NCERT Exemplar Class 12 Chemistry Ch 1 Solutions - Download PDF

Strategic angle (definition-driven). If you can recall Raoult's law in symbol form, the answer is one glance: only x appears in the law linking p and composition. Match option, done.

Concept used. Vapour pressure of a solvent above an ideal solution falls in proportion to the mole fraction of solvent: p_A = x_A p^∘_A. For a binary volatile solution p = x_A p^∘_A + (1-x_A)p^∘_B, again pure mole-fraction language. Henry's law for a gas dissolved in a liquid is p = K_H x, also mole fraction. Every classical "p vs composition" law uses x.

1. Write the equation in symbols. For a non-volatile solute the relative lowering is p^∘_A - p_Ap^∘_A = x_B = n_Bn_A + n_B. The right side is dimensionless and counts moles, not mass.
2. Why mass-fraction is the wrong family. Mass-percent depends on molar masses of solute and solvent, which Raoult's law does not contain. Swapping a solute of M_w = 60 for one of M_w = 180 at the same mass-% changes x_B threefold; p tracks x_B, not the mass.
3. Alternative approach (dimensional analysis). p has units of pressure. Raoult's law sets p ∝ p^∘ × (dimensionless), so the concentration variable must be dimensionless. Among the four options only x is a number ratio; ppm is a mass ratio.
4. Limiting case. As x_B → 0 (pure solvent), p → p^∘_A; as x_B → 1 (pure solute, non-volatile), p → 0. Both extremes are recovered only with x.

Option (i): mole fraction.

Quick reading. Two independent levers control dissolution rate: surface area (particle size) and temperature (kinetic energy). Maximise both. Option (iv) does. Done.

Concept used. Dissolution is a surface phenomenon: solvent molecules collide with solute particles at the interface and pull them into solution via ion-dipole or H-bond interactions. The collision rate per unit time scales with surface area A; the fraction of energetic events scales as e^-E_a/RT.

1. Surface-area argument quantified. For a cube of side , S/V = 6/. Crushing a 1 cm cube into 1 mm cubes increases S/V tenfold (6/0.1 = 60 vs 6). Going to 0.1 mm fine powder gives a 100-fold increase. So powdered sugar dissolves ∼ 10 to 100 times faster than crystals.
2. Temperature argument quantified. If E_a ≈ 20 kJ/mol, the rate ratio between hot (T₁ = 323 K, 50 ^∘C) and cold (T₂ = 298 K, 25 ^∘C) is k_hotk_cold = exp[ E_aR(1T₂-1T₁) ] ≈ 2.1. Going to 100 ^∘C boosts this to about 5.
3. Combined factor. Surface-area boost (∼ 30×) times temperature boost (∼ 2-5×) gives roughly 100 times faster dissolution for option (iv) over option (i).

Option (iv): Powdered sugar in hot water.

Picture-first. Picture a beaker of sugar water with extra crystals on the bottom. Watch a single sugar molecule: sometimes it's pulled into solution, sometimes it locks back onto a crystal face. At equilibrium these two visits balance out.

Concept used. A solution is saturated when it sits in dynamic equilibrium with undissolved solid. "Dynamic" = molecules keep moving both ways across the interface; "equilibrium" = the two flows match. The macroscopic concentration is constant precisely because R_d = R_c.

1. Definition recap. At phase equilibrium R_fwd = R_rev. Here fwd = dissolution, rev = crystallisation. Both rates are nonzero; their equality keeps [solute] constant.
2. Why "zero" (iv) is wrong. A radioactive-tracer experiment with ¹⁴C-labelled sugar crystals in unlabelled solution shows the label gradually appears in solution. So dissolution continues - it is not zero.
3. Why (i) and (ii) are wrong. If R_d > R_c, the amount in solution rises and the system is not yet at equilibrium (unsaturated). If R_c > R_d, the solid grows and the solution depletes (supersaturated). Only R_d = R_c matches stable saturated equilibrium.

Option (iii): equal to the rate of crystallisation.

Strategic angle (decision-tree). The three saturation states each have a distinct response to "add more solute":

• unsaturated → added crystal dissolves fully,
• saturated → added crystal stays undissolved,
• supersaturated → added crystal triggers precipitation of A.

Concept used. Supersaturation is a metastable state. The Gibbs free energy of the dissolved excess is higher than that of the solid; the system is kinetically trapped by the absence of a nucleation site. The added crystal supplies that surface, so the excess deposits almost instantly.

1. Quantitative picture. Let c_sat be the saturation concentration at T. A supersaturated solution has c > c_sat. After seeding, the excess (c - c_sat) deposits as solid and c falls to c_sat.
2. How is supersaturation achieved? Slow cooling: dissolve solute in hot solvent, then cool undisturbed. Solubility usually falls with T, but with no nucleus the excess stays in solution. Examples: sodium acetate "hot ice" hand warmers, rock candy from cooling sugar syrup.
3. Why not "concentrated"? "Concentrated" is relative; a 10% NaCl is dilute (saturation ∼ 36 g/100 mL), but a 1% CaSO₄ is already supersaturated. The cue is response to seeding, not absolute amount.

Option (ii): supersaturated.

Strategic angle (Le Chatelier filter). Solubility is a thermodynamic equilibrium. Its variables are dictated by what shifts that equilibrium. For solid-in-liquid: T and natures of solute and solvent matter; P does not.

Concept used. The pressure-dependence of any equilibrium constant is ∂(ln K)∂ P = -Δ VRT. For solid-to-dissolved equilibria in condensed phases, Δ V is 5 cm³/mol - about 10^-5 times the Δ V of gas-liquid equilibria - so a 1 atm pressure change shifts solubility by less than 10^-4 %.

1. Quantify the pressure effect. Take Δ V ≈ 5× 10^-6 m³/mol. ∂(ln K)∂ P = -5× 10^-68.314 × 298 = -2× 10^-9 Pa^-1. For Δ P = 1 atm = 10⁵ Pa, Δ(ln K) ≈ -2× 10^-4, i.e. ∼ 0.02% change in solubility - negligible.
2. Contrast with gas-in-liquid. For O₂ dissolving in water, Δ V ≈ -22.4 L/mol (gas volume lost at STP) - a billion times stronger pressure dependence. This is exactly Henry's law.
3. Why T is not the answer. ∂(ln K)/∂ T = Δ H/RT². Most ionic solids show 20–50% solubility change per 25 ^∘C; (i) is ruled out.
4. Why natures matter. Like dissolves like: dipole- dipole / H-bond / ion-dipole compatibility determines Δ H_sol. So (ii) and (iv) are ruled out.

Option (iii): Pressure.

Strategic angle. The physiological cue "anoxia at altitude" points to Henry's law: p ↓ ⇒ x ↓ ⇒ tissues are oxygen-starved. Don't be fooled by temperature - lower T alone helps oxygen solubility slightly.

Concept used. Gas-in-liquid solubility scales linearly with the gas's partial pressure at fixed T. Body temperature is essentially constant (37 ^∘C) for a living person, so the only physiologically relevant variable with altitude is atmospheric pressure.

1. Henry's law for blood oxygenation. Let K_H be Henry's constant for O₂ in blood plasma at body temperature. Then [O₂]_diss ∝ p_O2.
2. Quantify altitude. Barometric formula: p_atm(h) ≈ p₀ e^-h/H, H ≈ 8 km. At Everest base camp (5400 m), p ≈ 760 × e^-5.4/8 = 387 mmHg ≈ 0.51 p₀, so dissolved O₂ halves.
3. Why body T does not change. Internal T stays at 37 ^∘C even at altitude. So ambient cold is irrelevant to dissolved O₂ in blood. The hint in option (i) is a trap.

Option (ii): low atmospheric pressure.

Structural angle. Look at where H-bonds exist in the pure liquids and whether mixing creates or destroys them. Positive deviation = pure-liquid H-bonds get disrupted without being adequately replaced by mixed-pair H-bonds. Negative deviation = new H-bonds appear in the mixture that did not exist in either pure liquid.

Concept used. The deviation sign is set by Δ H_mix = H_mixture - (H_A + H_B). Δ H_mix > 0 ⇒ endothermic mixing ⇒ weaker A–B ⇒ positive deviation. Δ H_mix < 0 ⇒ exothermic mixing ⇒ stronger A–B ⇒ negative deviation.

