95 = 0.01579 mol. Volume at 0.1 M: V = 0.01579/0.1 = 0.1579 L = 157.9 mL.

Why this matters. Stoichiometry on a mixture is the same calculation as on a single substance, but you have to weight each contribution by its mole fraction. The ``average molar mass'' trick generalises to any mixture if you know the molar fractions. JEE Main 2019 (Apr 9 shift 1) recycled this exact mixture problem with different masses.

V_HCl ≈ 157.9 mL.

Strategic angle. Mixing two solutions is a balance problem. Conserve solute mass and solvent mass independently, then take the ratio.

1. Solute balance: 75 + 160 = 235 g.
2. Solvent balance: solvent in first =225 g, in second =240 g, total =465 g.
3. Total mass =235 + 465 = 700 g.
4. Mass % of solute =235/700 × 100 = 33.57%; solvent = 465/700 × 100 = 66.43%.

Alternative check via weighted average. w%_new = (w₁ m₁ + w₂ m₂)/(w₁+w₂) = (300 · 25 + 400 · 40)/700 = 23500/700 = 33.57%. Same answer, fewer arithmetic steps. The same logic gives the new mole fraction or molarity when two like solutions are blended.

Concept linkage. Two-stream blending is a special case of macroscopic species balance: solute in = solute out, solvent in = solvent out. For miscible solutions of the same solute the volume is approximately conserved as well, so C₁V₁ + C₂V₂ = CV also works on a volume basis (Q24 uses this in disguise).

Why this matters. The two-stream mixing balance is the basic operation in industrial blending and in lab dilution by addition (not just by water). The shortcut formula above is the dilute-mixture analogue of C₁V₁ + C₂V₂ = C V. JEE Main 2018 (Apr 16 shift 1) posed a mass-blending problem identical in form to this one.

Solute 33.57%, solvent 66.43%.

Structural observation. The single difference between molality and molarity is the denominator: mass of solvent versus volume of solution. Solute moles are common to both.

1. Compute moles of ethylene glycol once: 222.6/62 = 3.59 mol.
2. Divide by mass of solvent in kg: 3.59/0.200 = 17.95 m. Note: this is a very concentrated solution; ∼ 53 % by mass.
3. Divide by volume of solution in L. Volume from density: V = 422.6/1.072 = 394.2 mL. M = 3.59/0.3942 = 9.11 mol L^-1.
4. Compare: molarity ≪ molality here because at high concentration the solvent makes up a smaller fraction of the solution by mass, but the solute itself occupies real volume.

Why this matters. For dilute solutions m ≈ M. For concentrated ones (this is ∼ 9 M, ∼ 18 m) they differ a lot and the choice of unit matters. JEE Main 2024 (Jan 30 shift 2) had a near-identical antifreeze/ethylene-glycol problem.

m = 17.95 mol kg^-1; M = 9.11 mol L^-1.

Quick reading. For trace pollutants in water the dilute approximation w_water ≈ w_solution is essentially exact (the solute contributes nothing measurable to the total mass). That collapses the two parts of the problem to two short calculations.

1. Mass percentage of CHCl3: (15/10⁶) × 100 = 1.5 × 10^-3 %.
2. Molality. Take 1 kg of water; CHCl3 present =0.015 g. Moles = 0.015/119.5 = 1.255 × 10^-4. m = 1.255 × 10^-4 mol kg^-1.

Numerical cross-check. Order-of-magnitude: 15 ppm chloroform in water ∼ 15 × 10^-6 mass fraction. Multiplied by water molality scale (∼ 55.5 mol per kg water) and divided by M_w(CHCl3) = 119.5, we get 15 × 10^-6 × 55.5 / 119.5 × 1000 ≈ 7 × 10^-3 mol/kg –- wait, that overshoots. Redo carefully: 15 ppm by mass means 15 μg per gram of solution, = 15 mg per kg solution ≈ 15 mg per kg water. So moles = 0.015/119.5 = 1.26 × 10^-4 –- consistent with the steps above.

Concept linkage. ppm, ppb, mass-%, mole fraction and molality are all linear in mass for dilute traces. The conversion factors are fixed by molar masses; remember ppm → molality via m = ppm_mass × 10^-3/M_w (mol per kg water).

Why this matters. Drinking-water standards specify limits in μg L^-1 or ppb, which are the same kind of ratio at a smaller scale (1 ppb = 10^-3 ppm = 10^-7 %). Knowing how to flip between them is a public-health basic. JEE Main 2022 and NEET frequently reuse ``contaminant in ppm → molality'' problems.

1.5 × 10^-3 % w/w; m ≈ 1.25 × 10^-4 mol kg^-1.

Picture-first. Imagine pure water as a tightly woven mat of H-bonds. Slip ethanol molecules into the mat: the ethyl tail can only push the water molecules apart (van der Waals forces alone), while the ethanol O-H still wants to H-bond. The net effect is fewer, looser H-bonds in the mixture.

1. Compare the three pairwise interactions: water-water (very strong H-bonds, ∼ 21 kJ mol^-1), alcohol-alcohol (strong H-bonds), water-alcohol (slightly weaker than water-water, because the bulky -C2H5 disrupts the geometry).
2. Because Ē_cross < 12(E_ww+E_aa), mixing costs energy. Therefore _mixH > 0 and _mixV > 0 (small).
3. Vapour pressure shows positive deviation from Raoult's law: p_obs > p^∘_A x_A + p^∘_B x_B. The ethanol-water mixture even forms a low-boiling azeotrope at 95.6 % ethanol, exactly because of this.

Alternative reasoning via thermodynamics. At constant temperature, the sign of _mixG = _mixH - T_mixS governs whether mixing is spontaneous. Entropy of mixing is always positive (_mixS > 0), so even an endothermic mixing process goes forward as long as TΔ S exceeds Δ H. Ethanol and water mix because the entropy gain dominates the (small positive) enthalpy cost –- exactly the situation flagged by positive deviation from Raoult's law.

Concept linkage.

• Three force model. For any A–B binary mixture compare E_AB with the average of E_AA and E_BB.
• E_AB < average ⇒ positive deviation, _mixH > 0, _mixV > 0, minimum-boiling azeotrope possible (ethanol–water, b.p. 78.15 ^∘C at 95.6% ethanol).
• E_AB > average ⇒ negative deviation (acetone–chloroform, see Q1.37).
• E_AB = average ⇒ ideal solution (benzene–toluene, n-hexane–n-heptane).

