Solutions

Solutions

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your JEE Main attempt.

A solution is a homogeneous mixture of two or more components, classified by physical state into solid, liquid, or gaseous solutions. The solute is the component present in smaller amount; the solvent is the medium of dissolution present in larger amount.

Concentration units you must convert between:

Unit	Formula	Depends on
Molarity (M)	mol solute / L solution	Volume (T-dependent)
Molality (m)	mol solute / kg solvent	Mass (T-independent)
Mole fraction (x)	mol component / total mol	Dimensionless, Σx = 1

Raoult’s Law (volatile component): P_A = p_A° · x_A; for a non-volatile solute, relative lowering of vapour pressure Δp/p° equals x_solute. Colligative properties (ΔT_b, ΔT_f, π) depend only on particle count and are corrected by the van’t Hoff factor (i).

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Raoult’s Law and Vapour Pressure

For an ideal binary liquid solution, Raoult’s Law gives the partial pressure of each component as its pure vapour pressure multiplied by its mole fraction: p_A = p_A° · x_A and p_B = p_B° · x_B. By Dalton’s Law, P_total = p_A°x_A + p_B°x_B.

When A–B intermolecular forces are weaker than A–A and B–B, the solution shows positive deviation (e.g., ethanol + cyclohexane) and forms a minimum-boiling azeotrope. When A–B interactions are stronger, it shows negative deviation (e.g., chloroform + acetone, HNO₃ + H₂O) and forms a maximum-boiling azeotrope. Azeotropes cannot be separated by simple distillation.

Exam tip: JEE frequently frames an assertion-reason around “all positive-deviation solutions have minimum-boiling azeotropes” — true, but the reverse is the easier trap.

Colligative Properties

Four properties depend only on solute particle concentration, not identity:

1. Relative lowering of vapour pressure: (p° − p_s)/p° = x_solute
2. Elevation of boiling point: ΔT_b = K_b · m
3. Depression of freezing point: ΔT_f = K_f · m
4. Osmotic pressure (van’t Hoff equation): π = MRT

van’t Hoff Factor (i)

For electrolytes, i accounts for dissociation (NaCl → i ≈ 2; CaCl₂ → i ≈ 3; K₃[Fe(CN)₆] → i ≈ 4). For associated solutes (benzoic acid in benzene), i < 1.

• Degree of dissociation: α = (i − 1)/(n − 1)
• Degree of association: α = (1 − i)/(1 − 1/n)

Trap alert: Osmotic pressure numericals with an “abnormal molar mass” (e.g., dimerising solute) expect M_observed = i · M_theoretical used inverted from the molar-mass formula — read the question stem twice.

Osmotic Phenomena and Henry’s Law

Isotonic solutions share the same π. Reverse osmosis applies external pressure > π to desalinate seawater. Henry’s Law (p = K_H · x) governs gas solubility; K_H rises with temperature, so gas solubility in water typically drops on heating.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Edge Cases in Concentration Conversions

Molarity and molality interconversion uses solution density (ρ, g mL⁻¹):

$$M = \frac{m \cdot \rho \cdot 1000}{1000 + m \cdot M_{solute}}$$

For dilute aqueous solutions ρ ≈ 1.0 g mL⁻¹, so M ≈ m numerically — but in H₂SO₄, NaOH, or non-aqueous solvents, this approximation fails and JEE questions explicitly provide density. A 1.5 M H₂SO₄ solution (ρ = 1.10 g mL⁻¹) has molality ≈ 1.63 m, a ~9% difference.

Worked Micro-Example (van’t Hoff + Osmotic Pressure)

A 0.5% (w/v) solution of a protein has osmotic pressure 0.02 atm at 27 °C. Estimate molar mass assuming i = 1.

Using π = (w · R · T)/(M · V) with w/V = 0.5 g / 0.1 L = 5 g L⁻¹:

$$M = \frac{wRT}{\pi V} = \frac{5 \times 0.0821 \times 300}{0.02} \approx 6157 \text{ g mol}^{-1}$$

If the protein dimerises at this concentration (i = 0.5), M_observed = i · M_true, so true molar mass ≈ 12,314 g mol⁻¹ — a classic JEE two-step numerical.

Adjacent Topics and Strategy

Solutions connects directly to Electrochemistry (Kohlrausch’s law uses limiting molar conductivity from infinite dilution), Chemical Kinetics (pseudo-first-order reactions in large solvent excess), and Equilibrium (partition coefficient, distribution law). In JEE Main (~3% weightage, typically 1 question), expect a single MCQ or numerical worth +4 marks, solvable in under 90 seconds if you recall the i-correction formula. Skip lengthy azeotropy derivations — focus on numerical fluency with K_b (water = 0.52 K kg mol⁻¹), K_f (water = 1.86 K kg mol⁻¹), and K_H (O₂ in water at 298 K ≈ 34.86 k bar).

Common Mistakes Recap

• Confusing M and m → wrong answer in density-given problems.
• Forgetting i for electrolytes → answer is off by factor of 2 or 3.
• Applying Raoult’s Law with x_solute instead of x_solvent for Δp/p°.

Time strategy: If a question pairs a colligative property with a dissociation constant K_a or K_sp, expect α → i conversion first, then plug into ΔT_f or π.

Practice Prompts

1. Numerical: A solution of 1.2 g of a non-electrolyte in 50 g water freezes at −0.74 °C. Calculate molar mass. (K_f = 1.86; answer ≈ 60.3 g mol⁻¹)
2. Conceptual: Justify why a mixture of 95.6% ethanol + 4.4% water (the ethanol–water azeotrope) cannot yield 100% ethanol by fractional distillation.

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