1. Methanol-acetone H-bond ledger. Pure methanol: each O–H donates to a neighbour's lone pair - strong H-bond network. Pure acetone: dipole-dipole only, no H-bond donor. Mixed: methanol O–H weakly H-bonds to acetone C=O. Net: methanol–methanol H-bonds break; new methanol–acetone H-bonds form, but weaker. Δ H_mix > 0, Δ V_mix > 0, p_tot > p_Raoult. Forms a minimum-boiling azeotrope at x_MeOH ≈ 0.79, T_b ≈ 55 ^∘C.
2. Chloroform-acetone (negative example). CHCl₃ has an unusually acidic C–H (three EWG Cl's). It H-bonds strongly to acetone C=O: Cl₃C–H ⋯ O=C(CH₃)₂. Pure chloroform and pure acetone lack this H-bond. Mixing creates a new H-bond ⇒ A–B stronger ⇒ negative deviation, max-boiling azeotrope.
3. HNO₃ + water and phenol + aniline. HNO₃ + H₂O forms very strong A–B H-bonds (max-boiling azeotrope at 68 mass-%, ∼ 393.5 K). Phenol + aniline: phenol O–H donates to aniline N lone pair - strong A–B H-bond. Both negative.

Strategic angle (etymological). The word "colligative" encodes the answer: from Latin colligere, "to bind together by counting". These properties care about counts, not identity.

Concept used. Colligative behaviour arises because adding any non-volatile solute reduces the mole fraction of solvent on the liquid surface; the resulting entropy change is the same for any solute at equal x_B. Identity does not enter the entropy of mixing in the dilute limit.

1. Microscopic explanation. Vapour pressure depends on the rate of solvent molecules leaving the surface. If a fraction x_B of surface sites is occupied by solute, the escape rate falls by x_B regardless of solute identity.
2. Mathematical check. The four formulas Δ p/p^∘ = x_B, Δ T_b = K_b m, Δ T_f = K_f m, Π = CRT contain only solvent constants (K_b, K_f, p^∘) and a count of solute (x_B, m, C). No identity term.
3. Identity matters indirectly via i. For electrolytes, the Van't Hoff factor i converts formula units to particles: NaCl i ≈ 2, BaCl₂ i ≈ 3. i is itself a count, not chemical identity.

• Urea (non-electrolyte): Δ T_f = 1.86 × 0.1 = 0.186 K.
• NaCl (i = 2): Δ T_f = 1.86 × 0.1 × 2 = 0.372 K.

Option (ii): number of solute particles in solution.

Strategic angle (count the ions). Boiling-point elevation is colligative, so count particles. Na2SO4 gives three (two Na⁺ + one SO₄^2-); the other three salts give two each.

Concept used. For dilute strong-electrolyte solutions, Δ T_b = i K_b m ≈ i K_b c (since m ≈ c for dilute aqueous solutions with density ≈ 1). K_b is a solvent property (water: 0.512 K·kg/mol). So Δ T_b ∝ i at fixed c.

1. Compute Δ T_b for each.
   • NaOH: Δ T_b = 2 × 0.512 × 1 = 1.02 K.
   • Na₂SO₄: Δ T_b = 3 × 0.512 × 1 = 1.54 K.
   • NH₄NO₃: Δ T_b = 1.02 K.
   • KNO₃: Δ T_b = 1.02 K.
2. Pick the highest. Na₂SO₄ gives 1.54 K, beating the others by 0.52 K.
3. Realistic i caveat. At 1 M, i values fall below the ideal limit due to ion pairing. Literature: NaOH ≈ 1.83, Na₂SO₄ ≈ 2.36, NH₄NO₃ ≈ 1.83, KNO₃ ≈ 1.74. Ranking unchanged.

Strategic angle (dimensional analysis). Start from Δ T_b = K_b m and isolate K_b. Pure unit algebra - no chemistry needed once you remember the defining equation.

Concept used. K_b is a solvent-only constant linking Δ T_b (K) to m (mol/kg solvent). It depends only on the solvent's molar mass and enthalpy of vaporisation via K_b = R M_A T_b²1000 Δ H_vap.

1. Dimensional algebra. [K_b] = [Δ T_b][m] = Kmol/kg = K^-1. Numerical for water: K_b = 0.512 K·kg·mol^-1.
2. Cross-check from derived formula. Plug units into K_b = R M_A T_b² / (1000 Δ H_vap): R in J/(mol·K), M_A in g/mol, T_b² in K², Δ H_vap in J/mol. Numerator units: J·g·K·mol^-2. Divided by J/mol with the 1000 factor converting g → kg gives K·kg·mol^-1.
3. Eliminate wrong options. (ii) inverts molality; that would be m/Δ T_b. (iii) inverts both K and mol/kg. (iv) puts molality multiplicatively instead of dividing.

Option (i): K kg mol^-1 or K (molality)^-1.

Strategic angle. Two equal-concentration solutions differ in Δ T_f only through their Van't Hoff factors. Glucose: 1 particle per formula unit. MgCl2: 3 particles per formula unit. Ratio 3:1.

Concept used. The Van't Hoff factor i is the count of particles per formula unit after dissolution. For a strong 1:n electrolyte M^m+(X^-)_n, i = 1 + n (assuming complete dissociation). For MgCl2: 1 Mg + 2 Cl = 3 particles.

1. Write the depression for both solutes. (Δ T_f)_glu = i_glu K_f m = 1 × 1.86 × 0.01 = 0.0186 K. (Δ T_f)_Mg = i_Mg K_f m = 3 × 1.86 × 0.01 = 0.0558 K.
2. Take the ratio. (Δ T_f)_Mg(Δ T_f)_glu = 0.05580.0186 = 3.0.
3. Realistic correction. At 0.01 M, measured i for MgCl2 is slightly below 3 (∼ 2.7) due to ion pairing. Ratio drops to ∼ 2.7, still much closer to 3 than to 2 or 6. Hence "about three times" is correct.

Picture-first. Picture a mango cell as a balloon with a semi-permeable wall. Inside: dilute cell sap. Outside: dense salt brine. Water flees the dilute side for the dense side (osmosis), the balloon collapses, and the mango shrivels.

Concept used. The osmotic-pressure inequality _brine > _cell drives water out of the mango cells. Reverse osmosis would require an externally applied pressure greater than Π pushing water back into the cell - no such external pressure exists here.

1. Compare osmotic pressures. _brine = i C R T. For a 1 M NaCl pickle brine at 298 K: Π = 2 × 1 × 0.0821 × 298 ≈ 49 atm. Mango cell sap Π ≈ 2–5 atm. So a 50 atm pressure gradient pulls water out.
2. Verify direction. Water always flows from low-Π to high-Π side in normal osmosis. Brine has higher Π, so water leaves the cell.
3. Rule out reverse osmosis. Reverse osmosis requires external pressure P_ext > Π pushing solvent from concentrated to dilute side. In a passive pickle jar there is no external pressure: gravity and atmospheric pressure affect both sides equally.

Option (iv): it loses water due to osmosis.

Strategic angle (direct proportionality). Π = CRT is a linear law. Doubling C doubles Π. Done.

Concept used. Osmotic pressure is the external pressure required to stop osmotic inflow of solvent into a solution. The higher the solute concentration, the greater the chemical-potential deficit of solvent, hence the greater the pressure needed to counter it. The Van't Hoff equation has the same algebraic form as the ideal-gas law.

1. Numerical example. 0.01 M glucose at 298 K: Π = 0.01 × 0.0821 × 298 ≈ 0.245 atm. 0.1 M glucose at 298 K: Π ≈ 2.45 atm. Tenfold concentration ⇒ tenfold Π.
2. Analogy. Π = nRT/V mirrors P = nRT/V for an ideal gas: solute particles in a solvent behave like gas molecules in a vacuum at the same number density.
3. Why (iv) is wrong. Solutions of the same solute at the same T can always be compared via Π/C = RT, a constant. Different solutes or different T would require care; here both are fixed.

Option (i): higher than that of a dilute solution.

Strategic angle. Find the statement that contradicts a solvent-property fact. Statement (i) ignores that K_f varies by solvent. The others are textbook truths.

Concept used. Cryoscopic constant K_f has the molecular formula K_f = R M_A T_f²1000 Δ H_fus, where M_A is the solvent's molar mass, T_f its freezing point, Δ H_fus its enthalpy of fusion. None of these are universal; they depend on the solvent.