Numerical anchor. Pure water has ∼ 4 H-bonds per molecule, ethanol ∼ 2. In a 50:50 mixture by mole the average drops because the alcohol O-H makes the H-bond network sparser. The resulting mixing enthalpy is positive but small (∼ 1 kJ/mol), which is why ethanol and water mix in all proportions despite the positive deviation.

Why this matters. The same logic explains why ethanol-water cannot be separated into pure ethanol by simple distillation, and why absolute alcohol is made using a chemical drying agent or azeotrope breaker. NEET 2020 and JEE Main 2023 (Jan 24) both probed this exact ethanol–water reasoning chain.

Mixing weakens the H-bond network: positive deviation, _mixH > 0.

Strategic angle. Treat dissolution as a reversible chemical ``reaction'' between gas and solvent, then read off the temperature dependence from the sign of Δ H.

1. Identify the sign of Δ H. When gas molecules move from the gas phase (free, high kinetic energy) into the liquid (held by weak intermolecular forces), they lose kinetic energy as heat. So Δ H_soln < 0 (exothermic).
2. Apply Le Chatelier qualitatively: heating an exothermic equilibrium drives it backward, releasing gas.
3. Quantitatively, the van't Hoff equation dln KdT = Δ H^∘RT² with Δ H^∘ < 0 gives K decreasing with T.

Alternative reasoning: kinetic theory. Above a liquid, dissolved gas molecules and gas-phase molecules exchange continuously. Raising T increases the kinetic energy of both populations, but the liquid is held together by intermolecular forces while the gas is essentially free. So heating preferentially helps dissolved molecules escape the weak solute-solvent attraction, shifting the equilibrium toward the gas side. Quantitatively, the fraction of molecules with E > E_escape rises as e^-E/(kT) in a Maxwell-Boltzmann distribution, so solubility falls roughly exponentially with 1/T.

Concept linkage. Le Chatelier and the van't Hoff equation dln K_H^-1dT = -Δ H_solnRT², give the same prediction: with Δ H_soln < 0 the solubility constant decreases as T rises. The same equation governs the temperature dependence of all equilibrium constants and is the quantitative form of the qualitative Le Chatelier shift.

Why this matters. The same logic is why a warm soda goes flat faster than a cold one, and why beer foams more when poured cold into a warm glass. Real-world consequence: thermal pollution of rivers (from power-plant cooling water) lowers dissolved O2 levels and suffocates fish. NEET 2018 and JEE Main 2021 (Feb 24 shift 1) both reused this concept.

Because dissolution is exothermic, gas solubility decreases as temperature rises.

Strategic angle. Henry's law is the dilute-solute limit of the vapour-pressure law. It is essentially a thermodynamic statement that for low concentrations the chemical potential of a dissolved gas is a linear function of its mole fraction.

1. Statement. At constant T, the partial pressure of a gas in equilibrium with its solution is proportional to the mole fraction of gas in solution: p = K_H x.
2. Behaviour of K_H. Increases with T. The larger the K_H, the lower the solubility.
3. Four applications worth knowing. bottling of soft drinks; scuba diving (helium-oxygen mixtures to avoid bends); respiration at high altitude (anoxia); aquatic life dependent on dissolved O2.

Alternative forms of Henry's law. Henry's law can be written in three equivalent linear forms (all proportional to mole fraction x, molality m, or molar concentration C in the dilute regime):

• p = K_H x (mole-fraction form, NCERT default);
• p = K'_H m (molality form, K'_H in bar kg mol^-1);
• C = k_H,c p (concentration form, used in environmental chemistry; k_H,c in mol L^-1 atm^-1).

Concept linkage. Henry's law is the dilute-solute limit of the chemical-potential statement _{gas in soln} = μ^*_gas + RTln x. The ``constant'' K_H encodes the difference between standard chemical potentials of the gas in the two phases. As such, K_H behaves like an equilibrium constant: ln K_H = -Δ G^∘/RT.

Temperature dependence (numerical anchor). For O2 in water: K_H = 3.30 × 10⁴ atm at 293 K, ∼ 4.27 × 10⁴ atm at 298 K, ∼ 5.18 × 10⁴ atm at 310 K. The trend (rising K_H with T) is consistent with the exothermic dissolution of O2 in water.

Quantitative comparison of solubilities. At p = 1 atm and 298 K: O2 (K_H = 4.27 × 10⁴ atm) gives x = 2.34 × 10^-5. CO2 (K_H = 1.67 × 10³ atm) gives x = 6 × 10^-4, about 26 times more soluble. N2 (K_H = 8.68 × 10⁴ atm) gives the lowest aqueous solubility of the three.

When Henry's law fails. (i) When the gas reacts with the solvent (HCl, NH3, CO2 partial reaction with water); (ii) at high partial pressures where the linear regime breaks down; (iii) for dissociating solutes like HCl where the dissolved species splits into ions.

Why this matters. Henry's law is the bridge between gas-phase data (p, T) and aqueous-phase outcomes (how much gas dissolves). It is used in pollution dispersion, lung physiology, brewing, and the design of gas absorbers in chemical plants. NEET 2017 and JEE Main 2020 (Sept 6 shift 2) both placed Henry's law applications in their short-answer pools.

p = K_H x. Applies to dilute, non-reacting gas-liquid systems; underlies fizz in drinks, the bends, altitude sickness and dissolved-oxygen levels in water.

Quick reading. Same gas, same solvent, same temperature: only the amount of dissolved gas changes. So K_H is the same in both states, and the ratio of pressures equals the ratio of solubilities.

1. Write Henry's law for both: p₁ = K_H x₁, p₂ = K_H x₂. Divide: p₂/p₁ = x₂/x₁.
2. Solubility ratio (dilute): x ∝ w. So p₂ = p₁ · (w₂/w₁).
3. p₂ = 1 × (0.0500/0.00656) = 1 × 7.62 = 7.62 bar.

Numerical cross-check. p₂/p₁ = w₂/w₁ = 0.05/0.00656 = 7.62. Both pressures lie in the linear regime (a few bar over water at room temperature is well within the Henry's-law range). Order-of-magnitude is sensible: ∼ 8 × more dissolved gas needs ∼ 8 × higher pressure.

Why this matters. The proportionality form of Henry's law turns this from a thermodynamic statement into a one-line ratio you can use without ever computing K_H explicitly. JEE Main has reused this exact ``ratio of two Henry states'' setup repeatedly (e.g. 2020 Sept 6 shift 1 with N2 in blood).

p₂ ≈ 7.62 bar.

Strategic angle. A deviation from Raoult's law is just a mismatch between A-B forces in solution and the A-A, B-B forces in the pure liquids. The mismatch shows up simultaneously in three quantities: vapour pressure, enthalpy of mixing, and volume of mixing.

