Solutions and Colligative Properties

Physical Chemistry Weightage: 2-3 Questions (8-12 Marks) JEE Unit 5

“Welcome back! These notes assume you have already read the NCERT chapter and know what a solution is, how Raoult's and Henry's laws are stated and what the four colligative properties are, so they skip the textbook story and keep what JEE Main actually asks. Solutions gives two or three questions per paper and almost all are short calculations: converting concentration units, vapour pressure of a mixture, a van't Hoff factor with a degree of dissociation, or the osmotic pressure of a mixed solution. Learn the unit conversions, the vapour-composition relation, the i-factor table and the worked patterns below and this becomes one of the fastest scoring chapters.”

— SCORECHEM ACADEMIC TEAM

1. Concentration Terms and Conversions

Fig. 1: Concentration terms. Only molarity and normality change with temperature.

Molarity from mass percent: M = 10 × d × (% w/w) / M_w, with d in g/mL.
Molality from molarity: m = 1000 M / (1000 d − M M_w). For a dilute aqueous solution (d about 1) m is almost equal to M.
Mole fraction from molality (water): x_solute = m / (m + 55.5). Worked: 4.44 m urea gives x = 4.44/(4.44 + 55.5) = 0.074, i.e. 74 × 10⁻³.
Mass percent from mole ratio: 1 mol ethanol in 9 mol water has mass 46 g in 46 + 162 = 208 g, so mass % = 46/208 × 100 = 22.
Mixing two solutions of the same solute: add moles and add masses (or volumes); never average the percentages. Worked: 100 g of 98% H₂SO₄ plus 100 g of 49% H₂SO₄ holds 98 + 49 = 147 g acid (1.5 mol) and 200 − 147 = 53 g water (2.94 mol), so x(H₂SO₄) = 1.5/4.44 = 0.337.
Mixing solutions of different salts: find the moles of each ion and divide by the total volume. 20 mL of 0.5 M Na₂SO₄ + 50 mL of 0.2 M H₂SO₄ + 30 mL of 0.4 M Al₂(SO₄)₃ gives 100 mL with [Na⁺] = 0.2 M, [H⁺] = 0.2 M, [Al³⁺] = 0.24 M and [SO₄²⁻] = (10 + 10 + 36)/100 = 0.56 M.

⚠️ JEE Trap: Volume is not additive. Mixing two liquids does not always give V₁ + V₂; a non-ideal pair contracts or expands. Use mass and density when the question gives them. Also note that mass % uses the mass of the solution, not the solvent: 10% urea is 10 g in 90 g of water.

2. Henry's Law and Gas Solubility

Law: p = K_H x, where x is the mole fraction of the dissolved gas and K_H has pressure units. Larger K_H means lower solubility at the same pressure. K_H rises with temperature, so gases are less soluble in hot water.
Each gas in a mixture obeys the law with its own partial pressure. Worked: air at 10 atm is 80% N₂, so p(N₂) = 8 atm = 6080 mmHg. With K_H = 6.5 × 10⁷ mmHg, x(N₂) = 6080/(6.5 × 10⁷) = 9.35 × 10⁻⁵.
Graph shape: p against x is a straight line through the origin with slope K_H; so is log p against log x with slope 1 and intercept log K_H. The line for the gas with the larger K_H is steeper.
K_H depends on the gas, the solvent and the temperature. It is constant with concentration only while the solution is ideally dilute.
Limits: the law fails at very high pressure, at very low temperature, and for gases that react with or dissociate in the solvent (NH₃, HCl in water).
Uses: carbonated drinks are sealed at high pressure; scuba divers breathe He–O₂–N₂ mixtures to avoid the bends; low O₂ pressure at altitude causes anoxia.

⚠️ JEE Trap: K_H is not a universal constant. Statement: "K_H is the same for a solute in different solvents" is false, while "K_H is constant with concentration in the ideally dilute range" is true. Also convert atm to mmHg (or the other way) before dividing.