1. Show different K_f at common molality. For m = 0.1 sucrose in water: Δ T_f = 1.86 × 0.1 = 0.186 K. In benzene: Δ T_f = 5.12 × 0.1 = 0.512 K. In camphor: Δ T_f = 40 × 0.1 = 4.0 K. Three solvents, three different depressions ⇒ statement (i) is false.
2. Why K_f differs. Substituting into the formula: camphor has T_f ≈ 452 K (very high) and small Δ H_fus, giving K_f ≈ 40. Water has T_f = 273 K and large Δ H_fus, giving K_f = 1.86.
3. Verify (iii) more rigorously. BaCl₂: i ≈ 2.7 at 0.01 M. KCl: i ≈ 1.9. CH₃COOH (weak acid, slight dissociation): i ≈ 1.01. Sucrose: i = 1. So Π ranks BaCl₂ > KCl > CH₃COOH > sucrose ⇒ statement true.

Quick reading. For strong electrolytes, count the ions per formula unit. KCl and NaCl are 1:1 salts (2 ions each). K₂SO₄ is a 2:1 salt (3 ions).

Concept used. The Van't Hoff factor i corrects colligative formulas for the actual particle count. For a fully ionised strong electrolyte M_a X_b, dissociation gives a cations and b anions ⇒ i = a + b. Weak electrolytes give i < (a+b) partly because dissociation is incomplete; ion-pairing at finite C similarly reduces i below the ideal limit.

1. KCl. K⁺ + Cl^-: i = 1 + 1 = 2. At 0.01 M, experimental i ≈ 1.97.
2. NaCl. Na⁺ + Cl^-: i = 1 + 1 = 2. Identical behaviour to KCl (both 1:1).
3. K₂SO₄. 2 K⁺ + SO₄^2-: i = 2 + 1 = 3. At 0.01 M, experimental i ≈ 2.84.
4. Cross-check using Δ T_f. 0.01 m K₂SO₄ in water gives Δ T_f ≈ 0.054 K, vs Δ T_f ≈ 0.0186 K for 0.01 m glucose (i = 1). Ratio ≈ 2.9, matching i = 3.

Strategic angle (direction-check). Reverse osmosis reverses the normal direction of solvent flow. Normal osmosis: dilute → concentrated. Reverse osmosis: concentrated → dilute. Statement (ii) describes normal osmosis, mislabelled as reverse ⇒ false.

Concept used. In normal osmosis, solvent flows from low- solute to high-solute side, driven by the chemical-potential gradient of the solvent. In reverse osmosis, an external pressure P_ext > Π flips the gradient: solvent is pushed from high-solute to low-solute side, producing pure solvent on the low side. This is the operating principle of desalination plants.

1. Direction analysis.
   • Normal osmosis: _{solvent, dilute} > _{solvent, concentrated}, so solvent flows dilute → concentrated. Mathematically, _A = _A^* + RTln x_A; higher x_A ⇒ higher _A.
   • Reverse osmosis: apply P_ext > Π to the concentrated side. This raises _A on the concentrated side above its dilute counterpart, reversing the flow.
2. Statement (ii) audit. It says "lower C ⇒ higher C" - that is normal osmosis direction. Reverse osmosis is the opposite. So mislabelled ⇒ false.
3. Verify other statements quantitatively.
   • (i) Π has units of Pa (or atm); same as atmospheric pressure.
   • (iii) K_f formula explicit in solvent properties (M_A, T_f, Δ H_fus).
   • (iv) Δ p / p^∘ = pressure/pressure = dimensionless.

Option (ii): the false statement.

Strategic angle (Le Chatelier on dissolution). Dissolution of a gas in a liquid is exothermic. Heating drives the equilibrium backward (back to gas), so dissolved fraction x falls. Since p is fixed externally, K_H = p/x rises.

Concept used. Henry's law constant follows a van't Hoff-type temperature dependence d ln K_HdT = Δ H_{vap, gas-liq}RT² > 0 for typical gas-liquid systems where Δ H_sol < 0 (or equivalently Δ H for the reverse process > 0). So K_H rises with T.

1. Empirical data. For O₂ in water:
   • T = 273 K: K_H ≈ 2.5 × 10⁴ bar.
   • T = 298 K: K_H ≈ 4.4 × 10⁴ bar.
   • T = 323 K: K_H ≈ 5.9 × 10⁴ bar.
   K_H nearly doubles between 0 and 50 ^∘C.
2. Why this is the case mechanistically. Heating gives dissolved gas molecules enough kinetic energy to overcome solvation forces and escape into the gas phase. Solubility falls ⇒ K_H = p/x rises.
3. Aquatic-life consequence. This is why fish are stressed in warm summer water - dissolved O₂ drops.

Strategic angle (inverse-relationship reading). Henry's constant sits in the denominator when you solve for solubility: x = p/K_H. Larger denominator ⇒ smaller x. So K_H and solubility are inversely related.

Concept used. Henry's law expresses the equilibrium gas partial pressure above a solution in terms of dissolved mole fraction. K_H is the proportionality constant; its inverse 1/K_H plays the role of "solubility per unit pressure".

1. Show with numbers. At 298 K, for the same partial pressure p = 1 atm:
   • N₂: K_H = 76.48 kbar → x = 1/76480 ≈ 1.3 × 10^-5.
   • O₂: K_H = 44.05 kbar → x = 2.3 × 10^-5.
   • CO₂: K_H = 1.67 kbar → x = 6.0 × 10^-4.
   CO₂ has the lowest K_H and is the most soluble. N₂ has the highest K_H and is least soluble. Inverse relation confirmed.
2. Where does inversion come from? The equilibrium condition equates chemical potentials of dissolved gas and gaseous gas: _gas = _aq. Mathematically this gives μ^∘_gas + RTln(p/p^∘) = μ^∘_aq + RTln x. Rearranging, p = K_H x where K_H = exp[(μ^∘_aq - μ^∘_gas)/RT]. Highly soluble gases (favourable solvation) have small μ^∘_aq and thus small K_H.
3. Eliminate other options. (i) inverts the relationship. (iii) "constant for all gases" - wrong; each gas has its own K_H. (iv) "not related" - wrong; they are inversely related.

Strategic angle (pressure-against-osmotic). Identify which side is concentrated (B), recall the rule "P > Π on the concentrated side reverses flow", and match it to option (ii).

Concept used. The osmotic pressure Π is the precise external pressure that, applied on the concentrated side, exactly balances solvent flow. Anything less than Π leaves the flow going (A) → (B) (normal osmosis, but at reduced rate). Anything greater reverses the flow to (B) → (A).

1. Three-zone analysis of P_{ext, B}:
   • P_ext = 0: full osmosis (A) → (B).
   • 0 < P_ext < Π: reduced osmosis (A) → (B).
   • P_ext = Π: dynamic equilibrium, no net flow.
   • P_ext > Π: reverse osmosis (B) → (A).
   Option (ii) corresponds to the fourth zone.
2. Quantitative check. If the solution in (B) is 2 M NaCl at 25 ^∘C, Π = i CRT = 2 × 2 × 0.0821 × 298 ≈ 98 atm. So a piston pressure of 100 atm on (B) would just barely reverse the flow.
3. Why (iv) is wrong. P on piston (A) compresses the pure water and pushes more water into the SPM toward (B), i.e. same direction as normal osmosis. It does not give reverse osmosis (which requires squeezing the concentrated side, not the dilute side).

Option (ii): water moves (B) → (A) when P_{ext, B} > Π.

Strategic angle (NCERT key vs reality). The NCERT answer key uses the idealised model: NaCl is a strong electrolyte, i = 2 at every concentration. In reality i rises slightly as C falls (more complete dissociation in dilute solutions). The question's phrasing aligns with the idealised model.

Concept used. A strong electrolyte's i-factor approaches the stoichiometric maximum ν = a + b in the infinite-dilution limit. At finite concentration, i < ν due to ion-pairing and incomplete ionisation. For NaCl in water:

• Infinite dilution: i → 2.
• 0.001 M: i ≈ 1.97.
• 0.01 M: i ≈ 1.95.
• 0.1 M: i ≈ 1.87.

1. Idealised treatment. Treat NaCl as fully dissociated at all three C: i = 2 for all ⇒ (iii).
2. Realistic treatment. Account for ion-pairing. Higher C = more pairing = lower i. So i_A < i_B < i_C ⇒ (i).
3. Which to pick? NCERT's answer key gives (iii). This reflects the assumption commonly used in board exams. If asked for the more chemically accurate trend, (i) is correct.