• If A-B is weaker than average → molecules escape more easily → vapour pressure too high (positive deviation); heat must be absorbed to break the stronger pure-liquid bonds, so Δ H > 0; molecules also pack a little looser, so Δ V > 0.
• If A-B is stronger than average → molecules are held more tightly → vapour pressure too low (negative deviation); Δ H < 0 (energy released); Δ V < 0.

1. State Raoult's law for an ideal mixture and identify the ideal straight lines in p vs x.
2. Sketch the typical curves: positive deviation lies above the line and may show a vapour-pressure maximum (minimum-boiling azeotrope); negative deviation lies below and may show a minimum (maximum-boiling azeotrope).
3. Associate signs: positive deviation ⇔ Δ H > 0 and Δ V > 0. Negative deviation ⇔ Δ H < 0 and Δ V < 0.

Alternative criterion: vapour-pressure curve shape. Plot p_tot vs x_A from 0 to 1. The ideal line is a straight chord from p^∘_B (at x_A=0) to p^∘_A (at x_A=1). The shape of the experimental curve gives the diagnosis immediately:

• Straight chord ⇒ ideal, Δ H = 0, Δ V = 0. Examples: n-hexane + n-heptane, benzene + toluene, chlorobenzene + bromobenzene.
• Bulge above ⇒ positive deviation. Examples: ethanol + water, acetone + CS2, acetone + benzene, CCl4 + chloroform.
• Dip below ⇒ negative deviation. Examples: acetone + chloroform, HCl + water, HNO3 + water.

Concept linkage: molecular cause.

1. Positive deviation. A–B forces weaker than the average of A–A and B–B. The molecules of each component escape more easily, p_tot rises above ideal. Heat is required to break the stronger A–A and B–B bonds (Δ H > 0, endothermic). The looser packing causes a small volume expansion (Δ V > 0).
2. Negative deviation. A–B forces stronger than the average of A–A and B–B. The mixture forms new attractions (often a H-bond between an O-H acceptor and a C-H or C=O donor). Heat is released (Δ H < 0, exothermic) and volume shrinks (Δ V < 0).
3. Ideal. A–B forces equal the average; no net enthalpy or volume change on mixing.

Azeotrope formation. A sufficiently strong deviation creates a stationary point on the p_tot-x curve, hence on the T_b-x curve (boiling point at fixed total pressure).

• Strong positive deviation → maximum p_tot at some intermediate x → minimum boiling point → minimum-boiling azeotrope (e.g. 95.6% ethanol + 4.4% water, b.p. 78.15 ^∘C).
• Strong negative deviation → minimum p_tot at some intermediate x → maximum boiling point → maximum-boiling azeotrope (e.g. 68% HNO3 + 32% water, b.p. 120 ^∘C; 20.2% HCl + 79.8% water, b.p. 108.5 ^∘C).

Numerical anchor. For acetone + chloroform near x_ac = 0.4 the experimental p_tot is ∼ 220 mmHg, well below the Raoult prediction (∼ 280 mmHg) –- a textbook negative deviation (see Q1.37).

Why this matters. Azeotropes (constant-boiling mixtures) are exactly the result of extreme positive or negative deviation. They limit how pure you can make a binary distillate by simple distillation. JEE Main 2023 (Apr 6 shift 1), JEE Advanced 2019, and NEET 2018 each featured deviation-and-azeotrope MCQs based on this classification.

Positive deviation ⇔ _mixH > 0; negative deviation ⇔ _mixH < 0.

Strategic angle. The relative lowering of vapour pressure is an exact colligative property: it depends only on the mole fraction of the solute, not on its identity. So one measurement of p_s at known solute mass gives the molar mass directly.

1. Set the dilute approximation x_solute ≈ n_solute/n_water.
2. Substitute: 0.009/1.013 = (2/M) / (98/18).
3. Solve: M = (2 × 18) / (98 × 0.009/1.013) = (36)/(0.8709) = 41.35 g mol^-1.

Alternative form: direct substitution. Solve p_s = p^∘ x_w for M. With x_w = n_w/(n_w + n_s) and n_s = 2/M, n_w = 98/18 = 5.444: 1.004 = 1.013 × 5.444/(5.444 + 2/M). Rearranging: 5.444 + 2/M = 5.444 × (1.013/1.004) = 5.4928. So 2/M = 0.0488 ⇒ M = 40.98 ≈ 41 g mol^-1. The slight numerical difference from the dilute-form answer (41.35 vs 40.98) reflects the dilute-approximation error.

Concept linkage. Relative lowering of vapour pressure is exact (it follows from Raoult's law without extra assumptions on the solute, beyond non-volatility). The dilute form (denominator ≈ n_w) is an approximation for small x_solute. For x_s 0.05 (typical 2-5% solutions) the error is below 1%; for larger x_s, use the exact form.

Why this matters. This is exactly how the molar mass of a new, non-volatile compound is measured in the lab: dissolve a known mass, measure the vapour-pressure lowering, and invert the formula. The Ostwald–Walker dynamic method exploits this. JEE Main 2022 (June 25 shift 2) and NEET 2019 both pulled vapour-pressure molar-mass MCQs.

M ≈ 41.35 g mol^-1.

Strategic angle. Raoult's law in the linear form p_tot = p^∘_B + (p^∘_A - p^∘_B) x_A shows that total vapour pressure varies linearly between the two pure-liquid values as composition changes from pure B to pure A.

1. Compute x_hep = (26/100)/[(26/100)+(35/114)] = 0.260/0.567 = 0.459 and x_oct = 0.541.
2. Apply the linear form: p_tot = 46.8 + (105.2 - 46.8) × 0.459 = 46.8 + 58.4 × 0.459 = 46.8 + 26.79 = 73.59 kPa.
3. Same answer, fewer multiplications.

Numerical cross-check. p_tot = 46.8 + (105.2-46.8) · x_hep = 46.8 + 58.4 · 0.459 = 73.6 kPa, matches the longer-route answer. Order-of-magnitude sanity: p_tot must lie between 46.8 and 105.2 kPa (pure-liquid endpoints) –- our 73.6 kPa sits cleanly between them.

Why this matters. For ideal mixtures, p_tot varies linearly with composition, which gives the iconic straight p-x line of Raoult's law and is the starting point for understanding non-ideal deviations. JEE Main 2020 (Jan 8) and NEET 2019 each used a benzene–toluene or heptane–octane ideal-mixture vapour-pressure problem.

p_total ≈ 73.6kPa.