3. Raoult's Law and Vapour Composition

Fig. 2: Total vapour pressure against composition at constant T. Dashed line = Raoult's law.

Partial and total pressure: p_A = x_AP_A°, p_B = x_BP_B°, P = p_A + p_B = P_B° + (P_A° − P_B°)x_A. A plot of P against x_A is a straight line.
Vapour composition: y_A = p_A/P, and 1/P = y_A/P_A° + y_B/P_B°. The vapour is richer in the more volatile component (the one with larger P°).
Worked pattern 1 (vapour composition): benzene P° = 80 torr, methylbenzene P° = 24 torr, equimolar liquid. p = 40 and 12, so y(methylbenzene) = 12/52 = 0.23.
Worked pattern 2 (find x from y): P°_A = 55, P°_B = 15, y_A = 0.8. Then 55x = 0.8[55x + 15(1 − x)], so 55x = 12 + 32x and x_A = 12/23 = 0.5217.
Worked pattern 3 (find P° from two mixtures): 3 mol A + 1 mol B gives 500 mm Hg and adding 1 mol A raises it by 20 mm Hg. Then 3P_A° + P_B° = 2000 and 4P_A° + P_B° = 2600, so P_A° = 600 and P_B° = 200 mm Hg. Convert each mixture to a linear equation in the unknown pressures.
Clausius–Clapeyron: ln(P₂/P₁) = (ΔH_vap/R)(1/T₁ − 1/T₂). Vapour pressure depends only on the liquid and T, not on the amount of liquid or the container shape.

⚠️ JEE Trap: Liquid versus vapour mole fraction. Raoult's law uses x (liquid). Using the vapour fraction to compute a partial pressure, or forgetting that x_A + x_B = 1 in the liquid, breaks every mixture problem. Also remember that when the acetone–CS₂ pair is plotted, both partial-pressure curves lie ABOVE the ideal line (positive deviation).

4. Non-Ideal Solutions, Azeotropes, Immiscible Liquids

Fig. 3: Boiling-point (T-x) diagrams. Dark curve: vapour composition; coloured curve: liquid composition (the two touch at the azeotrope).

Property	Positive deviation	Negative deviation
A–B forces	weaker than A–A, B–B	stronger than A–A, B–B
V.P. of mixture	higher than Raoult	lower than Raoult
ΔH_mix, ΔV_mix	positive (cools, expands)	negative (warms, contracts)
Azeotrope	minimum boiling	maximum boiling
Examples	ethanol + water, ethanol + acetone, CCl₄ + CHCl₃	acetone + chloroform, HNO₃ + water, HCl + water

Azeotrope: a constant-boiling mixture with x_liquid = y_vapour. It cannot be separated by fractional distillation. Ethanol–water is 95.6% ethanol; HNO₃–water is 68% HNO₃.
Ideal solutions (benzene + toluene, n-hexane + n-heptane): ΔH_mix = 0 and ΔV_mix = 0. In every solution, ideal or not, ΔS_mix > 0 and ΔG_mix < 0 for spontaneous mixing.
Immiscible liquids: each exerts its own vapour pressure, so P = P_A° + P_B° and the mixture boils below either pure liquid. The vapour mole ratio n_A/n_B = P_A°/P_B°, so the mass ratio W_A/W_B = P_A°M_A/(P_B°M_B). This is the basis of steam distillation of aniline and nitrobenzene.

⚠️ JEE Trap: Deviation and azeotrope type. Positive deviation gives a MINIMUM-boiling azeotrope (highest vapour pressure, lowest boiling point); negative deviation gives a MAXIMUM-boiling one. Students often flip the pairing because "positive" sounds like "high boiling".

5. The Four Colligative Properties

Fig. 4: The four colligative properties. All depend on the number of particles, not their identity.

