NCERT Solutions for Class 12 Chemistry Chapter 1: Solutions
Complete NCERT Class 12 Chemistry Chapter 1 solutions — all 12 intext questions and 41 exercises answered in topper-style detail, with full revision notes on concentration terms, Henry’s law, Raoult’s law, ideal and non-ideal solutions, azeotropes, colligative properties, osmosis and the van’t Hoff factor.
Chapter at a glance — complete revision notes
Introduction
- In normal life we rarely meet pure substances; most materials are mixtures of two or more pure substances, and their utility depends on their composition — brass (copper + zinc) has properties very different from pure copper or zinc; a trace of fluoride (1 ppm) in drinking water prevents tooth decay but higher concentrations make teeth mottled and sodium fluoride is rat poison; intravenous injections must be dissolved in water containing salts at particular ionic concentrations. This unit studies solutions — their formation and properties like vapour pressure and the colligative properties. Almost all processes in the body occur in liquid solutions.
1.1 Types of solutions
- A solution is a homogeneous mixture of two or more components — its composition and properties are uniform throughout. The component present in the largest quantity is the solvent (it determines the physical state of the solution); the others are the solutes. We consider binary solutions (two components); each may be solid, liquid or gas — so nine types (Table 1.1): gas in gas (air: oxygen in nitrogen), liquid in gas (chloroform vapour in nitrogen), solid in gas (camphor in nitrogen); gas in liquid (CO₂ in water — soda), liquid in liquid (ethanol in water), solid in liquid (glucose in water); gas in solid (hydrogen in palladium), liquid in solid (amalgam of mercury with sodium), solid in solid (copper in gold — alloys).
1.2 Expressing concentration of solutions
- Mass percentage (w/w) = (mass of component/total mass of solution) × 100 (e.g. 10% glucose: 10 g glucose + 90 g water; commercial bleaching solution 3.62 mass % sodium hypochlorite). Volume percentage (V/V) = (volume of component/total volume) × 100 (35% v/v ethylene glycol antifreeze lowers the freezing point of water to 255.4 K (−17.6 °C)). Mass by volume percentage (w/V) = mass of solute per 100 mL of solution (used in medicine and pharmacy). Parts per million (ppm) = (parts of the component/total parts of the solution) × 10⁶ — for trace quantities (dissolved oxygen in 1 L of sea water ≈ 5.8 × 10⁻³ g, 5.8 ppm; pollutants in water or the atmosphere).
- Mole fraction (x) of A in a binary mixture x_A = n_A/(n_A + n_B); the sum of all mole fractions in a solution = 1; useful for vapour pressure relations. Molarity (M) = moles of solute per litre of solution (mol L⁻¹; 0.25 M NaOH = 0.25 mol in 1 L) — depends on temperature (volume changes). Molality (m) = moles of solute per kilogram of solvent (mol kg⁻¹) — independent of temperature. Mass %, ppm, mole fraction and molality are temperature-independent; molarity is not.
- Worked examples: mole fraction of ethylene glycol in a 20 % (mass) aqueous solution = 0.068 (and water 0.932); molarity of 5 g NaOH in 450 mL = 0.278 M; molality of 2.5 g ethanoic acid in 75 g benzene = 0.556 mol kg⁻¹.
1.3 Solubility
- Solubility is the maximum amount of solute that dissolves in a specified amount of solvent at a given temperature (and pressure). “Like dissolves like“: polar solutes dissolve in polar solvents (NaCl, sugar in water) and non-polar solutes in non-polar solvents (naphthalene, anthracene in benzene) — a solute dissolves if the intermolecular interactions in the solute and solvent are similar to those between them.
- Dynamic equilibrium: dissolution (solute enters solution) and crystallisation (solute separates) occur at the same rate in a saturated solution (in equilibrium with undissolved solute — concentration = solubility); an unsaturated solution can dissolve more solute.
1.3.1 Solubility of a solid in a liquid
Obeys Le Chatelier’s principle: if the dissolution is endothermic (Δ_sol H > 0) solubility increases with temperature; if exothermic (Δ_sol H < 0) it decreases. Pressure has no significant effect (solids and liquids are practically incompressible).1.3.2 Solubility of a gas in a liquid
Solubility increases with pressure (Fig. 1.1: compressing the gas increases the number of gas particles per unit volume above the solution and the rate at which they enter it until a new equilibrium is reached). Henry’s law: the partial pressure of the gas in the vapour phase is proportional to the mole fraction of the gas in the solution: p = K_H x (Fig. 1.2: a plot of p against x is a straight line through the origin with slope K_H) — the higher the K_H at a given pressure, the lower the solubility. K_H increases with temperature (Table 1.2: N₂, O₂ …): solubility of gases decreases with rise in temperature (dissolution of a gas is like condensation — exothermic — so by Le Chatelier’s principle solubility falls on heating); aquatic species are more comfortable in cold water. Applications: (i) soft drinks/soda water are sealed under high pressure of CO₂ to increase its solubility; (ii) bends in scuba divers — at depth, high pressure increases the solubility of atmospheric gases in blood; on ascent the pressure falls and nitrogen bubbles release in the blood, blocking capillaries (painful, dangerous); tanks are filled with air diluted with helium (11.7 % He, 56.2 % N₂, 32.1 % O₂) to avoid bends and the toxic effect of N₂; (iii) at high altitudes the partial pressure of oxygen is lower, so the oxygen concentration in the blood and tissues of climbers is low — causing anoxia (weakness, inability to think clearly). Worked example: N₂ at 293 K and 0.987 bar (K_H = 76.48 kbar): x = 1.29 × 10⁻⁵, i.e. 0.716 millimol per litre of water.