Quick reading. The molality fixes the ratio of solute moles to solvent kilograms once and for all. The mole fraction of water in any 1-molal aqueous solution of a non-electrolyte is x_w = 55.56/56.56 = 0.9823, independent of the solute's identity.

1. Compute x_w once: 55.56/56.56 = 0.9823.
2. Multiply by p^∘ = 12.3 kPa: p = 12.08 kPa.
3. The drop is about 1.8%, consistent with a dilute solution.

Why this matters. The same shortcut applies to any non-electrolyte in water at moderate concentration; the relevant parameter is n_solute/n_water, which for 1 m aqueous is 1/55.56. JEE Main 2018 (Apr 15) tested this exact ``1-molal aqueous vapour pressure'' setup.

p ≈ 12.08kPa.

Strategic angle. A reduction to ``80 %'' of the original is a large effect, so be careful to use the exact form of the mole-fraction relation, not the linearised approximation.

1. Mole fraction of solute equals relative lowering: x_s = 0.20.
2. Set x_s = n_s/(n_s + n_oct) = 0.20 with n_oct = 1. Solve: n_s = 0.20/(1 - 0.20) = 0.25 mol.
3. Mass = 0.25 × 40 = 10 g.

Alternative algebraic route. From x_s = 0.20, x_solv = 0.80, so n_s/n_solv = 0.20/0.80 = 0.25. With n_solv = 1 mol, n_s = 0.25 mol ⇒ w_s = 10 g.

Numerical cross-check. Reverse-substituting: x_s = 0.25/(0.25+1) = 0.20 . Then p_s = p^∘ · (1-0.20) = 0.80 p^∘ .

Why this matters. The molality-based dilute approximation can under- or over-estimate by a factor of 1.25 for x_s = 0.20. Always check whether the lowering is small (x_s 0.05) before using the simplified form. JEE Main 2017 had a vapour-pressure problem with a comparable x_s ≈ 0.2 that punished the dilute approximation.

w_solute = 10 g.

Structural observation. The relative lowering form lets us write each measurement as a ratio of the same two unknowns. Eliminate one unknown by taking the ratio of the two equations; back-substitute to get the other.

1. Write both equations using (p^∘ - p_s)/p_s = n_s/n_w: (p^∘-2.8)/2.8 = 6/M and (p^∘-2.9)/2.9 = 5/M.
2. Ratio eliminates M: (p^∘-2.8)/(p^∘-2.9) · (2.9/2.8) = 6/5, leading to p^∘ = 3.53 kPa.
3. Back-substitute: 0.73/2.8 = 6/M, so M = 23.0 g mol^-1. (Recognise this as the molar mass of sodium, suggesting the solute is something like NaOH or a Na salt –- but the problem only asks for the number.)

Alternative algebraic check. Plug M = 23 and p^∘ = 3.53 back into both original equations:

• State 1: (3.53-2.8)/2.8 = 0.2607. 6/M = 6/23 = 0.2609. Match .
• State 2: (3.53-2.9)/2.9 = 0.2172. 5/M = 5/23 = 0.2174. Match .

Concept linkage. Two-measurement systems with two unknowns appear throughout physical chemistry: determining K_a and unknown initial concentration from two pH readings, deducing Δ H^∘ and Δ S^∘ from K at two temperatures, finding rate constant and order from two rate measurements. The technique generalises far beyond colligative properties.

Numerical anchor. The literature value of p^∘_w at 298 K is ≈ 3.17 kPa (∼ 23.8 mmHg). Our computed 3.53 kPa is slightly higher, consistent with the data being given to only 2 – 3 significant figures. The solute molar mass ≈ 23 matches the atomic mass of sodium.

Why this matters. The same two-state technique is used to determine K_a, K_b, equilibrium constants and other unknown ``intercepts'' whenever a parameter appears linearly in an experimentally measured relation. JEE Main 2021 (Sept 1 shift 1) used this exact ``two-vapour-pressure measurements'' setup.

p^∘_w = 3.53 kPa, M_solute ≈ 23 g mol^-1.

Strategic angle. Since the same solvent (water) is used in both experiments, K_f is the same. Two ratios of molality give one ratio of depressions, which transfers the cane-sugar result to glucose in one step.

1. Same mass percentage means the same mass of solute in the same mass of water, so molality is inversely proportional to molar mass: m₂/m₁ = M₁/M₂ = 342/180 = 1.9.
2. Hence depressions are in the same ratio: Δ T_f,2 = Δ T_f,1 × (M₁/M₂) = 2.15 × (342/180) = 2.15 × 1.9 = 4.085 K.
3. Freezing point of glucose solution = 273.15 - 4.085 = 269.07 K.

Numerical cross-check. Using the literature K_f^lit(water) = 1.86 K kg/mol would predict Δ T_f,1 = 1.86 × 0.1538 = 0.286 K, far smaller than the given 2.15 K. So the problem's data implicitly use a different K_f value (effective 13.98), which we extracted from the cane-sugar measurement and reused for glucose –- yielding the self-consistent 269.07 K answer.

Why this matters. For a same-solvent comparison, the colligative property is inversely proportional to the molar mass of the solute (at fixed mass percentage). That ``molar-mass inverse'' relation is the heart of cryoscopy and ebullioscopy. NEET 2020 and JEE Main 2022 (June 25) both included same-solvent freezing-point comparisons.

Freezing point ≈ 269.07 K.

Strategic angle. Two compounds of two elements give two molar masses, which give two linear equations in the unknown atomic masses. Solve simultaneously.

1. Apply M = 1000 K_f w/(Δ T_f · w₁) twice. M_AB2 = 5100/46 = 110.87, M_AB4 = 5100/26 = 196.15.
2. Linear system: a + 2b = 110.87, a + 4b = 196.15.
3. Subtract: 2b = 85.28, so b = 42.64 g mol^-1. Substitute: a = 25.59 g mol^-1.

Numerical cross-check. Plug a = 25.59, b = 42.64 back into the formulas: AB2: a + 2b = 25.59 + 85.28 = 110.87 . AB4: a + 4b = 25.59 + 170.56 = 196.15 . Both molar masses recover the cryoscopic data exactly.

Concept linkage. The atomic mass of B (∼ 42.6) is close to no real element –-this is a synthetic problem, but the technique of solving a linear pair to find atomic masses generalises: e.g. for a Group I + Group VII pair, one cryoscopic measurement plus one b.p. measurement gives both atomic masses.

Why this matters. Combining colligative-property measurements of two related compounds with the same elements is a classic ``mol-by-mol'' linear-algebra trick used to deduce formulas of unknown compounds. JEE Main 2018 (Apr 16) and NEET 2017 each used a two-compound cryoscopy linear-system problem.

a_A ≈ 25.6, a_B ≈ 42.6.