Strategic angle (interaction comparison). Read each information line as setting the inequality between A–B and the average of A–A / B–B. Equal ⇒ ideal. A–B weaker ⇒ positive. A–B stronger ⇒ negative.

Concept used. Raoult's law derives from assuming all binary contacts have the same enthalpy. Bromoethane and chloroethane are both small saturated halides of nearly identical polarity and size, so all three contact types are roughly equal - close to ideal.

1. Solution (A): bromoethane + chloroethane. Both are haloalkanes with similar dipole moments and sizes. Halogen atoms (Br, Cl) have similar electronegativities; C–Br and C–Cl bonds have similar polarisabilities. So A–A ≈ B–B ≈ A–B. Ideal solution; Raoult holds.
2. Solution (B): ethanol + acetone. A–A H-bonds in ethanol are strong (O–H ⋯ O); B–B dipole-dipole in acetone is weaker. A–B (ethanol H-bond to acetone O) is weaker than ethanol's own H-bond network. So A–A > A–B ⇒ positive deviation. The information line stating "A–A or B–B stronger than A–B" matches positive deviation.
3. Solution (C): chloroform + acetone. Cl₃C–H makes a new H-bond to acetone O=C, stronger than the dipole-dipole of either pure liquid. A–B > A–A or B–B ⇒ negative deviation. Forms maximum-boiling azeotrope.
4. Match the options. Only (ii) reflects information (A) correctly. The other three options invert the deviation sign for solutions (B) and (C).

Alternative approach (Δ H_mix sign). Δ H_mix = 0: ideal. > 0: positive deviation (endothermic). < 0: negative deviation (exothermic). Solution (A) matches the Δ H = 0 case.

Strategic angle (Raoult applied directly). Pure water on one side, dissolved NaCl on the other. Raoult's law says any solute lowers the solvent's vapour pressure. So pure water (A) necessarily has the higher v.p.

Concept used. Adding any non-volatile solute reduces the mole fraction of solvent on the liquid surface; vapour pressure scales with that mole fraction. NaCl is non-volatile (it does not escape into vapour at this T) and dissociates into Na⁺ + Cl^-, doubling the solute particle count.

1. Quantify x_B in beaker (B). 2 M NaCl in 400 mL = 0.8 mol NaCl = 1.6 mol of dissolved particles (after complete dissociation). Moles of water in 400 mL ≈ 400/18 = 22.2. So x_B = 1.622.2 + 1.6 = 1.623.8 ≈ 0.067.
2. Compute v.p. of solution (B). p_B = (1 - 0.067) p^∘_water = 0.933 p^∘_water. At 25 ^∘C, p^∘_water = 23.76 mmHg, so p_B ≈ 22.17 mmHg. Pure water vapour pressure p_A = 23.76 mmHg. p_A - p_B = 1.59 mmHg, so the v.p. drop is about 6.7%.
3. Why not (iv) "twice"? A solute does not double or halve the vapour pressure unless x_B ≈ 0.5, i.e. for very concentrated solutions. In dilute solutions the lowering is small.

Strategic angle (be cautious with the NCERT key). The question asks about minimum-boiling azeotropes (low-boiling mixtures), which arise from positive deviation. Positive deviation requires A–B weaker than A–A/B–B. NCERT's answer key lists (i), which conflicts with the chemistry but is the marked answer.

Concept used. An azeotrope is a constant-boiling mixture that cannot be separated by fractional distillation. Two types:

• Minimum-boiling azeotrope: mixture's b.p. is below the lower of the two pure b.p.'s; corresponds to positive Raoult deviation; A–B weaker than A–A/B–B. Example: ethanol-water (95.6 mass-% ethanol, b.p. 78.2 ^∘C).
• Maximum-boiling azeotrope: mixture's b.p. is above the higher of the two pure b.p.'s; corresponds to negative Raoult deviation; A–B stronger than A–A/B–B. Example: HNO₃-water (68 mass-%, b.p. 120 ^∘C).

1. Identify the deviation. Minimum-boiling = lower b.p. than expected = higher v.p. than Raoult ⇒ positive deviation.
2. Translate to A–B interactions. Positive deviation means molecules in the mixture experience less attractive force from their A–B neighbours than they did in the pure phase. So A–B < A–A and A–B < B–B. This is option (iv).
3. Why option (ii) is partly right. Statement (ii) says vapour pressure increases because more molecules escape from the solution. That is exactly the description of positive deviation, and it is consistent with minimum- boiling azeotrope. (ii) is correct in spirit.
4. The NCERT-key issue. NCERT's official answer key marks (i), which actually describes maximum-boiling azeotrope (negative deviation). This is generally accepted as an error in the key; many publishers' solutions select (iv) or (ii). For board exams, choose (i) to align with the key.

Per NCERT key: option (i). Chemically, the correct option is (iv).

Strategic angle (mass conservation of solute). Total moles of NaCl are conserved: M₁ V₁ = moles_{after dilution}. Then divide by the new solvent mass to get molality.

Concept used. Dilution does not change the amount of solute - only the amount of solvent grows. For dilute aqueous solutions, 1 L ≈ 1 kg of water (the small mass of dissolved solute can be neglected if we are only after 3-significant-figure molality).

1. Initial moles. n = M₀ V₀ = 0.02 × 4 = 0.08 mol NaCl.
2. New solvent mass. Original solution had 4 L ≈ 4 kg of water; we add 1 L ≈ 1 kg water. Total ≈ 5 kg. Subtle correction: the original 4 L of 0.02 M NaCl already contains 0.08 mol × 58.44 g/mol = 4.7 g NaCl. If solution density is 1.0 g/mL, the original solution has total mass ≈ 4000 g; subtracting solute gives solvent ≈ 3995 g ≈ 3.995 kg, not exactly 4 kg. After adding 1 L water, solvent = 4.995 kg. The molality is then 0.08 / 4.995 = 0.01601 mol/kg - still rounds to 0.016.
3. Final molality. m = 0.085.00 = 0.016 mol/kg.
4. Why other options are wrong.
   • (i) 0.004 would be (0.02 × 4)/(20) - misuse of formula.
   • (ii) 0.008 would be 0.04/5 - dropping a factor.
   • (iii) 0.012 would be using 6.67 kg solvent - wrong.

Strategic angle (read the hint). The information sentence tells us that A–A H-bonds break upon mixing. That single fact implies: A–A > A–B, hence weaker A–B, hence positive deviation, hence minimum-boiling azeotrope. The full chain runs from a one-line observation to the answer.

Concept used. A minimum-boiling azeotrope is a composition at which the mixture boils at a temperature lower than either pure component. This happens when the mixture has higher vapour pressure than expected from Raoult - i.e. when A–B forces are weaker than the pure-liquid forces.

1. Build the chain.
   • H-bonds break (hint) ⇒ A–A interaction lost on mixing.
   • A–B (methanol O–H ⋯ O=C of acetone) is weaker than A–A (methanol O–H ⋯ O–H of methanol).
   • Net cohesion in the mixture is reduced.
   • Molecules find it easier to escape into vapour.
   • Total vapour pressure > Raoult prediction.
   • Curve has a maximum (peak in p vs x); boiling point has a minimum.
2. Sketch the p vs x curve. Above the Raoult straight line, with a peak somewhere in the middle (composition of the azeotrope).
3. Identify the azeotrope composition. Literature: methanol-acetone forms a minimum-boiling azeotrope at x_methanol ≈ 0.79, T_b ≈ 55 ^∘C (vs methanol b.p. = 64.7 ^∘C and acetone b.p. = 56 ^∘C; the azeotrope is below both, just barely below acetone's).

Alternative approach (Δ H_mix sign). The breaking of H-bonds requires energy input. So Δ H_mix > 0 (endothermic mixing). Endothermic mixing ⇒ positive deviation ⇒ minimum-boiling azeotrope. Same conclusion via energy bookkeeping.

Strategic angle (inverse ordering). Henry's law: solubility ∝ 1/K_H. So order solubility by inverting K_H. Smallest K_H → most soluble.

Concept used. K_H values span six orders of magnitude across these four gases, reflecting vastly different solvation energies. HCHO (formaldehyde) is very hydrophilic - it reacts with water to form a hydrate (H₂C(OH)₂). Ar is noble and weakly solvated.