1.4 Vapour pressure of liquid solutions
1.4.1 Vapour pressure of liquid–liquid solutions — Raoult’s law
For two volatile liquids 1 and 2 in equilibrium with vapour (p_total = p₁ + p₂, Dalton’s law), Raoult’s law (François-Marie Raoult, 1886): the partial vapour pressure of each component is directly proportional to its mole fraction: p₁ = x₁p₁⁰, p₂ = x₂p₂⁰, so p_total = p₁⁰ + (p₂⁰ − p₁⁰)x₂ — linear in x₂ (Fig. 1.3); the mole fraction of component i in the vapour phase y_i = p_i/p_total; the vapour is richer in the more volatile component. Example 1.5: CHCl₃ (200 mm Hg) and CH₂Cl₂ (415 mm Hg) at 298 K: 25.5 g CHCl₃ + 40 g CH₂Cl₂ gives p_total = 347.9 mm Hg (x_CHCl₃ = 0.296, x_CH₂Cl₂ = 0.704); vapour mole fractions 0.17 (CHCl₃) and 0.83 (CH₂Cl₂).1.4.2 Raoult’s law as a special case of Henry’s law
For a gas (volatile) solute p = K_H x; Raoult’s law p₁ = x₁p₁⁰ has the same form — Raoult’s law is a special case of Henry’s law with K_H = p₁⁰.1.4.3 Vapour pressure of solutions of solids in liquids
The vapour pressure of the solvent decreases when a non-volatile solute is dissolved (Fig. 1.4) — because the solute particles occupy part of the surface, reducing the number of solvent molecules escaping; the decrease depends on the quantity (mole fraction) of the solute, not its nature (1 mol of sucrose or of urea in 1 kg water lowers it by about the same amount). For a non-volatile solute: p₁ = x₁p₁⁰ and the relative lowering of vapour pressure (p₁⁰ − p₁)/p₁⁰ = x₂.
1.5 Ideal and non-ideal solutions
1.5.1 Ideal solutions
Obey Raoult’s law over the entire range of concentration; Δ_mix H = 0 and Δ_mix V = 0 (no heat absorbed or evolved; volume additive). At the molecular level the A–A, B–B and A–B interactions are nearly equal (e.g. n-hexane + n-heptane, bromoethane + chloroethane, benzene + toluene) — perfectly ideal solutions are rare.1.5.2 Non-ideal solutions
Do not obey Raoult’s law over the entire range (Fig. 1.6). Positive deviation: vapour pressure higher than predicted; A–B interactions weaker than A–A and B–B; molecules escape more easily; Δ_mix H > 0 (endothermic), Δ_mix V > 0 — ethanol + acetone (acetone breaks the hydrogen bonds of ethanol), carbon disulphide + acetone, ethanol + water. Negative deviation: vapour pressure lower; A–B interactions stronger than A–A and B–B; Δ_mix H < 0, Δ_mix V < 0 — phenol + aniline (stronger H-bonds), chloroform + acetone (hydrogen bonding between CHCl₃ and the carbonyl oxygen), nitric acid + water.1.5.3 Azeotropes
Binary mixtures having the same composition in liquid and vapour phases that boil at a constant temperature (cannot be separated by fractional distillation). Minimum boiling azeotrope — large positive deviation (ethanol–water ≈ 95 % ethanol by volume, 351.15 K); maximum boiling azeotrope — large negative deviation (nitric acid–water ≈ 68 % HNO₃, 32 % water by mass, boiling at 393.5 K).
1.6 Colligative properties and determination of molar mass
- Colligative properties depend on the number of solute particles (relative to the total number of particles) and not on their identity: (1) relative lowering of vapour pressure, (2) elevation of boiling point, (3) depression of freezing point, (4) osmotic pressure. They are used to determine the molar mass of the solute.
1.6.1 Relative lowering of vapour pressure
(p₁⁰ − p₁)/p₁⁰ = x₂ = n₂/(n₁ + n₂); for dilute solutions n₂ ≪ n₁: = n₂/n₁ = (w₂M₁)/(M₂w₁) so M₂ = w₂M₁p₁⁰/[w₁(p₁⁰ − p₁)]. Example 1.6: 30 g of solute in 90 g of water … (molar mass determination).1.6.2 Elevation of boiling point
The vapour pressure of the solution is lower, so it must be heated to a higher temperature to boil at the same external pressure (Fig. 1.7: vapour pressure curves of pure solvent and solution; ΔT_b = T_b − T_b⁰). For dilute solutions ΔT_b = K_b m (K_b = molal elevation constant (ebullioscopic constant), K kg mol⁻¹); M₂ = 1000 K_b w₂/(ΔT_b w₁) (w in g). K_b (water) = 0.52 K kg mol⁻¹.1.6.3 Depression of freezing point
The freezing point is the temperature at which the solid and the liquid have the same vapour pressure; the solution’s vapour pressure curve intersects the solid’s curve at a lower temperature (Fig. 1.8) — ΔT_f = K_f m (K_f = cryoscopic constant), M₂ = 1000 K_f w₂/(ΔT_f w₁); K_f (water) = 1.86 K kg mol⁻¹ (Table 1.3 lists K_b and K_f of common solvents: e.g. benzene K_f 5.12, K_b 2.53; acetic acid K_f 3.90, K_b 3.07). Ethylene glycol is used as an antifreeze in car radiators; Example 1.9: 45 g ethylene glycol in 600 g water: ΔT_f = 2.2 K, freezing point 270.95 K.1.6.4 Osmosis and osmotic pressure
Osmosis — the spontaneous flow of solvent molecules through a semipermeable membrane (SPM; e.g. cellulose acetate, pig’s bladder, parchment, cell walls) from pure solvent to solution (or from a dilute to a concentrated solution) (Fig. 1.9). The osmotic pressure π is the excess pressure that must be applied to the solution side to stop osmosis (Fig. 1.10). For dilute solutions π = CRT = (n₂/V)RT (Jacobus van’t Hoff), so M₂ = w₂RT/(πV); π ∝ concentration. Advantages of osmotic pressure for molar mass determination: measured at room temperature, uses molarity not molality, and a large effect — suited to polymers and biomolecules (proteins) that are unstable at higher temperatures. Isotonic solutions have the same osmotic pressure (0.9 % (mass/V) NaCl = normal saline); a cell placed in a hypotonic solution swells, in a hypertonic one shrinks.1.6.5 Reverse osmosis and water purification
If a pressure larger than the osmotic pressure is applied to the solution side, the solvent flows from the solution to the pure solvent through the SPM — reverse osmosis (Fig. 1.11); used for desalination of sea water (cellulose acetate membrane).