Quick reading. Osmotic pressure scales linearly with concentration at fixed temperature. So we transfer the 36 g/L data point to the new osmotic pressure by simple proportion.

1. ₁ = 4.98 at C₁ = 0.200 M. We want C₂ at ₂ = 1.52 bar.
2. C₂/C₁ = ₂/₁ = 1.52/4.98 = 0.3052.
3. C₂ = 0.200 × 0.3052 = 0.0610 M.

Why this matters. The van't Hoff form π = CRT makes osmosis behave like an ``ideal gas of solute particles'' in the solvent. The same ratio technique determines unknown concentrations from osmotic-pressure measurements in biology and pharmacology. JEE Main 2019 (Apr 9 shift 2) used an osmotic-ratio problem of identical form.

C₂ ≈ 0.061 M.

Structural observation. Three quick filters decide the interaction type for any pair: ``ion present?'' (then ion-dipole wins); ``O-H or N-H present and a lone-pair O/N/F partner?'' (then H-bond wins); ``both polar?'' (then dipole-dipole); else, ``both non-polar?'' (then dispersion).

• Pair (i): non-polar + non-polar ⇒ dispersion.
• Pair (ii): non-polar + non-polar ⇒ dispersion.
• Pair (iii): ionic + polar (water) ⇒ ion-dipole.
• Pair (iv): methanol's O-H + acetone's C=O lone pair ⇒ hydrogen bond.
• Pair (v): polar + polar, no O-H/N-H ⇒ dipole-dipole.

Alternative method: dipole-moment table. A quick reference: μ(H2O) = 1.85 D, μ(CH3OH) = 1.69 D, μ(CH3CN) = 3.92 D, μ(C3H6O) = 2.88 D, μ(CCl4) = 0, μ(I2) = 0, μ(n-hexane) = 0. Two species with μ > 0 but no O-H/N-H interact mainly by dipole-dipole; both with μ = 0 rely on dispersion; an ionic solid plus a polar liquid gives ion-dipole; an O-H/N-H plus a lone-pair acceptor gives H-bonding.

Concept linkage. The four canonical intermolecular forces (dispersion < dipole-dipole < H-bond < ion-dipole) span several orders of magnitude in energy (∼ 1 to ∼ 100 kJ/mol). The dominant force in a pair sets miscibility, boiling-point elevations, solubility class, and ultimately phase behaviour. The same ladder explains why NaCl dissolves in water but not in cyclohexane, and why I2 dissolves in CCl4 but not in water.

Why this matters. The same logic predicts whether two liquids will be miscible: the dominant intermolecular force tells you whether ``like dissolves like'' is satisfied. JEE Main 2024 (Jan 24 shift 1), JEE Main 2023 (Apr 10 shift 2), and NEET 2022 each featured pair-by- pair intermolecular-force MCQs of this exact form.

(i) dispersion, (ii) dispersion, (iii) ion-dipole, (iv) H-bond, (v) dipole-dipole.

Strategic angle. Rank solutes by the strength of their internal cohesive forces relative to dispersion. The stronger the solute-solute attraction (ion-ion ≫ H-bond ≫ dipole-dipole ≫ dispersion), the more energy is needed to disperse it into a non-polar solvent, and the lower its solubility.

• Cyclohexane: dispersion only, matches octane perfectly.
• CH3CN: dipole-dipole, modest mismatch.
• CH3OH: H-bond network, larger mismatch.
• KCl: ion-ion lattice, biggest mismatch.

Alternative ranking by dipole moment. Quick polarity scoreboard: μ(KCl) → ionic lattice (infinite effective polarity), μ(CH3OH) = 1.69 D, μ(CH3CN) = 3.92 D, μ(cyclohexane) = 0. Reading right-to-left (decreasing polarity) gives the increasing-solubility order in n-octane: cyclohexane > CH3CN > CH3OH > KCl. Yes, methanol's dipole moment is smaller than acetonitrile's, but methanol also has an O-H that brings strong H-bonding among methanol molecules; that extra cohesive cost makes methanol less soluble in octane than acetonitrile.

Concept linkage. The general rule is: the more energy required to break the solute-solute attractions, the less soluble the solute in a non-polar solvent. Ion-ion (KCl) costs the most, H-bond (methanol) is next, dipole-dipole (acetonitrile) is third, and pure dispersion (cyclohexane) costs the least.

Why this matters. Polarity rules govern phase-transfer chemistry, extraction protocols, and chromatography. Cyclohexane washes are used to remove non-polar impurities; ionic salts are washed out with water. JEE Main 2022 (June 24 shift 2) and NEET 2021 both probed this ``arrange by solubility in non-polar solvent'' format.

KCl < CH3OH < CH3CN < cyclohexane.

Structural observation. For each candidate, mark (a) presence of an O-H/N-H, (b) size of any hydrocarbon part, (c) ability to ionise. Water solubility then follows from a simple balance of these three.

• Insoluble: toluene (pure hydrocarbon).
• Partially soluble: phenol (O-H but big ring), chloroform (no O-H, weak H-bond donor), pentanol (O-H balanced by long alkyl chain).
• Highly soluble: formic acid (small, two H-bond sites, weak acid), ethylene glycol (two O-H, small carbon skeleton).

Alternative method: ``hydroxyl-to-carbon ratio''. A rough rule of thumb: an alcohol or acid is freely miscible with water if it has at least one O-H per ∼ 4 carbons. Ethylene glycol (2 O-H per 2 C) and formic acid (1 O-H + 1 C=O per 1 C) sit at the high end. Methanol/ethanol/propanol are still miscible (1 O-H per ≤ 3 C). Pentanol (1 O-H per 5 C) is at the threshold and is only partially soluble. Octanol is essentially insoluble.

Concept linkage. Water dissolves what it can H-bond to and ionic species. Pure hydrocarbons (toluene, hexane) are insoluble. Partly-polar species with bulky hydrocarbon tails sit in the middle. Small, multiply hydroxylated species (sugars, glycols, glycerol) are miscible. Carboxylic acids are special because they can also ionise in water, adding ion-dipole on top of H-bonding.

Why this matters. This polarity-vs-size balance explains why methanol and ethanol are miscible with water but octanol is not; why short-chain carboxylic acids are miscible and long-chain ones (stearic acid) are not. JEE Main 2023 (Jan 31 shift 2) and NEET 2018 both used multi-compound water-solubility classification problems.

Toluene: insoluble; phenol, chloroform, pentanol: partially soluble; formic acid, ethylene glycol: highly soluble.