Dilute-solution shortcut: for a very dilute solution, (P° − P_s)/P_s = i m M_solvent/1000, so RLVP is proportional to molality.
Worked pattern 4 (RLVP from K_b): 1.5 g X in 150 g of solvent (M = 300) raises the boiling point by 0.5 K with K_b = 5. m = 0.5/5 = 0.1, so n_X = 0.015 mol and n_solvent = 0.5 mol. RLVP = 0.015/0.515 = 0.029, i.e. 3 × 10⁻².
Worked pattern 5 (two compounds of the same elements): P and Q form PQ and PQ₂. 1 g of PQ in 50 g of solvent gives ΔT_b = 1.176 K and 1 g of PQ₂ gives 0.689 K, with K_b = 5. M = K_bw/(ΔT_b × 0.05): M(PQ) = 85 and M(PQ₂) = 145, so Q = 60 and P = 25.
Freezing-point work: only pure solvent freezes out, so the remaining solution gets more concentrated and the freezing point keeps falling. K_f = RT_f°²M/(1000 ΔH_fus). Worked: 2.7 kg each of water and acetic acid; water (150 mol) is the solvent, so m = (2700/60)/2.7 = 16.7 and ΔT_f = 1.86 × 16.7 = 31, i.e. −31 °C.
Vapour pressure from boiling elevation: ΔT_b = 2 K with K_b = 0.52 gives m = 3.85, and P = P°(1 − m × 18/1000) = 760(1 − 0.0692) = 707 mm Hg.
Comparing solutions: compute i × concentration for each. Worked: 2.2 g glucose in 125 mL (0.098), 1.9 g CaCl₂ in 250 mL (0.0685 × 3 = 0.205), 9 g urea in 500 mL (0.30) and 20.5 g Al₂(SO₄)₃ in 750 mL (0.08 × 5 = 0.40). Boiling point order: glucose < CaCl₂ < urea < Al₂(SO₄)₃.

⚠️ JEE Trap: The solvent choice and the units of ΔT. The solvent is the component with more moles, not the one you read first. And ΔT is a difference, so in K it equals the same number in °C. Report a freezing point as T_f° − ΔT_f (a negative number for water).

6. Van't Hoff Factor: Dissociation and Association

Fig. 5: van't Hoff factor. i = 1 + (n - 1)α for dissociation; i = 1 + (1/n - 1)β for association.

Dissociation: i = 1 + (n − 1)α, so α = (i − 1)/(n − 1). Association into an n-mer: i = 1 + (1/n − 1)β. Full dimerisation gives i = 0.5.
i from colligative data: i = (observed ΔT)/(calculated ΔT) = M_calculated/M_observed.
Worked pattern 6 (find α): 10 g of AB₂ (M = 200) in 100 g of water boils at 100.52 °C. m = 0.5, so ΔT_b = 0.52 = i(0.52)(0.5) and i = 2, giving 2 = 1 + 2α and α = 0.5 (AB₂ gives three ions).
Worked pattern 7 (weak acid): 5 mL acetic acid (d = 1.2 g/mL) in 1 L water is 6 g = 0.1 mol. α = √(6.25 × 10⁻⁵/0.1) = 0.025, so i = 1.025 and ΔT_f = 1.025 × 1.86 × 0.1 = 0.19, freezing at −19 × 10⁻² °C. For osmotic pressure of 0.03 M HX (K_a = 1.2 × 10⁻⁵): α = 0.02, C_total = 0.03 × 1.02 and π = 0.0306 × 0.083 × 300 = 0.76 bar.
Coordination compounds: π = iCRT gives i, and i = 1 + (n − 1)α gives the number of ions. For 0.01 M solution with π = 0.984 atm at 300 K and 75% ionisation: i = 4, so n = 5 and the formula is Ba₃[Co(CN)₅]₂.

⚠️ JEE Trap: What counts as a particle. Count the ions the salt actually gives: K₄[Fe(CN)₆] gives 5, but [Co(NH₃)₆]Cl₃ gives 4 (the complex ion stays intact). And i < 1 means association, i > 1 dissociation, never the reverse.