1.7 Abnormal molar masses and van’t Hoff factor
- When the solute dissociates (electrolytes: KCl → K⁺ + Cl⁻) the number of particles increases, so the colligative property is larger and the observed molar mass is lower than the normal; when the solute associates (ethanoic acid in benzene forms dimers) the number of particles decreases, the colligative property is smaller and the observed molar mass is higher. van’t Hoff factor i = (normal molar mass)/(observed molar mass) = (observed colligative property)/(calculated colligative property) = (total moles of particles after association/dissociation)/(moles before). i > 1 for dissociation, i < 1 for association, i = 1 when no change. Modified formulae: Δ_p/p₁⁰ = i x₂, ΔT_b = i K_b m, ΔT_f = i K_f m, π = i CRT. Degree of dissociation α = (i − 1)/(n − 1) for an electrolyte giving n ions (e.g. i = 1 + α for AB; i = 1 + 2α for AB₂ …); degree of association for dimerisation i = 1 − α/2. (Table 1.4: values of i for NaCl, KCl, MgSO₄, K₂SO₄ at different concentrations.)
Key diagrams from the chapter
Summary (NCERT)
- A solution is a homogeneous mixture; solid, liquid and gaseous solutions; concentration as mole fraction, molarity, molality and percentages; Henry’s law for gases in liquids; Raoult’s law for the relative lowering of vapour pressure of a solvent by a non-volatile solute and its general form p_total = p₁⁰x₂… (for two volatile liquids); ideal solutions and positive/negative deviations; azeotropes; colligative properties and molar masses; reverse osmosis; solutes that dissociate show lower and those that associate show higher molar mass than the actual, expressed by the van’t Hoff factor i.
Intext 1.1 — Calculate the mass percentage of benzene (C₆H₆) and carbon tetrachloride (CCl₄) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
Concept. Mass percentage (w/w) of a component = (mass of that component ÷ total mass of the solution) × 100. It is independent of temperature because masses do not change with temperature, and the percentages of all components must add up to 100.
Step 1 — total mass of solution. Mass of solution = mass of benzene + mass of CCl₄ = 22 g + 122 g = 144 g.
Mass % of C₆H₆ = (22 g ÷ 144 g) × 100 = 15.28 %
Mass % of CCl₄ = (122 g ÷ 144 g) × 100 = 84.72 %
Check. 15.28 + 84.72 = 100.00 — the two percentages must sum to 100, which confirms the arithmetic. Note that CCl₄ is the solvent (larger amount) and benzene the solute.
Extra depth. The mass percentage is one of the four temperature-independent concentration terms (the others are ppm, mole fraction and molality) because it involves only masses. In practice it is the form used for commercial solutions such as 3.62 % NaOCl bleach or 10 % glucose in hospitals; always report the solvent as well as the solute because the percentages of the two components must total 100 %.
Conclusion: The solution contains 15.28 % benzene and 84.72 % carbon tetrachloride by mass.
Intext 1.2 — Calculate the mole fraction of benzene in a solution containing 30% by mass in carbon tetrachloride.
Concept. A solution “30 % by mass” means 30 g of benzene in every 100 g of solution; so we take a convenient 100 g sample: 30 g C₆H₆ and 70 g CCl₄. Mole fraction x_A = n_A ÷ (n_A + n_B).
Step 1 — molar masses. C₆H₆ = 6 × 12 + 6 × 1 = 78 g mol⁻¹; CCl₄ = 12 + 4 × 35.5 = 154 g mol⁻¹.
Step 2 — moles.
n(C₆H₆) = 30 g ÷ 78 g mol⁻¹ = 0.3846 mol
n(CCl₄) = 70 g ÷ 154 g mol⁻¹ = 0.4545 mol
Step 3 — mole fraction.
x(C₆H₆) = 0.3846 ÷ (0.3846 + 0.4545) = 0.3846 ÷ 0.8391 = 0.458 ≈ 0.459
Consequently x(CCl₄) = 1 − 0.459 = 0.541. Mole fractions of all components always add to 1.
Extra depth. Mole fraction is dimensionless and is the quantity that appears directly in Raoult’s law and Henry’s law, which is why it is preferred for vapour-pressure problems. Notice that 30 % by mass benzene is only 0.459 by moles, because benzene (78 g mol⁻¹) is lighter than CCl₄ (154 g mol⁻¹) and therefore contributes more particles per gram.
Conclusion: The mole fraction of benzene is ≈ 0.459 (and that of CCl₄ ≈ 0.541).
Intext 1.3 — Calculate the molarity of each of the following solutions: (a) 30 g of Co(NO₃)₂·6H₂O in 4.3 L of solution (b) 30 mL of 0.5 M H₂SO₄ diluted to 500 mL.
Concept. Molarity (M) = moles of solute ÷ volume of solution in litres. For dilution, the number of moles of solute remains constant, so M₁V₁ = M₂V₂.
(a) Co(NO₃)₂·6H₂O. Molar mass = 59 + 2(14 + 48) + 6(18) = 59 + 124 + 108 = 291 g mol⁻¹.
n = 30 g ÷ 291 g mol⁻¹ = 0.103 mol
M = 0.103 mol ÷ 4.3 L = 0.024 mol L⁻¹ (0.024 M)
(b) Dilution. Moles of H₂SO₄ in 30 mL of 0.5 M = 0.5 mol L⁻¹ × 0.030 L = 0.015 mol. This is now present in 500 mL (0.5 L).
M₂ = (0.5 M × 30 mL) ÷ 500 mL = 0.03 M
Dilution to a volume 16.7 times larger reduces the concentration by the same factor: 0.5 ÷ 16.67 = 0.03 M.