Quick reading. Convert mass to moles, add water mass, divide by density to get volume, then divide moles by litres.

1. n = 92/23 = 4 mol Na+.
2. Total mass =1092 g; volume =1092/1.25 = 873.6 mL.
3. M = 4/0.8736 = 4.58 mol L^-1.

Numerical cross-check. Reverse: at 4.58 M and density 1.25 g/mL, mass of Na+ per L = 4.58 × 23 = 105.3 g; mass of solution per L = 1250 g; mass of water per L = 1250 - 105.3 = 1144.7 g = 1.145 kg; mass of Na+ per kg water = 105.3/1.145 = 91.97 g. Matches the given 92 g/kg .

Why this matters. The same calculation is the way municipal water plants convert their ``hardness'' or ``salinity'' numbers from mg/L of Ca^2+/Mg^2+ into proper molar concentrations. JEE Main 2020 (Sept 5 shift 1) had a brine-salinity molarity problem.

M ≈ 4.58 mol L^-1.

Strategic angle. For a 1:1 salt the solubility-product relation reduces to K_sp = s² and the answer is a single square root.

1. Write K_sp = s² for the 1:1 salt CuS.
2. s = √6 × 10^-16 = √6 × 10^-8 = 2.449 × 10^-8 M.
3. Interpret: this is the solubility, also the maximum molarity of dissolved CuS.

Numerical cross-check. s² = (2.449 × 10^-8)² = 6.0 × 10^-16 . In mass terms: s × M_w(CuS) = 2.449 × 10^-8 × 95.6 = 2.34 × 10^-6 g/L = 2.34 μg/L –- on the order of micrograms per liter, vanishingly insoluble.

Why this matters. For 2:1 salts (e.g. CaF2), the exponent changes: K_sp = 4s³. For 3:1 salts (AlCl3 as sparingly soluble) it would be 27 s⁴. Always count the stoichiometry first. JEE Main 2022 (June 26 shift 1), NEET 2019, and JEE Advanced 2018 each tested K_sp-to-solubility conversions with different salt types.

s ≈ 2.45 × 10^-8 M.

Quick reading. A single division.

1. Total mass = 6.5 + 450 = 456.5 g.
2. Mass % = 6.5/456.5 × 100 = 1.424 %.

Why this matters. Pharmacopoeial preparations are routinely specified as mass percentages; computing them correctly is the first step in any dosage calculation. JEE Main 2019 (Jan 10 shift 2) used this exact aspirin-in-acetonitrile problem.

≈ 1.42 % w/w.

Strategic angle. Compute moles of drug per gram of solvent from the molality, then scale up to deliver the required dose.

1. Moles per 1 kg water at 1.5 × 10^-3 m: that is the molality itself, 1.5 × 10^-3 mol per 1000 g water.
2. Mass of drug per 1000 g water: 311 × 1.5 × 10^-3 = 0.4665 g. So the solution is 0.4665 g drug per 1000 g water.
3. Mass of solution that contains 1.5 mg of drug: 1.5 × 10^-3 / 0.4665 × 1000 ≈ 3.22 g.

Why this matters. Pharmaceutical molalities are typically 10^-3 m or lower; computing the mass of carrier needed to deliver a milligram dose is a basic pharmacology calculation. JEE Main 2017 and NEET 2019 each used drug-dose molality problems.

Required mass ≈ 3.22 g.

Quick reading. Two multiplications.

1. Moles = MV = 0.15 × 0.250 = 0.0375 mol.
2. Mass = n × M_w = 0.0375 × 122 = 4.575 g.

Numerical cross-check. At 0.15 M and V = 0.250 L, expected solute moles = 0.0375. Mass = 0.0375 × 122 = 4.575 g . The volumetric flask needs to be filled to exactly 250 mL after dissolution.

Why this matters. This is the basic recipe for any standard-solution preparation: ``weigh out M V · M_w grams, dissolve and make up to the mark.'' JEE Main 2018 (Apr 16) used a near-identical benzoic-acid recipe problem.

≈ 4.575 g.

Strategic angle. Colligative properties of weak electrolytes depend on i, which in turn depends on K_a. So rank by K_a and the order of Δ T_f follows.

1. K_a: CH3COOH = 1.8 × 10^-5, CCl3COOH ≈ 0.2, CF3COOH ≈ 0.59. Order: CH3COOH ≪ CCl3COOH < CF3COOH.
2. Strength rises because of the -I effect: three electronegative halogens pull electron density from the O-H bond, weaken it, and stabilise the conjugate base.
3. More dissociation → more particles per mole of acid → higher i → higher Δ T_f = i K_f m.

Alternative approach: estimate α. For each acid, α = √K_a/C at the same molality C: α(CH3COOH) = √1.8 × 10^-5/C –- very small. α(CCl3COOH) and α(CF3COOH) –- both near 1 because K_a > 0.1. So i = 1 + α goes from ≈ 1 to ≈ 2 across the series, giving a depression ratio of nearly two between the weakest and the strongest.

Concept linkage. Periodic-trend angle: F (period 2) gives a shorter, stronger C-F bond than C-Cl, but the C-F bond is also more polar because F is the most electronegative element. Net effect: F shows a stronger inductive (-I) effect than Cl in a saturated chain, even though the bond is shorter. The -I effect of halogens also falls off with distance –- effective up to  3 carbons.

Why this matters. This problem is a classic bridge between organic chemistry (inductive effects) and physical chemistry (colligative properties). The same logic explains why HF is weaker than HCl as a halogen acid (different reasoning, but again about the stability of the conjugate base). JEE Advanced 2019, JEE Main 2023 (Jan 30 shift 1), and NEET 2020 all linked -I effect to colligative depression in this exact way.

Δ T_f: CH3COOH < CCl3COOH < CF3COOH (due to increasing -I effect of halogens, larger K_a, larger i).

Strategic angle. Three small computations: molality, α from K_a, then Δ T_f = (1+α) K_f m.

1. Molality: 10/122.5 = 0.0816 mol in 0.250 kg water, so m = 0.3266 mol kg^-1.
2. α = √K_a/C = √1.4× 10^-3/0.3266 = 0.0655.
3. i = 1.0655, Δ T_f = 1.0655 × 1.86 × 0.3266 = 0.647 K.

Numerical cross-check. If we (wrongly) assumed full dissociation i=2: Δ T_f = 2 · 1.86 · 0.3266 = 1.215 K, almost twice our 0.647 K. The small K_a keeps α low (∼ 6.5%) so the depression is only slightly higher than for an undissociated solute (K_f m = 0.607 K). Our answer (0.647 K) sits cleanly between the two limits.