1. Compute solubilities at p = 1 atm. x = 1K_H (in same pressure units).
   • Ar: x = 1/40.39 = 2.5 × 10^-2 (in whatever units K_H is - relative ordering only).
   • CO2: x = 1/1.67 = 0.60.
   • CH4: x = 1/0.413 = 2.42.
   • HCHO: x = 1/(1.83 × 10^-5) = 5.46 × 10⁴.
   HCHO is highly soluble (essentially miscible) because of its hydrate-formation chemistry; Ar is least soluble.
2. Increasing solubility order. Ar (least) < CO2 < CH4 < HCHO (most). Match with option (iii).
3. Why this order makes chemical sense.
   • Ar: noble gas, only dispersion forces in water, lowest solubility.
   • CO2: weakly polar, partly reacts with water (H2O + CO2 -> H2CO3), moderate solubility.
   • CH4: nonpolar but slightly larger and more polarisable than Ar, slightly more soluble.
   • HCHO: carbonyl group hydrogen-bonds and hydrates with water → "infinite" solubility.

Option (iii): Ar < CO2 < CH4 < HCHO.

Strategic angle (constraint reading). Each option fixes one variable and lists the remaining ones. So we must identify, for each fixed variable, which of a, b, c still matter.

Concept used. Henry's law p = K_H(T) x, where K_H is a function of T and identity of the gas. At fixed T, only p and identity affect solubility. At fixed P, only T and identity matter. Nature of the gas always matters because K_H is gas- specific.

1. Test option (i). "At constant T, (a) and (c) affect solubility." With T fixed, K_H is fixed for each gas, but K_H still differs across gases. So nature of gas (a) matters, and pressure (c) matters. (b) is held constant. So (i) is correct.
2. Test option (ii). "At constant P, (a) and (b) affect solubility." With P fixed, varying T changes K_H for each gas; varying identity changes K_H as well. (c) is held constant. So (ii) is correct.
3. Test option (iii). "(b) and (c) only" - ignores nature of solute, which always matters. Wrong.
4. Test option (iv). "(c) only" - ignores both T and identity. Wrong.

Options (i) and (ii).

Strategic angle (ideal-solution criteria check). The question gives the defining condition of an ideal solution. Then it asks which statements are not true. Eliminate true statements (Δ H = Δ V = 0) and accept false ones (azeotrope, non- ideal).

Concept used. An ideal solution is defined by Raoult-obeying behaviour at all compositions: equal A–A, A–B, B–B interactions ⇒ Δ H_mix = 0 (no net energy released or absorbed) ⇒ Δ V_mix = 0 (no contraction/ expansion) ⇒ no azeotrope. Benzene-toluene is the canonical real-world example: both are nonpolar aromatics of similar size and polarisability.

1. (i) _mix H = 0. For an ideal solution this is by definition true. So (i) is true, not a not-true statement.
2. (ii) _mix V = 0. Same reasoning; ideal solutions have zero volume change. So (ii) is true, not a not-true statement.
3. (iii) Minimum-boiling azeotrope. Ideal solutions do not form azeotropes. Benzene-toluene mixtures distil cleanly across all compositions. So (iii) is false - making (iii) one of the not-true statements.
4. (iv) Not ideal. The condition given (equal intermolecular forces) is exactly the definition of ideal ⇒ the mixture IS ideal. So (iv) is false - making (iv) another not-true statement.

Not-true statements: (iii) and (iv).

Strategic angle (definition-driven). Colligative = counting. Each option either says "counting" (correct) or "depends on nature" (wrong).

Concept used. Relative lowering of vapour pressure: p^∘ - pp^∘ = x_B where x_B is mole fraction of solute. For dilute electrolyte solutions: x_B becomes effective x_B = i · x_formula. Neither expression contains a solute-identity term.

1. Statement (i). For a non-electrolyte (i = 1), relative lowering depends on concentration via x_B. Two non-electrolytes of different identity at the same x_B give the same lowering. So "depends on c, not on nature" - true.
2. Statement (ii). For an electrolyte, i matters. Two electrolytes with the same i at the same c give the same lowering. "Depends on number of particles, not on nature" - true.
3. Statement (iii). Says it depends on nature for non- electrolytes. False.
4. Statement (iv). Says it depends on nature for both types. False.

Correct options: (i) and (ii).

Strategic angle (definitions check). Two equivalent forms of the Van't Hoff factor: observed/calc colligative, or normal/abnormal M_w. Both are correct; their reciprocals are wrong.

Concept used. Colligative properties are proportional to the number of particles. If a solute associates, fewer particles than expected ⇒ lower Δ T_f (or smaller other property) ⇒ apparent molar mass higher than true (because we'd divide a given mass by fewer-than-expected moles). If a solute dissociates, more particles than expected ⇒ higher colligative property ⇒ apparent molar mass lower than true.

1. Derive form (iii) directly. Colligative property ∝ i · c. So observed = i · c times the per-particle factor; calculated (assuming no dissociation) = c times the per-particle factor. Ratio: i = observed/calculated. Statement (iii) is correct.
2. Derive form (i) from (iii). Molar mass is calculated from colligative property using M = w · K / (Δ T · w_solvent). The observed Δ T gives "abnormal" M; the expected Δ T (no dissociation) gives "normal" M. Since Δ T and M are inversely related, i = normal M/abnormal M. Statement (i) is correct.
3. Why (ii) and (iv) are wrong. They are the reciprocals - they would give 1/i, not i. For NaCl, i = 2, so reciprocal would give 0.5, not 2.

Strategic angle (isotonic implies equal colligative). If two solutions have the same Π at the same T, they have the same effective particle concentration iC. All colligative properties depend on iC, so all colligative properties match between isotonic solutions.

Concept used. Isotonicity is defined by equal osmotic pressure. Mathematically: ₁ = ₂ ⇔ i₁ C₁ = i₂ C₂. From this single equation, all four colligative properties (relative lowering of v.p., Δ T_b, Δ T_f, Π) are equal for the two solutions, since each is proportional to iC (or im, which is ≈ iC for dilute aqueous solutions).

1. Map equal Π to equal Δ T_b. ₁ = ₂ ⇒ i₁ C₁ RT = i₂ C₂ RT ⇒ i₁ C₁ = i₂ C₂. For dilute aqueous solutions C ≈ m, so i₁ m₁ = i₂ m₂. Then Δ T_b = K_b i m is identical for both.
2. Density is not forced. Two isotonic solutions can have different densities (e.g. glucose vs sucrose at the same osmolarity). The NCERT key's choice of (ii) "density" is debatable; option (iv) would also be defensible. Use the NCERT-marked answer (ii)+(iii) on the board.
3. Solute is not forced. 0.3 M urea and 0.15 M NaCl are both isotonic with blood, but the solutes differ. (i) is wrong.

Strategic angle (identify non-ideal mixtures). The question boils down to "which mixtures form azeotropes?". Azeotropic composition is where liquid and vapour have the same composition.

Concept used. For an ideal mixture obeying Raoult, the vapour is always enriched in the more volatile component, so y_A ≠ x_A except at the pure-component endpoints. For a non-ideal mixture with strong positive or negative deviation, the p-curve forms an extremum; at that point y = x - an azeotrope.

1. Benzene-toluene. Ideal (textbook example). y_A ≠ x_A at all intermediate compositions. So fractional distillation works cleanly. Not the answer.
2. Hexane-heptane. Both nonpolar alkanes with similar structures and polarisabilities. Ideal. Same as benzene- toluene. Not the answer.
3. Water-HNO₃. HNO₃ + H₂O form strong A–B H-bonds; negative deviation; max-boiling azeotrope at 68 mass-% HNO₃, T_b = 393.5 K. At this composition, y = x ⇒ option (ii) correct.
4. Water-ethanol. Ethanol-water O–H⋯O are weaker in mixture than in pure ethanol or pure water; positive deviation; min-boiling azeotrope at 95.6 mass-% ethanol, T_b = 78.2 ^∘C. At this composition, y = x ⇒ option (iii) correct.

• Ideal: benzene-toluene, hexane-heptane, methanol-ethanol.
• Min-boiling: ethanol-water, methanol-acetone.
• Max-boiling: water-HNO₃, water-HCl, water-H₂SO₄.

Correct options: (ii) and (iii).

Strategic angle (literal definition). Isotonic just means "same Π". Anything that follows from ₁ = ₂ is allowed; constraints on solute/solvent identity are not.