Extra depth. Molarity depends on temperature because the volume of a solution expands on warming, so a 0.5 M solution at 298 K is slightly more dilute at 350 K. The relation M₁V₁ = M₂V₂ holds only for dilution (the moles of solute stay constant), not when two solutions of different solutes are mixed.
Conclusion: (a) 0.024 M; (b) 0.03 M.
Intext 1.4 — Calculate the mass of urea (NH₂CONH₂) required in making 2.5 kg of 0.25 molal aqueous solution.
Concept. 0.25 molal means 0.25 mol of urea in 1 kg (1000 g) of water (the solvent). The 2.5 kg given is the mass of the solution (solute + solvent), so the solvent mass must be found first.
Step 1 — molar mass of urea. NH₂CONH₂ = CH₄N₂O = 12 + 4 + 28 + 16 = 60 g mol⁻¹. Mass of urea in 1 kg of water = 0.25 × 60 = 15 g.
Step 2 — mass of solution containing 15 g urea per 1000 g water. Solution mass = 1000 g + 15 g = 1015 g.
Step 3 — proportion. 1015 g of solution contains 15 g urea, so 2500 g of solution contains:
Mass of urea = (15 g ÷ 1015 g) × 2500 g = 36.946 g ≈ 37 g
Extra depth. This problem separates the mass of the solution from the mass of the solvent — the commonest source of error in molality calculations. If one wrongly takes 2.5 kg as the solvent mass the answer would be 37.5 g instead of 36.95 g, which shows how a small conceptual slip alters the final value.
Conclusion: About 36.95 g (≈ 37 g) of urea is required; the rest (≈ 2463 g) is water.
Intext 1.5 — Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL⁻¹.
Concept. 20 % (w/w) means 20 g KI in 100 g solution, i.e. 20 g KI + 80 g water. Molar mass of KI = 39 + 127 = 166 g mol⁻¹. Moles of KI = 20 ÷ 166 = 0.1205 mol.
(a) Molality = moles of solute ÷ kg of solvent:
m = 0.1205 mol ÷ 0.080 kg = 1.506 ≈ 1.5 mol kg⁻¹
(b) Molarity needs the volume of the solution: V = mass ÷ density = 100 g ÷ 1.202 g mL⁻¹ = 83.19 mL = 0.08319 L.
M = 0.1205 mol ÷ 0.08319 L = 1.448 ≈ 1.45 mol L⁻¹
(c) Mole fraction. n(H₂O) = 80 ÷ 18 = 4.444 mol.
x(KI) = 0.1205 ÷ (0.1205 + 4.444) = 0.1205 ÷ 4.5649 = 0.0264 ≈ 0.0263
| Concentration term | Formula used | Value |
|---|---|---|
| Molality | n(KI) ÷ kg water | 1.5 m |
| Molarity | n(KI) ÷ L solution (needs density) | 1.45 M |
| Mole fraction | n(KI) ÷ (n(KI) + n(H₂O)) | 0.0263 |
Extra depth. Molality (1.5 m) and molarity (1.45 M) are almost equal here because the solution is relatively dilute and its density (1.202 g mL⁻¹) is close to that of a typical aqueous solution; in concentrated solutions the two differ greatly. The small mole fraction (0.0263) shows that KI is a minor component by number of particles even at 20 % by mass.
Conclusion: Molality = 1.5 m, molarity = 1.45 M, mole fraction of KI = 0.0263.
Intext 1.6 — H₂S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H₂S in water at STP is 0.195 m, calculate Henry’s law constant.
Concept. Henry’s law: p = K_H · x, where x is the mole fraction of the gas in solution. At STP (as used in this NCERT problem) the pressure of H₂S is taken as 0.987 bar. Solubility 0.195 m means 0.195 mol H₂S per 1 kg of water.
Step 1 — mole fraction. n(H₂S) = 0.195 mol; n(H₂O) = 1000 ÷ 18 = 55.5 mol.
x(H₂S) = 0.195 ÷ (0.195 + 55.5) = 0.195 ÷ 55.695 = 0.0035
Step 2 — K_H.
K_H = p ÷ x = 0.987 bar ÷ 0.0035 = 282 bar
A large K_H value means the gas is only slightly soluble at that pressure — consistent with H₂S being sparingly soluble in water.
Extra depth. The value K_H = 282 bar is large, which tells us H₂S is only sparingly soluble in water at 0.987 bar; compare it with CO₂ (K_H = 1.67 × 10⁸ Pa = 1670 bar in intext 1.7) — a smaller K_H means a more soluble gas. H₂S is also toxic, so its solubility data matter for laboratory ventilation and for qualitative analysis (Group II precipitation).
Conclusion: Henry’s law constant K_H = 282 bar.
Intext 1.7 — Henry’s law constant for CO₂ in water is 1.67 × 10⁸ Pa at 298 K. Calculate the quantity of CO₂ in 500 mL of soda water when packed under 2.5 atm CO₂ pressure at 298 K.
Concept. p(CO₂) = K_H · x(CO₂). Convert the pressure to Pa: 2.5 atm × 1.01325 × 10⁵ Pa/atm = 2.533 × 10⁵ Pa.
x(CO₂) = p ÷ K_H = 2.533 × 10⁵ ÷ 1.67 × 10⁸ = 1.517 × 10⁻³
Step 2 — moles of CO₂. 500 mL of water ≈ 500 g = 27.8 mol of water (500 ÷ 18). Because CO₂ is present in a very small quantity, n(CO₂) ≪ n(H₂O), so:
x(CO₂) = n ÷ (n + 27.8) ≈ n ÷ 27.8 ⇒ n = 1.516 × 10⁻³ × 27.8 = 0.0421 mol
Step 3 — mass. Molar mass of CO₂ = 44 g mol⁻¹; mass = 0.0421 × 44 = 1.85 g.
Extra depth. Soda water bottled at 2.5 atm holds roughly 1.85 g of CO₂ per 500 mL;; on opening the bottle the pressure above the liquid falls and the excess dissolved gas comes out as bubbles. The same calculation explains why warm soda fizzes more: K_H rises with temperature, so solubility falls.