Concept linkage. Chloro-substituted carboxylic acids are stronger than acetic acid by 2-4 orders of magnitude in K_a due to the -I effect of Cl. Compare 2-chlorobutanoic acid here (K_a = 1.4 × 10^-3) with acetic acid (K_a = 1.8 × 10^-5): about 80 times more acidic.

Why this matters. Weak-acid colligative properties are the basis of measuring K_a by cryoscopy. The same data can be inverted to compute K_a if α is known (Q1.33 is exactly this inverse). JEE Main 2022 (July 26 shift 2) had a chloro-acid cryoscopy problem of identical structure.

Δ T_f ≈ 0.65 K.

Quick reading. Three computations: m from masses, i from Δ T_f, K_a from α.

1. m = 0.250/0.500 = 0.500 mol kg^-1 for 19.5 g (M_w = 78).
2. Expected Δ T_f for i=1: 1.86 × 0.500 = 0.93 K. Observed: 1.00 K. So i = 1.00/0.93 = 1.0753.
3. α = i - 1 = 0.0753.
4. K_a = C α²/(1-α) = 0.5 × 0.005667 / 0.9247 = 3.07 × 10^-3.

Numerical cross-check. Reverse: with K_a = 3.07 × 10^-3 and C = 0.500 M, α = √K_a/C = √6.14 × 10^-3 = 0.0784. Close to our 0.0753 (small discrepancy from the 1-α correction). Then Δ T_f = (1+0.0784) · 1.86 · 0.5 = 1.003 K, almost exactly the measured 1.00 K .

Concept linkage. Fluoroacetic acid is more acidic than acetic acid (K_a = 1.8 × 10^-5) by about 170 times. The F atom's strong -I effect stabilises the conjugate base CH2FCOO-. Compare with trifluoroacetic acid (K_a ≈ 0.59), about ∼ 30 000 times more acidic than acetic acid –- three fluorines reinforce the -I effect.

Why this matters. Cryoscopic measurement of K_a predates modern pH meters. It also avoids the small-pH measurement problems for moderately strong acids where the activity correction is large. JEE Advanced 2018 and JEE Main 2023 (Jan 25 shift 2) used the inverse cryoscopy-to-K_a technique.

i ≈ 1.075; K_a ≈ 3.07 × 10^-3.

Quick reading. Use p_s = p^∘ · n_w/(n_w + n_s) directly.

1. n_w/n_tot = 25.0/25.139 = 0.9945.
2. p_s = 17.535 × 0.9945 = 17.438 mmHg.

Alternative form: relative lowering. x_solute = n_s/(n_s+n_w) = 0.1389/25.139 = 0.005525. (p^∘ - p_s)/p^∘ = x_solute = 0.005525, so p_s = 17.535(1 - 0.005525) = 17.535 - 0.097 = 17.438 mmHg .

Numerical cross-check. The lowering is Δ p = 0.097 mmHg (∼ 0.55% of p^∘). For 25/180 = 0.139 mol solute against 450/18 = 25 mol water, x_s ≈ 0.0055 –- consistent with the percentage drop. Reverse: p_s/p^∘ = 17.438/17.535 = 0.9945 = x_w .

Why this matters. A small mole fraction of solute lowers vapour pressure by a small but measurable amount. The lowering is the universal indicator that any non-volatile species is present in the liquid. JEE Main 2021 (Feb 24 shift 2) and NEET 2018 each used glucose-in-water vapour-pressure problems.

p_s ≈ 17.44 mmHg.

Quick reading. One division.

1. Henry's law: x = p/K_H.
2. x = 760/(4.27× 10⁵) = 1.78× 10^-3.

Why this matters. At atmospheric pressure (760 mmHg) only ∼ 0.18 % of the liquid is dissolved methane (by mole fraction). That's enough to be technologically relevant (gas-solvent extraction, natural-gas processing) but small enough that Henry's law (linear regime) applies. JEE Main 2020 (Jan 9 shift 1) used a Henry's-law gas-solubility computation of this type.

x_CH4 ≈ 1.78 × 10^-3.

Strategic angle. The system has one unknown (p^∘_A), which can be reached in two steps: subtract p_B from p_total, then divide by x_A.

1. Compute mole fractions: x_A = 0.1139, x_B = 0.8861.
2. p_B = 500 × 0.8861 = 443.1 torr.
3. p_A = 475 - 443.1 = 31.9 torr.
4. p^∘_A = 31.9/0.1139 = 280 torr.

Numerical cross-check. With p^∘_A = 280.3 and x_A = 0.1139: p_A = 280.3 × 0.1139 = 31.93 torr . p_B = 500 × 0.8861 = 443.05 torr . p_tot = 31.93 + 443.05 = 474.98 ≈ 475 torr .

Why this matters. This is how you measure the vapour pressure of a pure compound when only the mixture's total vapour pressure is convenient to measure (e.g. when the pure liquid is hard to obtain or unstable). JEE Main 2019 (Apr 12 shift 2) used this exact technique on an A–B binary.

p_A^soln ≈ 31.9 torr, p^∘_A ≈ 280 torr.

Picture-first. Plot what Raoult predicts (three straight lines), overlay the experimental data, observe where the data sits relative to the lines. Below the lines means negative deviation.

1. Three ideal straight lines: p_ac^id = 741.8 x, p_chl^id = 632.8 (1-x), p_tot^id = 632.8 + 109 x.
2. Experimental data: at every interior composition, both partial pressures and the total are less than the Raoult prediction.
3. Deviation: negative. The acetone-chloroform H-bond holds more molecules in the liquid than Raoult predicts.

Tabular cross-check (experimental vs ideal totals). At each tabulated composition, sum the experimental partial pressures and compare to the Raoult prediction p_tot^id = 632.8 + 109 x_ac:

• x_ac = 0.118: exp 54.9 + 548.1 = 603.0; ideal = 632.8 + 12.86 = 645.7. Below by 42.7 mmHg.
• x_ac = 0.234: exp 110.1 + 469.4 = 579.5; ideal = 658.3. Below by 78.8.
• x_ac = 0.360: exp 202.4 + 359.7 = 562.1; ideal = 672.0. Below by 109.9 –- biggest gap so far.
• x_ac = 0.508: exp 322.7 + 257.7 = 580.4; ideal = 688.2. Below by 107.8 –- still very negative.
• x_ac = 0.582: exp 599.5; ideal = 696.2. Below by 96.7.
• x_ac = 0.645: exp 615.3; ideal = 703.1. Below by 87.8.
• x_ac = 0.721: exp 641.8; ideal = 711.4. Below by 69.6.