Concept used. The isotonic condition ₁ = ₂ constrains only the effective particle concentration iC, not the chemical identity. So:

• Same solute / same solvent / same C ⇒ isotonic (trivially).
• Different solutes / same solvent / different C ⇒ isotonic possible if i₁ C₁ = i₂ C₂.
• Different solvents are uncommon in textbook problems but theoretically allowed.

1. (ii) Direct definition. Same Π is what "isotonic" means. Correct.
2. (iii) Logical implication. The definition imposes no constraint on identity. So solute and solvent may or may not coincide. Correct.
3. (i) Over-constrains. If solute and solvent must both match, then "isotonic" would just mean "same concentration" - a different concept. Wrong.
4. (iv) Over-constrains differently. Says solute always the same - wrong for the same reason.

Correct options: (ii) and (iii).

Strategic angle (linearity test). Raoult's law makes p_tot a linear function of mole fraction. The graphs that are straight lines correspond to ideal solutions. The curved graphs correspond to positive (concave up) or negative (concave down) deviation.

Concept used. For an ideal mixture, total pressure is a weighted average of the pure-component pressures, with weights x₁ and x₂ that sum to 1. This is mathematically equivalent to a linear interpolation between p^∘₂ (at x₁ = 0) and p^∘₁ (at x₁ = 1).

1. Algebraic check. p = x₁ p^∘₁ + (1-x₁) p^∘₂ = p^∘₂ + (p^∘₁ - p^∘₂) x₁. Linear in x₁, with intercept p^∘₂ and slope p^∘₁ - p^∘₂.
2. Identify ideal curves. A graph is linear iff it plots as a straight line. Curves (i) and (iv) are straight; the other two are curved.
3. Slope sign. If p^∘₁ > p^∘₂ (component 1 more volatile), p rises with x₁. If p^∘₁ < p^∘₂, p falls with x₁. Both up-slopes and down- slopes are valid ideal representations - that's why both (i) and (iv) are correct.

Strategic angle (which one volatile, which non-volatile?). Colligative properties are defined for non-volatile solute + volatile solvent. Identify which option matches that template.

Concept used. Classical colligative formulas assume:

• Solvent contributes to vapour phase (p^∘_solv).
• Solute does not contribute (non-volatile).

1. Test (i) solid in liquid. Glucose (solid, T_b → ∞ practically) in water (volatile). Classical colligative. Correct.
2. Test (ii) non-volatile liquid in volatile liquid. Glycerol (T_b = 290 ^∘C, effectively non-volatile) in water. Same template. Correct.
3. Test (iii) gas in non-volatile liquid. A gas dissolved in a non-volatile liquid is a Henry's-law problem. Vapour phase contains only the gaseous solute. No classical colligative behaviour of the solvent. Wrong.
4. Test (iv) volatile in volatile. Both contribute to vapour. Raoult's law for two volatiles. Vapour pressure is x₁ p^∘₁ + x₂ p^∘₂, not the colligative lowering x_B p^∘_A. So this is not the classical colligative case. Wrong.

Strategic angle (phase-diagram reading). On a binary T-x phase diagram, fractional distillation walks the system along the upper (vapour) and lower (liquid) curves toward the more volatile component. If those curves meet at an interior point (extremum), the trajectory is trapped - that's the azeotrope.

Concept used. An azeotrope is the composition at which the p-x curve has a local maximum (positive deviation, min-boiling azeotrope) or local minimum (negative deviation, max-boiling azeotrope). At an extremum the slope of the p-x curve is zero, which forces y = x. Mathematically: the Gibbs–Duhem relation at the extremum becomes x₁ dp₁ + x₂ dp₂ = 0 with dp/dx = 0, giving y_i = x_i.

1. Pre-azeotrope phase. If pure A boils at T_A and pure B at T_B with T_A < T_B, then vapour above an A–B mixture is initially richer in A. Repeated condensation and re-vapourisation (fractional distillation) progressively enriches the distillate in A.
2. Approach to azeotrope. As distillation proceeds, the remaining liquid (and the distillate) drift toward the azeotropic composition. The trajectory in the phase diagram is bounded by the azeotrope.
3. At the azeotrope. The phase-diagram curves of liquid and vapour touch. Boiling now produces vapour of the same composition as liquid. Both components come over together in the distillate.
4. Confirm with a specific example. Ethanol-water: azeotrope at 95.6% ethanol. Cannot distil to pure ethanol by simple fractional distillation - at 95.6% the vapour is also 95.6% ethanol, so condensing it just gives back 95.6% liquid.

The mixture reached its azeotropic composition.

Strategic angle (volatility classification). The single distinguishing variable is whether the solute itself is volatile. Non- volatile → classical colligative → T_b goes up. Volatile → Raoult-binary → total p goes up at every T → mixture boils at lower T.

Concept used. Boiling point is the temperature at which the liquid's vapour pressure equals atmospheric pressure (typically 1 atm). Adding a non-volatile solute reduces the liquid's p at any T, so you must heat further to reach 1 atm. Adding a volatile solute that is more volatile than the solvent raises total p at any T, so the mixture reaches 1 atm at a lower T.

1. Quantify the NaCl effect. Δ T_b = K_b m i. With K_b = 0.512 K·kg/mol, m ≈ 1 mol/kg, i = 2: Δ T_b = 0.512 × 1 × 2 = 1.02 K. So 1 M NaCl in water boils at ∼ 101 ^∘C.
2. Quantify the methanol effect (qualitatively). Methanol has p^∘_MeOH(65^∘C) ≈ 760 mmHg. At 25 ^∘C, p^∘_MeOH ≈ 132 mmHg vs water's 23.76 mmHg - about 5.5× as volatile. Adding 1 mol methanol (x ≈ 0.018) to 1 L water raises total p by: Δ p = x_MeOH (p^∘_MeOH - p^∘_water) ≈ 0.018 × 108 ≈ 1.9 mmHg. Total p ≈ 25.6 mmHg at 25 ^∘C, vs pure water 23.76 mmHg. At any T, p is higher for the mixture, so T_b is lower. Methanol-water actually forms a minimum-boiling azeotrope (p_b ≈ 64.7 ^∘C at x_MeOH = 0.79).
3. Compare directions. NaCl Δ T_b > 0; methanol Δ T_b < 0 (at this concentration, mixture is past the azeotrope only if methanol is very dilute, but with 1 mol methanol in 1 L water, x = 0.018, no azeotrope reached but T_b is still below 100 ^∘C due to methanol vapour contribution.)

Strategic angle (energy balance). Dissolution is thermo- dynamically favourable when Δ G_sol = Δ H_sol - TΔ S_sol < 0. Entropy of mixing is always positive, so the key is whether Δ H_sol is small or negative. That depends on how well A–B interactions can replace A–A and B–B.

Concept used. The enthalpy of mixing splits into three contributions: Δ H_mix = -E_AA - E_BB + 2 E_AB, where E_AA, E_BB, E_AB are the respective pairwise interaction energies. If E_AB ≈ (E_AA + E_BB)/2, Δ H_mix ≈ 0 and the entropy of mixing drives dissolution. If E_AB ≪ (E_AA + E_BB)/2 (mismatch), Δ H_mix > 0 and dissolution may be unfavourable.

1. Polar-polar case (NaCl in water). Water hydrogen-bonds and dipole-dipole interacts strongly. NaCl's lattice energy is ∼ 787 kJ/mol, but the hydration enthalpy (Δ H_hyd ≈ -784 kJ/mol for Na⁺ + Cl^- combined) almost exactly compensates. So Δ H_sol ≈ +3.9 kJ/mol (slightly endothermic). Entropy of dissociation drives dissolution. Like dissolves like: ionic + polar solvent.
2. Non-polar / non-polar (I₂ in CCl₄). Only weak dispersion forces in both. Mixing replaces dispersion- only with dispersion-only - minimal energy change. Entropy of mixing drives dissolution. Like dissolves like: non-polar + non-polar.
3. Mismatch (oil in water). Water H-bonds are strong and cohesive; oil molecules can't H-bond. Inserting oil forces water to give up H-bonds without gaining replacement ⇒ large Δ H_mix > 0 AND entropy penalty (hydrophobic effect: water orders around the oil to minimise H-bond loss). Both terms unfavourable ⇒ no dissolution.

Strategic angle (mass vs volume invariance). Mass is conserved under heating; volume is not. Any concentration formula whose denominator is a mass or mole-count is T-invariant; any formula with volume in the denominator is T-dependent.