Conclusion: 500 mL of soda water holds about 0.0421 mol ≈ 1.85 g of CO₂.
Intext 1.8 — The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.
Concept. Raoult’s law for each component: p_A = x_A p_A°, p_B = x_B p_B°; total pressure p = p_A + p_B = p_A° + (p_B° − p_A°) x_B. Composition of vapour (Dalton’s law): y_i = p_i ÷ p_total.
Step 1 — liquid composition. 600 = 450 + (700 − 450) x_B ⇒ 250 x_B = 150 ⇒ x_B = 0.6; hence x_A = 0.4.
Step 2 — partial pressures.
p_A = 0.4 × 450 = 180 mm Hg; p_B = 0.6 × 700 = 420 mm Hg; total = 600 mm Hg ✔
Step 3 — vapour composition.
y_A = 180 ÷ 600 = 0.30; y_B = 420 ÷ 600 = 0.70
The vapour is richer in B (0.70 vs 0.60 in liquid) because B is the more volatile component (higher p°).
Extra depth. This is an ideal solution of two volatile liquids, so Dalton’s law (p_total = p_A + p_B) and Raoult’s law are used together. The calculation shows the separation principle of fractional distillation: the vapour is richer in the more volatile component B (0.70) than the liquid (0.60), so condensing and re-vaporising the vapour enriches B further.
Conclusion: Liquid: x_A = 0.4, x_B = 0.6. Vapour: y_A = 0.30, y_B = 0.70.
Intext 1.9 — Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH₂CONH₂) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
Concept. For a non-volatile solute, p₁ = x₁ p₁° and relative lowering (p₁° − p₁)/p₁° = x₂ (mole fraction of solute).
Moles. n(urea) = 50 ÷ 60 = 0.833 mol; n(water) = 850 ÷ 18 = 47.22 mol.
x(water) = 47.22 ÷ (47.22 + 0.833) = 47.22 ÷ 48.05 = 0.9827
p = 0.9827 × 23.8 = 23.4 mm Hg
Relative lowering = (23.8 − 23.4) ÷ 23.8 = x(urea) = 0.833 ÷ 48.05 = 0.0173
Extra depth. For a non-volatile solute the vapour-pressure lowering is purely a function of the number of solute particles: the relative lowering equals x(urea) = 0.0173 whatever the solute is. This is the fundamental colligative property from which ΔT_b, ΔT_f and π are all derived.
Conclusion: Vapour pressure of the solution ≈ 23.4 mm Hg; relative lowering ≈ 0.0173.
Intext 1.10 — Boiling point of water at 750 mm Hg is 99.63 °C. How much sucrose is to be added to 500 g of water such that it boils at 100 °C?
Concept. ΔT_b = K_b · m, with K_b (water) = 0.52 K kg mol⁻¹. Required elevation = 100 − 99.63 = 0.37 °C = 0.37 K.
m = ΔT_b ÷ K_b = 0.37 ÷ 0.52 = 0.7115 mol kg⁻¹
So 0.7115 mol of sucrose per kg of water; for 500 g (0.5 kg) water: n = 0.7115 × 0.5 = 0.3558 mol.
Molar mass of sucrose C₁₂H₂₂O₁₁ = 12 × 12 + 22 + 11 × 16 = 342 g mol⁻¹.
Mass = 0.3558 × 342 = 121.67 g
Extra depth. Boiling-point elevation is used to determine molar mass and to estimate how much solute is needed for a target boiling point. Using K_b = 0.52 K kg mol⁻¹ for water, the very large amount of sucrose (121.7 g in 500 g water) shows that the elevation of boiling point is a small effect — 0.37 K needs a molality of 0.71 m.
Conclusion: About 121.67 g of sucrose must be added.
Intext 1.11 — Calculate the mass of ascorbic acid (Vitamin C, C₆H₈O₆) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5 °C. K_f = 3.9 K kg mol⁻¹.
Concept. ΔT_f = K_f m = K_f × (w₂ ÷ M₂) × (1000 ÷ w₁).
Molar mass of C₆H₈O₆ = 72 + 8 + 96 = 176 g mol⁻¹.
m = 1.5 ÷ 3.9 = 0.3846 mol kg⁻¹
Moles required in 75 g (0.075 kg): n = 0.3846 × 0.075 = 0.02885 mol.
Mass = 0.02885 × 176 = 5.077 g ≈ 5.08 g
Extra depth. Acetic acid is chosen as the solvent because its K_f (3.9 K kg mol⁻¹) is larger than that of water (1.86), so the same molality produces a larger, more measurable depression. Ascorbic acid (vitamin C) is a non-electrolyte in this solvent, so no van’t Hoff correction is needed.
Conclusion: 5.08 g of ascorbic acid is required.
Intext 1.12 — Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37 °C.
Concept. π = CRT = (n ÷ V) RT, with n = w ÷ M. Use R = 8.314 kPa L K⁻¹ mol⁻¹ (= 8.314 × 10³ Pa L K⁻¹ mol⁻¹), V in litres, T in kelvin.
n = 1.0 g ÷ 185000 g mol⁻¹ = 5.41 × 10⁻⁶ mol; V = 0.450 L; T = 37 + 273 = 310 K.
π = (5.41 × 10⁻⁶ ÷ 0.450) × 8.314 × 10³ × 310 = 30.96 Pa ≈ 31 Pa
The osmotic pressure is tiny in absolute terms but can be measured accurately by a sensitive manometer — this is why osmotic pressure is the method of choice for large molar masses (polymers, proteins).
Extra depth. Osmotic pressure is the most sensitive colligative property for macromolecules: 1 g of a polymer of molar mass 185,000 gives only 31 Pa (0.3 mbar), yet this is easily measurable with a sensitive manometer, whereas the corresponding ΔT_f would be of the order of 10⁻⁵ K. That is why proteins and polymers are characterised by osmometry.
Conclusion: π ≈ 30.96 Pa.
Exercise 1.1 — Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.