7. Osmosis, Tonicity and Mixed Solutions

Fig. 6: Osmosis, tonicity and reverse osmosis.

π = iCRT (R = 0.0821 L atm K⁻¹ mol⁻¹, or 0.083 L bar K⁻¹ mol⁻¹ if the question gives it). Also π = ρgh for the height of a liquid column.
Isotonic solutions have equal i × C. Worked: a cell with π = 12 atm at 300 K needs NaCl with 12 = 2 × C × 0.08 × 300, so C = 0.25 M = 0.25 × 58.5 = 15 g/L.
Two non-electrolytes in one solution: add the moles. 0.3 g of A (M = 60) and 0.9 g of B (M = 180) in 100 mL water at 27 °C gives 0.01 mol in 0.1 L, so π = 0.1 × 0.082 × 300 = 2.46 atm.
Net osmotic pressure across a membrane: π_net = |ΣiC_inside − ΣiC_outside|RT. An artificial cell with 0.2 M glucose in 0.05 M NaCl: 0.2 − (2 × 0.05) = 0.1 M, so π = 0.1 × 0.083 × 300 = 2.49 bar = 24.9 × 10⁻¹ bar.
After a reaction: work out what is left. 200 mL of 0.2 M BaCl₂ + 500 mL of 0.1 M Na₂SO₄ leaves 0.01 mol Na₂SO₄ (3 particles) and 0.08 mol NaCl (2 particles) in 0.7 L, since BaSO₄ precipitates. π = (0.03 + 0.16)/0.7 × 0.0821 × 300 = 6.69 atm.
Reverse osmosis: P_ext > π pushes solvent out of the solution (desalination). Osmotic pressure is the best method for polymers and proteins because π is large even for tiny molality.

⚠️ JEE Trap: Using the wrong concentration in π. π needs molarity of particles (i × C), not molality of the compound, and it needs T in kelvin. For a dissolved gas or precipitate, include only species left in solution.

8. How JEE Frames Questions

Numerical (integer answer): concentration conversions, ΔT_b or ΔT_f with i, vapour pressure of a solution, π of a mixture, α from data. Watch the "nearest integer" scale factor (×10⁻¹ or ×10⁻²).
Compare and rank: boiling or freezing order of four solutions; osmotic pressure of glucose, NaCl, CaCl₂, Al₂(SO₄)₃ at the same molarity (i = 1, 2, 3, 5).
Two-mixture algebra: find P° values from two vapour pressure readings; find mole fractions from vapour composition.
Statements and graphs: Henry's law constant statements, P–x plots for acetone–CS₂ (positive deviation), log p against log x for a gas (slope 1).

⚠️ JEE Trap: Skipping units in R. With R = 0.0821 use litre and atm; with 0.083 use litre and bar; with 8.314 use joule. Also check whether a question says "assume complete dissociation" before you use a α formula.

9. Quick Sheet and Checklist

Idea	Rule
Molarity from % w/w	M = 10 d x / M_w
Mole fraction from molality (water)	x = m/(m + 55.5)
Henry's law	p = K_Hx (larger K_H, lower solubility)
Raoult	P = x_AP_A° + x_BP_B°
Vapour composition	1/P = y_A/P_A° + y_B/P_B°
RLVP (exact)	(P° − P_s)/P_s = i n/N
ΔT_b, ΔT_f	i K m
Osmotic pressure	π = i C R T
Dissociation / association	i = 1 + (n − 1)α; i = 1 + (1/n − 1)β
Immiscible liquids	W_A/W_B = P_A°M_A/(P_B°M_B)

Before the exam, check you can:

Convert between molarity, molality, mole fraction and mass percent in under a minute.
Find the vapour composition or the liquid composition of an ideal binary mixture, and find P° values from two mixtures.
Rank solutions by boiling point, freezing point or π using i × C.
Find α or β from a colligative property, including a weak acid with K_a.
Match deviation type, ΔH_mix and azeotrope type without hesitation.