Concept used. Thermal expansion: a liquid's volume rises roughly linearly with T at fixed mass. Water has _V ≈ 2.1 × 10^-4/K at 25 ^∘C, so heating 1 L of water from 25 to 50 ^∘C expands it to ∼ 1.005 L - a 0.5% volume increase, hence a 0.5% drop in any "per L" concentration. Mass and moles in 1 L are unaffected.

1. Quantify the molarity change. For a solution at T₁ vs T₂: M(T₂)M(T₁) = V(T₁)V(T₂) = 11 + _V (T₂ - T₁). With _V = 2.1 × 10^-4/K and Δ T = 25 K, M(T₂)/M(T₁) ≈ 1/(1 + 0.00525) ≈ 0.9948. So molarity drops by about 0.5% per 25 K rise.
2. Verify molality T-invariance. Both n_B (mole count) and the solvent mass in kg are T-independent. So m is T-independent.
3. Why this matters in calculations. Spectroscopy and electrochemistry typically work at fixed T where molarity is well defined. Colligative calculations (which often involve T changes, e.g. Δ T_b) prefer molality because it doesn't drift with the changing temperature.

Strategic angle (interpretation in three sentences). K_H tells you how much pressure is required to dissolve a given mole fraction of gas. High K_H ⇒ need lots of pressure to dissolve a little gas (insoluble gas). Low K_H ⇒ small pressure dissolves lots (soluble gas).

Concept used. The numerical value of K_H encodes the gas's affinity for the solvent at a specific T. It is set by the gas's enthalpy of solvation and the chemistry of its interaction with solvent molecules. Polar/reactive gases (HCl, CO₂, HCHO) have low K_H in water; noble/non-polar gases (Ar, He, N₂) have high K_H.

1. Solubility prediction. If K_H for O₂ in water is 44 kbar at 25 ^∘C, then at p = 0.21 atm (air), x = p/K_H = 0.21 atm/(44000 atm) = 4.8 × 10^-6. Translates to ∼ 8 mg/L dissolved oxygen.
2. Temperature dependence. The van't Hoff form dln K_H/dT = -Δ H_sol/RT² (with Δ H_sol < 0 for typical gases) gives K_H rising with T. Useful for predicting how dissolved gas changes with seasonal T.
3. Compound comparison. For a given solvent, the ratio of K_H values is the inverse ratio of solubilities. E.g. K_H(N₂) / K_H(O₂) = 76/44 ≈ 1.73, so O₂ is ∼ 1.73 × more soluble in water than N₂ at the same p.

Strategic angle (Le Chatelier on Henry's law). Gas-in-water dissolution is exothermic (Δ H_sol < 0). Le Chatelier ⇒ heating shifts the equilibrium back toward gas ⇒ less dissolved gas ⇒ aquatic respiration suffers in warm water.

Concept used. Aquatic respiration depends on dissolved O₂. The dissolved O₂ concentration is set by Henry's law at the air-water interface. Atmospheric p_O2 is essentially constant; the variable is K_H(T), which rises with T. So warm water carries less O₂.

1. Quantify. At 5 ^∘C, [O₂]_aq ≈ 12 mg/L. At 25 ^∘C, ≈ 8 mg/L. At 35 ^∘C, ≈ 7 mg/L. That is a 40% drop between cool stream water and warm summer pond.
2. Biological threshold. Most salmonid fish need [O₂] > 6 mg/L to thrive; survival becomes difficult below ∼ 3 mg/L. Warm water in summer pushes oxygen levels below threshold, causing fish kills.
3. Mechanism (molecular). At higher T, dissolved O₂ molecules have more kinetic energy and can escape back into the atmosphere; meanwhile fewer escape from the atmosphere into water. Net solubility falls.

Cold water has higher dissolved O₂ (lower K_H); aquatic species respire more easily.

Strategic angle (one law, three applications). All three phenomena trace to the same equation x = p/K_H. Bends: high p → low p rapidly. Altitude: low p from start. Soda: very high p in bottle → atmospheric p on opening. In each case, the fall in p above the solution forces excess gas out of solution.

Concept used. Henry's law equilibrium requires p = K_H x. If p drops below K_H x (i.e. solution is super- saturated for the new p), gas escapes as bubbles. The rate of escape depends on diffusion / nucleation kinetics.

1. Bends quantitatively. A diver at 40 m depth experiences p_total ≈ 5 atm; p_N2 ≈ 4 atm. Blood holds ∼ 4 × the surface dissolved N₂ (∼ 7 mL N₂/L blood vs 1.5 mL/L at sea level). On instant ascent, dissolved N₂ must equilibrate to ∼ 1.5 mL/L, releasing ∼ 5.5 mL of N₂ per L of blood. Without controlled ascent, this bubbles out in the small blood vessels, causing pain and tissue damage.
2. Altitude quantitatively. At Everest base camp (5400 m), p_O2 = 0.21 × 387 ≈ 81 mmHg vs 160 mmHg at sea level - a 50% drop. By Henry's law, the dissolved O₂ in blood plasma drops by the same factor (the hemoglobin-bound O₂ also falls, more sharply due to the steep oxygen-saturation curve). Result: dizziness, breathlessness, headache.
3. Soda fizz quantitatively. A sealed soda bottle has p_CO2 above the liquid ≈ 2.5 atm, dissolved [CO₂] ≈ 0.08 M. On opening, p_CO2 drops to ∼ 0.0004 atm (atmosphere is 400 ppm CO₂). Equilibrium [CO₂] at the new p: ∼ 10^-5 M. So nearly all the dissolved CO₂ must escape - that is the audible fizz and visible bubbling.

Strategic angle (surface-rate / Raoult). Vapour pressure reflects the dynamic equilibrium between molecules leaving and entering the surface. Adding non-volatile solute reduces the number of surface sites available to escape; entry rate from vapour is unchanged. Net escape rate drops; equilibrium p drops.

Concept used. Vapour pressure at equilibrium is set by the rates: r_escape ∝ x_solvent^surface and r_return ∝ p. Equilibrium: r_escape = r_return, giving p ∝ x_solvent. This is Raoult's law, with proportionality constant p^∘_solvent.

1. Calculate the lowering quantitatively. A 10% glucose solution: 100 g glucose (M_w = 180) in 1 L water ⇒ n_glu = 0.556 mol, n_water = 1000/18 = 55.56 mol. Mole fraction of water: x_water = 55.56/(55.56 + 0.556) = 0.990. So vapour pressure drops to 99% of pure water's p.
2. Validate with Raoult. At 25 ^∘C: p_soln = 0.990 × 23.76 = 23.52 mmHg vs pure water 23.76 mmHg. A 0.24 mmHg drop, measurable on a manometer.
3. Microscopic picture. The number of surface water molecules is reduced by ∼ 1% for 10% glucose; this reduces escape rate by the same factor. New equilibrium p is 99% of pure water's p.

Non-volatile glucose occupies surface sites; fewer water molecules escape; p drops.

Strategic angle (colligative + chain reaction). The salt itself doesn't melt the snow directly. It dissolves in the small amount of liquid water always present at the ice surface (even at -10 ^∘C, an "interfacial liquid layer" exists). The resulting brine, with a lower freezing point, melts adjacent ice, which dilutes the brine but adds more water, which dissolves more salt - chain reaction continues until either all salt is dissolved or the brine concentration falls below the saturation that gives adequate Δ T_f for ambient T.

Concept used. The freezing-point depression formula Δ T_f = iK_f m predicts that a saturated NaCl solution (about 5.4 M, i.e. 25 mass-%) freezes near -21 ^∘C - the eutectic point. Above this T, NaCl can keep highways clear.

1. Compute brine Δ T_f. For 1 mol NaCl per kg water (with i = 1.9 at high concentration due to ion pairing): Δ T_f = 1.86 × 1 × 1.9 ≈ 3.5 K. Concentrated brine (5 m): Δ T_f ≈ 18.6 K, freezing point ≈ -19 ^∘C.
2. Phase-diagram view. On the H₂O–NaCl phase diagram, the eutectic at T_e = -21.1 ^∘C, x_NaCl = 23 mass-% is the lowest temperature at which a NaCl-water mixture can remain liquid. Below the eutectic, even brine freezes.
3. CaCl₂ for colder conditions. CaCl₂ dissociates into 3 ions (i = 3), giving Δ T_f about 50% higher per mol/kg than NaCl. Saturated CaCl₂ brine remains liquid to ∼ -50 ^∘C. In Arctic conditions CaCl₂ is used instead of NaCl.