Definition. A solution is a homogeneous mixture of two or more substances, i.e. a mixture whose composition and properties are uniform throughout (no visible boundary between components, and the particles are of molecular/ionic size). The component present in the largest amount, which decides the physical state of the solution, is the solvent; the other component(s) are the solutes. In this chapter we deal with binary solutions (one solute + one solvent).
Types. Since solute and solvent can each be a gas, a liquid or a solid, nine types of binary solutions are possible (NCERT Table 1.1). They are grouped by the physical state of the solvent (which is also the state of the solution):
| Type of solution | Solute | Solvent | Common examples |
|---|---|---|---|
| Gaseous solutions | Gas | Gas | Mixture of oxygen and nitrogen gases (air) |
| Liquid | Gas | Chloroform mixed with nitrogen gas (mist, fog) | |
| Solid | Gas | Camphor in nitrogen gas | |
| Liquid solutions | Gas | Liquid | Oxygen dissolved in water; CO₂ in water (soda water) |
| Liquid | Liquid | Ethanol dissolved in water | |
| Solid | Liquid | Glucose or sugar dissolved in water | |
| Solid solutions | Gas | Solid | Solution of hydrogen in palladium |
| Liquid | Solid | Amalgam of mercury with sodium | |
| Solid | Solid | Copper dissolved in gold (alloys such as brass, bronze) |
Explanation of each group. (i) Gaseous solutions form because gases mix completely in all proportions — in air, N₂ (≈78 %) is the solvent and O₂ (≈21 %) and others are solutes. (ii) Liquid solutions are the most common and most important (aqueous solutions): gases like CO₂ dissolve in water under pressure, miscible liquids like ethanol mix in all proportions, and ionic/polar solids like NaCl or glucose dissolve with the solvent molecules surrounding the solute particles. (iii) Solid solutions are homogeneous solids, for instance alloys in which atoms of one metal occupy lattice positions or interstitial sites of the other; they have properties quite different from the pure metals (e.g. brass is harder than copper).
Extra depth. Table 1.1 organises the nine kinds of binary solutions; examples are from everyday life (air, soda water, brine, alloys such as brass, amalgams used in dentistry). A solution is homogeneous at the molecular level, unlike a colloid or suspension in which particles are large enough to scatter light or settle.
Conclusion: A solution is a homogeneous mixture; nine types of binary solutions exist depending on the physical states of solute and solvent (gas, liquid, solid).
Exercise 1.2 — Give an example of solid solution in which the solute is a gas.
Answer: a solution of hydrogen in palladium. Palladium metal is the solvent (solid), and hydrogen gas is the solute. Finely divided palladium can absorb up to about 900 times its own volume of hydrogen; the hydrogen molecules dissociate and the hydrogen atoms occupy the interstitial sites (holes) in the palladium lattice, giving a homogeneous solid solution (an interstitial hydride, PdHx) whose composition is the same throughout.
Why it is a solid solution. The physical state of the resulting material is solid, which is the state of the solvent; the solute is a gas, so it is a gas-in-solid solution (see Table 1.1). Because the absorption is reversible and depends on temperature and pressure, palladium is used for storing and purifying hydrogen — only hydrogen can diffuse through hot palladium, which is how very pure H₂ is made.
Extra depth. Palladium can absorb about 900 times its own volume of H₂; the hydrogen atoms occupy octahedral interstitial sites of the face-centred cubic lattice. This property makes Pd useful for hydrogen purification (hot Pd is permeable only to H₂), as a hydrogenation catalyst and in hydrogen-storage research.
Conclusion: Hydrogen dissolved in palladium is the standard example of a solid solution with a gaseous solute.
Exercise 1.3 — Define the following terms: (i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage.
(i) Mole fraction (x). The ratio of the number of moles of one component to the total number of moles of all components in the solution. For a binary solution of A and B:
x_A = n_A ÷ (n_A + n_B); x_B = n_B ÷ (n_A + n_B); x_A + x_B = 1
It is a pure number (no unit), is independent of temperature, and is the quantity used in Raoult’s and Henry’s laws.
(ii) Molality (m). The number of moles of solute dissolved per kilogram (1000 g) of solvent:
m = moles of solute ÷ mass of solvent in kg (unit: mol kg⁻¹)
It does not depend on temperature because it involves masses only; hence it is used in boiling point elevation and freezing point depression.
(iii) Molarity (M). The number of moles of solute dissolved per litre (1 dm³) of solution:
M = moles of solute ÷ volume of solution in litres (unit: mol L⁻¹)
It depends on temperature because the volume of a solution expands or contracts when heated or cooled.
(iv) Mass percentage (w/w). The mass of a component expressed as a percentage of the total mass of the solution:
Mass % of component = (mass of component ÷ total mass of solution) × 100
Example: a 10 % glucose solution has 10 g glucose dissolved in 90 g of water (100 g solution). It is independent of temperature.
| Term | Per what? | Depends on T? | Unit |
|---|---|---|---|
| Mole fraction | Total moles | No | None |
| Molality | kg of solvent | No | mol kg⁻¹ |
| Molarity | L of solution | Yes | mol L⁻¹ |
| Mass % | 100 g of solution | No | % |
Extra depth. The four terms form two groups: the temperature-independent ones (mole fraction, molality, mass %) are used in thermodynamic relations, while molarity is used for volumetric (titration) work. Converting between molality and molarity needs the density, which is why density data are given in problems such as Ex. 1.4, 1.5 and 1.8.
Conclusion: Mole fraction, molality and mass % are temperature-independent; molarity changes with temperature.
Exercise 1.4 — Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL⁻¹?
Approach. Take 100 g of the solution: it contains 68 g HNO₃ and 32 g water. Calculate the moles of HNO₃ and the volume of 100 g of solution using the density, then M = n ÷ V(L).
Step 1 — moles of HNO₃. Molar mass of HNO₃ = 1 + 14 + 48 = 63 g mol⁻¹. n = 68 ÷ 63 = 1.0794 mol.
Step 2 — volume of solution. V = mass ÷ density = 100 g ÷ 1.504 g mL⁻¹ = 66.49 mL = 0.06649 L.