Molecular origin.

• Pure acetone: weak dipole-dipole among C=O groups.
• Pure chloroform: weak dipole-dipole; the H on CHCl3 is slightly acidic because the three Cl atoms withdraw electron density.
• Mixed: the C=O lone pair of acetone forms a directional hydrogen bond with the acidic C-H of chloroform (a textbook example of a C-H⋯O hydrogen bond, ∼ 9 kJ/mol). The new A-B attraction is stronger than the average of pure-liquid forces, so molecules are held in the liquid more tightly than Raoult predicts.

Concept linkage to thermodynamics.

• _mixH < 0 (exothermic) –- mixing the two liquids releases heat (acetone–chloroform mixing has been measured at ∼ -1.5 kJ/mol at x_ac ≈ 0.5).
• _mixV < 0 –- a slight volume contraction.
• Strong negative deviation → minimum in p_tot-vs-x → maximum in T_b-vs-x → maximum-boiling azeotrope. Acetone–chloroform azeotrope: ∼ 64.5% chloroform, b.p. 64.5 ^∘C (above either pure component, but only marginally above chloroform's 61.2 ^∘C).

Why this matters. The negative deviation in this system is chemistry's textbook example of inter-component H-bonding raising the boiling point and lowering the vapour pressure. Industrial separation of acetone-chloroform mixtures requires extractive distillation because the azeotrope cannot be broken by ordinary distillation. JEE Advanced 2017 used acetone–chloroform deviation directly, and JEE Main 2022 (June 26) tested azeotrope/deviation pairs.

Negative deviation from Raoult's law throughout the composition range.

Strategic angle. Compute the two partial pressures, sum to get total, then divide.

1. x_ben = (80/78)/[(80/78)+(100/92)] = 1.0256/2.1126 = 0.4855.
2. p_ben = 0.4855 × 50.71 = 24.62 mmHg.
3. p_tol = 0.5145 × 32.06 = 16.49 mmHg. Total = 41.11 mmHg.
4. y_ben = 24.62/41.11 = 0.599.

Numerical cross-check. y_ben = x_ben · p^∘_ben/p_tot = 0.4855 · 50.71/41.11 = 0.4855 · 1.234 = 0.599 . The enrichment factor α = (p^∘_A/p^∘_B) = 50.71/32.06 = 1.58 is the ratio of pure-liquid vapour pressures and is the relative-volatility constant of the column.

Concept linkage. Vapour-phase enrichment y_A > x_A whenever p^∘_A > p^∘_B. The amount of enrichment is quantified by the relative volatility _A,B = p^∘_A/p^∘_B (here = 1.58). Higher α means easier separation. For close-boiling pairs, α is barely above unity and many distillation stages are required.

Why this matters. Successive vaporisations and condensations (the operations of a fractionating column) build up enrichment of the more volatile component until effectively pure benzene leaves the top of the column. JEE Main 2019 (Apr 9 shift 1) and NEET 2017 each used benzene–toluene vapour-composition problems.

y_ben ≈ 0.599.

Strategic angle. Treat oxygen and nitrogen separately, because Henry's law is linear and applies to each species in the dilute regime.

1. Total pressure in mmHg: 10 × 760 = 7600.
2. Partial pressures: 20 % O2 → 1520 mmHg, 79 % N2 → 6004 mmHg.
3. Mole fractions in water: x_O2 = 1520/(3.30 × 10⁷) = 4.61 × 10^-5; x_N2 = 6004/(6.51 × 10⁷) = 9.22 × 10^-5.

Why this matters. This is why scuba divers breathing air at depth (high P) accumulate large amounts of dissolved N2 that can bubble out on ascent (the bends). The risk scales as the partial pressure of N2. JEE Main 2023 (Jan 24 shift 2) and NEET 2019 each used multi-gas Henry's-law problems of this form.

x_O2 ≈ 4.61 × 10^-5, x_N2 ≈ 9.22 × 10^-5.

Strategic angle. Solve van't Hoff π = iCRT for C, multiply by volume to get moles, multiply by molar mass to get mass.

1. C = π/(iRT) = 0.75/(2.47 × 0.0821 × 300) = 0.0123 M.
2. Moles in 2.5 L: 0.0123 × 2.5 = 0.0308 mol.
3. Mass: 0.0308 × 111 = 3.42 g.

Numerical cross-check. With C = 0.01233 M, iCRT = 2.47 · 0.01233 · 0.0821 · 300 = 0.750 atm .

Concept linkage. The van't Hoff factor i corrects π = CRT for ion dissociation. Theoretical i_max(CaCl2) = 3 (one Ca^2+ + two Cl-); observed 2.47 falls short because of ion-pairing in solution. The same iCRT formula applies to Δ T_b, Δ T_f, and relative lowering of vapour pressure –- only RT changes to K_b, K_f, or 1.

Why this matters. Osmotic-pressure tonicity is the basis of intravenous saline preparations. Knowing i tells you what concentration of salt makes a solution isotonic with blood (π ≈ 7.7 atm). JEE Main 2022 (June 25 shift 2) and NEET 2018 both probed electrolyte π = iCRT problems.

≈ 3.42 g CaCl2.

Strategic angle. Three terms in π = iCRT: write each clearly, then multiply.

1. Moles = 0.025/174 = 1.437 × 10^-4.
2. C = 1.437 × 10^-4/2 = 7.184 × 10^-5 M.
3. i = 3 (complete dissociation, three ions per formula unit).
4. π = 3 × 7.184 × 10^-5 × 0.0821 × 298 = 5.27 × 10^-3 atm.

Numerical cross-check. π = iCRT: 3 · 7.184 × 10^-5 · 0.0821 · 298 = 3 · 1.757 × 10^-3 = 5.27 × 10^-3 atm . In Pa: 5.27 × 10^-3 · 1.013 × 10⁵ ≈ 534 Pa .

Concept linkage. Osmotic pressure is the most sensitive colligative property for very dilute solutions because R has a moderate value but T is large (in Kelvin). A 1-millimolar solution generates π ≈ 0.025 atm ∼ 19 mmHg of osmotic pressure –- easily measurable. By contrast, Δ T_f at the same C would be ∼ 0.002 K, below detection.

Why this matters. Osmotic-pressure measurements at ppm-level concentrations characterise polymers and proteins. The same equation underlies reverse osmosis for water purification and dialysis for kidney patients. JEE Main 2024 (Jan 31 shift 1) and NEET 2020 used electrolyte osmotic-pressure problems analogous to this one.

π ≈ 5.27 × 10^-3 atm (about 534 Pa).