Salt lowers water's freezing point via colligative depression Δ T_f = iK_f m.

Strategic angle (definition + examples + mechanism). A semipermeable membrane is defined by what it allows through (small solvent) and what it blocks (larger solute). The mechanism is size-selectivity at the pore scale.

Concept used. Membranes are characterised by their pore size distribution. A SPM has pores around 0.5–2 nm - large enough for a water molecule (kinetic diameter ∼ 0.27 nm) but small enough to block hydrated ions (Na⁺: 0.36 nm hydrated; Cl^-: 0.32 nm; sugars > 0.7 nm). The selectivity also depends on membrane chemistry: hydrophilic membranes preferentially solvate water; hydrophobic ones block water.

1. Categories of SPM.
   • Natural: cell membranes (phospholipid bilayer, ∼ 5 nm thick), parchment paper, animal bladders, plant cell walls.
   • Synthetic: cellulose acetate (used in RO desalination), polyamide (newer-generation RO), polysulfone (ultrafiltration), polyimide.
2. Physical mechanism. Pores act like molecular sieves. Solvent molecules (water: ∼ 0.27 nm) pass through ∼ 0.5–2 nm pores; solute molecules either too big to fit or kinetically excluded by the membrane chemistry stay behind.
3. Role in colligative chemistry. The SPM is what enables osmosis to occur in the first place. Without selectivity, both solute and solvent would diffuse and equilibrate without any pressure build-up.

Strategic angle (classical and modern materials). The NCERT answer is cellulose acetate. In practice, modern RO plants use polyamide (PA) thin-film composite membranes, which have higher salt rejection and higher water flux.

Concept used. A RO membrane must satisfy four design criteria simultaneously: (1) selectively pass water, (2) reject dissolved salts, (3) withstand 50–80 atm operating pressure, (4) resist chemical degradation and fouling. Material chemistry must balance hydrophilicity (water transport) with mechanical durability.

1. Cellulose acetate (CA). Acetylation of cellulose hydroxyl groups; the resulting polymer is hydrophilic (water passes through) but compact enough to reject salts. Standard industrial RO membrane until ∼ 1980. CA performance: salt rejection ∼ 95%, flux ∼ 10 L/m²/hr at 30 atm.
2. Polyamide (PA) thin-film composite. Developed in 1972 (Cadotte, Loeb). A thin (∼ 0.2 μm) PA selective layer on a porous polysulfone support. Salt rejection > 99.5%, flux ∼ 30 L/m²/hr at 50 atm. Now the industry standard.
3. Why NCERT mentions cellulose acetate. NCERT Class 12 Chemistry textbook (and Exemplar) cite cellulose acetate as the canonical RO membrane, reflecting the original Loeb-Sourirajan (1959) cellulose-acetate RO membrane patent. The Exemplar answer key reads "cellulose acetate".

Cellulose acetate (NCERT standard answer).

Strategic angle (defining-property pairing). Each Column I term has a single, unambiguous definition. Cross-match by reading each Column II text and asking "which solution is this defining?"

Concept used. The six solution-types in this question span the basic taxonomy of mixtures: by saturation (saturated), by component count (binary), by phase (solid solution), and by osmotic relationship (iso-, hypo-, hypertonic).

1. (i) Saturated. Definition: at equilibrium with undissolved solute, contains maximum dissolvable amount. Column II (d) is the verbatim definition. Pair (i)→(d).
2. (ii) Binary. "Bi-" = two. A solution with two components. Column II (c) matches directly. Pair (ii)→(c).
3. (iii) Isotonic. "Iso-" = same. Same osmotic pressure as another solution. Column II (a) matches. Pair (iii)→(a).
4. (iv) Hypotonic. "Hypo-" = under. Lower osmotic pressure than another. Column II (b) matches. Pair (iv)→(b).
5. (v) Solid solution. A solution in the solid phase (e.g. brass, bronze, steel). Column II (f) matches. Pair (v)→(f).
6. (vi) Hypertonic. "Hyper-" = over. Higher osmotic pressure than another. Column II (e) matches. Pair (vi)→(e).

Strategic angle (state-pair classification). Identify solute state and solvent state, then match. Trivial once both states are identified.

Concept used. Solutions exist in nine possible combinations of (solute state) × (solvent state). The most common combinations are solid-in-liquid (most aqueous chemistry), liquid-in- liquid (alcoholic beverages, organic solvent mixtures), and gas-in- liquid (carbonated drinks, dissolved O₂ in water). Less common but important: solid-in-solid (alloys), gas-in-solid (hydrogen storage), liquid-in-solid (dental amalgams).

1. Soda water. CO₂ (gas) is the solute; water (liquid) is the solvent. State pair: gas-in-liquid ⇒ (e).
2. Sugar solution. Sucrose (solid) is the solute; water is the solvent. State pair: solid-in-liquid ⇒ (c).
3. German silver. An alloy of Cu/Zn/Ni - all solid. State pair: solid-in-solid ⇒ (d).
4. Air. N₂, O₂, Ar, CO₂ etc. all in gaseous phase. State pair: gas-in-gas ⇒ (b).
5. H₂ in Pd. Palladium absorbs up to 900× its own volume of H₂ gas, the H₂ becoming dissolved into the Pd lattice. State pair: gas-in-solid ⇒ (a). This is the basis of laboratory H₂ purification and proposed hydrogen-storage technology.

Strategic angle (formula recall). Pair each law with its canonical formula. No interpretation needed - straight memory check.

Concept used. The five formulas of solution chemistry. Each defines a quantitative relationship between a concentration variable and an observable property. Together they form the toolkit of Class 12 solution chemistry.

1. Raoult's law. For a binary mixture of volatile liquids: p = x₁ p^∘₁ + x₂ p^∘₂. Pair (i)→(c). Derivation: each component contributes proportionally to its mole fraction.
2. Henry's law. For a gas dissolved in a liquid: p = K_H x where K_H is gas-specific. Pair (ii)→(e). Derivation: similar to Raoult but with a different proportionality constant.
3. Boiling-point elevation. Δ T_b = K_b m for a non-volatile solute. Pair (iii)→(d). Derived from Clausius-Clapeyron + Raoult's law.
4. Freezing-point depression. Δ T_f = K_f m. Pair (iv)→(a). Derived from solid-liquid equilibrium + Raoult.
5. Osmotic pressure (Van't Hoff equation). Π = CRT for dilute solutions. Pair (v)→(b). Same algebraic form as PV = nRT.

Strategic angle (formula recall). Each unit has a precise numerator/denominator. Match them.

Concept used. The five concentration units differ in their numerator (mass vs volume vs moles) and denominator (total mass vs total volume vs total moles vs solvent mass vs solution volume). Picking the right unit for a problem matters: molarity for titration, molality for colligative work, mole fraction for vapour pressure.

1. Mass-%. = (mass solute / mass solution) × 100. Denominator: total solution mass. Pair (i)→(d).
2. Volume-%. = (volume solute / volume solution) × 100. Used for liquid-liquid (e.g. ethanol-water ABV). Pair (ii)→(c).
3. Mole fraction. = (moles component) / (total moles). Dimensionless. Pair (iii)→(b).
4. Molality. = (moles solute) / (kg solvent). The "kg solvent" makes this T-independent. Pair (iv)→(e).
5. Molarity. = (moles solute) / (L solution). The "L solution" makes this T-dependent. Pair (v)→(a).

Strategic angle (causal link). Molarity has V in the denominator. V depends on T via thermal expansion. Therefore M depends on T through V. The reason is exactly the mechanism by which M acquires its T-dependence.

Concept used. Thermal-expansion coefficient _V = (1/V)(∂ V/∂ T)_P. For water at 25 ^∘C, _V ≈ 2.1 × 10^-4/K. A 25 ^∘C rise expands solution by ∼ 0.5%, dropping molarity by the same fraction (since n is fixed).

1. Quantify the effect. For 1.00 M aqueous solution: heating 25 → 50 ^∘C expands volume by ∼ 0.5%, so M drops to ∼ 0.995 M.
2. Causality. The temperature affects V first, which then affects M = n/V. So the reason (V changes with T) is the explanation for the assertion (M changes with T).
3. Contrast with molality. Molality m = n/m_solvent does not change with T because mass (kg) doesn't change. That's why colligative-property calculations prefer molality.