M = 1.0794 mol ÷ 0.06649 L = 16.23 mol L⁻¹
Extra depth. Concentrated nitric acid (68 %) is about 16 M; a 1 M working solution is made by diluting 62 mL to 1 L with M₁V₁ = M₂V₂. The density of 1.504 g mL⁻¹ makes it considerably denser than water, so 100 g of the acid occupies only 66.5 mL.
Conclusion: Molarity of concentrated nitric acid = 16.23 M.
Exercise 1.5 — A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL⁻¹, then what shall be the molarity of the solution?
Basis. 100 g solution = 10 g glucose (C₆H₁₂O₆, M = 180 g mol⁻¹) + 90 g water (M = 18 g mol⁻¹).
n(glucose) = 10 ÷ 180 = 0.0556 mol; n(water) = 90 ÷ 18 = 5.0 mol.
Molality:
m = 0.0556 mol ÷ 0.090 kg = 0.617 mol kg⁻¹
Mole fractions:
x(glucose) = 0.0556 ÷ (0.0556 + 5.0) = 0.011; x(water) = 5.0 ÷ 5.0556 = 0.989
Molarity (density given): V = 100 g ÷ 1.2 g mL⁻¹ = 83.33 mL = 0.0833 L.
M = 0.0556 mol ÷ 0.0833 L = 0.667 mol L⁻¹
| Quantity | Value |
|---|---|
| Molality of glucose | 0.617 m |
| x (glucose) | 0.011 |
| x (water) | 0.989 |
| Molarity | 0.667 M |
Extra depth. Glucose is a non-electrolyte, so its colligative effects follow the simple formulae. The mole fraction of glucose (0.011) is small although its mass percentage is 10 %, because glucose (180 g mol⁻¹) is ten times heavier than water; this is why a 10 % glucose solution lowers the freezing point of water by only about 1.1 K.
Conclusion: m = 0.617 mol kg⁻¹; x(glucose) = 0.011, x(water) = 0.989; M = 0.667 mol L⁻¹.
Exercise 1.6 — How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na₂CO₃ and NaHCO₃ containing equimolar amounts of both?
Concept. Both salts react with HCl: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂ (2 mol HCl per mole) and NaHCO₃ + HCl → NaCl + H₂O + CO₂ (1 mol HCl per mole). Let the mixture contain x mol of each.
Step 1 — find x. Molar masses: Na₂CO₃ = 106 g mol⁻¹, NaHCO₃ = 84 g mol⁻¹. 106x + 84x = 1 g ⇒ 190x = 1 ⇒ x = 0.005263 mol.
Step 2 — moles of HCl. From Na₂CO₃: 2 × 0.005263 = 0.010526 mol; from NaHCO₃: 0.005263 mol. Total = 0.015789 mol.
Step 3 — volume. V = n ÷ M = 0.015789 mol ÷ 0.1 mol L⁻¹ = 0.15789 L = 157.9 mL.
Extra depth. The titration of a carbonate–bicarbonate mixture with HCl is a classic double-indicator problem; here the mixture is equimolar so the stoichiometry (2 mol HCl per mol Na₂CO₃, 1 per mol NaHCO₃) fixes the total acid needed (0.0158 mol). Because 1 g is a small sample, the volume of 0.1 M acid (157.9 mL) is large, which is why a more concentrated acid would be used in practice.
Conclusion: 157.9 mL of 0.1 M HCl are required.
Exercise 1.7 — A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.
Concept. Masses of solute and of solution are additive when two solutions of the same solute are mixed.
Solute in the first solution = 25 % of 300 g = 75 g. Solute in the second = 40 % of 400 g = 160 g. Total solute = 235 g. Total solution = 300 + 400 = 700 g.
Mass % of solute = (235 ÷ 700) × 100 = 33.57 %
Mass % of solvent = 100 − 33.57 = 66.43 %. As a check, the answer lies between 25 % and 40 %, closer to 40 % because more of the 40 % solution was used.
Extra depth. Mixing solutions follows the “alligation” logic: the final mass percentage lies between the two original percentages (25 % and 40 %) and is closer to 40 % because more of the stronger solution is used. This is the mass-balance method used in industrial blending.
Conclusion: The mixture is 33.57 % solute and 66.43 % solvent by mass.
Exercise 1.8 — An antifreeze solution is prepared from 222.6 g of ethylene glycol (C₂H₆O₂) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL⁻¹, then what shall be the molarity of the solution?
Step 1 — moles. Molar mass of C₂H₆O₂ = 24 + 6 + 32 = 62 g mol⁻¹. n = 222.6 ÷ 62 = 3.59 mol.
Molality = 3.59 mol ÷ 0.200 kg = 17.95 mol kg⁻¹.
Step 2 — volume of solution. Mass of solution = 222.6 + 200 = 422.6 g. V = 422.6 ÷ 1.072 = 394.2 mL = 0.3942 L.
M = 3.59 mol ÷ 0.3942 L = 9.11 mol L⁻¹
Such a concentrated solution lowers the freezing point of water greatly (ΔT_f = K_f m ≈ 1.86 × 17.95 ≈ 33 K by the simple formula) and is why ethylene glycol is used as radiator antifreeze.
Extra depth. Ethylene glycol is miscible with water in all proportions and lowers the freezing point considerably (about 1.86 K per molal), so a 17.95 molal solution would freeze far below 0 °C in practice, much lower than the simple formula suggests. This is the principle of antifreeze in car radiators; a 35 % v/v mixture freezes at 255.4 K.
Conclusion: Molality = 17.95 m; molarity = 9.11 M.
Exercise 1.9 — A sample of drinking water was found to be severely contaminated with chloroform (CHCl₃), supposed to be a carcinogen. The level of contamination was 15 ppm (by mass): (i) express this in percent by mass (ii) determine the molality of chloroform in the water sample.
Concept. 15 ppm (by mass) means 15 parts of CHCl₃ in 10⁶ parts of the sample, i.e. 15 g of CHCl₃ in 10⁶ g of solution.
(i) Percent by mass = (15 ÷ 10⁶) × 100 = 1.5 × 10⁻³ %.
(ii) Molality. The solution is almost entirely water, so 10⁶ g of solution ≈ 10⁶ g (1000 kg) of water. Molar mass of CHCl₃ = 12 + 1 + 106.5 = 119.5 g mol⁻¹, so n = 15 ÷ 119.5 = 0.1255 mol.
m = 0.1255 mol ÷ 1000 kg = 1.255 × 10⁻⁴ mol kg⁻¹ ≈ 1.25 × 10⁻⁴ m
Extra depth. Parts per million is used for trace contaminants: 15 ppm of chloroform means 15 g in 10⁶ g of water. Such a level is a serious health risk because chloroform is a suspected carcinogen, and because water is ≈ 1 kg per litre, ppm is numerically equal to mg per litre (mg L⁻¹).
Conclusion: (i) 1.5 × 10⁻³ %; (ii) 1.25 × 10⁻⁴ m.
Exercise 1.10 — What role does the molecular interaction play in a solution of alcohol and water?
Pure components. In pure ethanol the molecules are held together by hydrogen bonds (R–O–H···O); in pure water there is an extensive hydrogen-bonded network. Both are strongly associated liquids.
In the mixture. When alcohol is dissolved in water, the alcohol molecules interpose between water molecules and break some of the existing hydrogen bonds in water (and in alcohol). The new alcohol–water H-bonds that form are weaker than the original water–water and alcohol–alcohol hydrogen bonds because the non-polar alkyl group of the alcohol interferes. Thus the A–B interactions are weaker than A–A and B–B interactions.
| Consequence | Observation |
|---|---|
| Escaping tendency of molecules | Increases — molecules leave the solution more easily |
| Vapour pressure | Higher than predicted by Raoult’s law — positive deviation |
| Δ_mix H | Positive — heat is absorbed (solution cools) |
| Δ_mix V | Positive — volume slightly greater than the sum |
| Special case | Ethanol–water forms a minimum-boiling azeotrope (≈ 95 % ethanol) |
Extra depth. The ethanol–water system is the standard example of positive deviation: the mixture is not ideal, absorbs heat (Δ_mix H > 0) and shows a slight volume expansion. Distillation of this mixture cannot give more than about 95 % ethanol because of the minimum-boiling azeotrope; absolute alcohol requires chemical drying (e.g. with CaO) or azeotropic distillation.
Conclusion: Alcohol–water solutions show positive deviation from Raoult’s law because the weakened hydrogen bonding makes A–B interactions weaker than A–A and B–B.
Exercise 1.11 — Why do gases always tend to be less soluble in liquids as the temperature is raised?
Reason. The dissolution of a gas in a liquid is an exothermic process: gas molecules lose their freedom of motion and are held by the liquid, as in condensation, so heat is evolved (Δ_sol H < 0). Gas + liquid ⇌ solution + heat.
Le Chatelier’s principle. When the temperature of an equilibrium system is raised, the system shifts in the direction that absorbs the extra heat, i.e. the endothermic direction. Therefore the equilibrium moves backwards — the dissolved gas molecules escape from the solution — and the solubility of the gas decreases. At higher temperature the molecules also have greater kinetic energy and easily break away from the liquid.
Henry’s law view. Henry’s law constant K_H increases with temperature (Table 1.2); at a given partial pressure, a higher K_H means a lower mole fraction of dissolved gas.
Everyday evidence. A fizzy drink loses its CO₂ faster when warm; dissolved O₂ in warm lakes is lower so aquatic species are more comfortable in cold waters (thermal pollution kills fish).
Extra depth. The effect has practical consequences: the “bends” in divers, the loss of fizz in warm soft drinks and the death of fish in thermally polluted water are all consequences of lower gas solubility at a higher temperature. Mathematically, d ln K_H/dT = Δ_sol H/RT², and since the solution of a gas is exothermic the K_H must increase with T.
Conclusion: Gas solubility falls on heating because dissolution of a gas is exothermic; by Le Chatelier’s principle the equilibrium shifts towards the gas phase.
Exercise 1.12 — State Henry’s law and mention some important applications.
Statement. At a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas. Equivalently (Henry’s law in terms of mole fraction): the partial pressure of the gas in the vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution.
p = K_H · x
K_H is Henry’s law constant: it depends on the nature of the gas and temperature. A higher K_H means lower solubility at a given pressure; K_H rises as temperature increases. A plot of p against x is a straight line through the origin with slope K_H (NCERT Fig. 1.2); for a given pressure, the gas with the larger K_H (N₂ vs O₂ in Table 1.2) is less soluble.
Important applications.
- Soft drinks and soda water: CO₂ is dissolved under high pressure (sealed bottles) to increase its solubility; opening the bottle releases the pressure and the gas fizzes out.
- Scuba diving — bends: at depth, high pressure increases the solubility of atmospheric gases (especially N₂) in blood. On sudden ascent the pressure falls and dissolved N₂ comes out as bubbles that block capillaries — a painful and dangerous condition. To avoid bends, divers use tanks of air diluted with helium (11.7 % He, 56.2 % N₂ and 32.1 % O₂), because helium is far less soluble in blood.
- High altitude — anoxia: at high altitude the partial pressure of O₂ is lower, so the concentration of O₂ in blood and tissues of climbers and people living there is low; they feel weak and cannot think clearly (symptoms of anoxia).
- Deep-sea / oxygen therapy and gas storage: the same relation is used in estimating the amount of gas that can be dissolved when a gas is stored under pressure.
Extra depth. Henry’s law holds only for dilute solutions at moderate pressures and for gases that do not react with the solvent; HCl and NH₃, which react with water, do not obey it. The law is used in industrial carbonation, in oxygen therapy and in explaining why cold, high-altitude lakes have different dissolved oxygen levels.
Conclusion: Henry’s law (p = K_H x) explains carbonation, the bends in scuba divers and anoxia at high altitude.
Exercise 1.13 — The partial pressure of ethane over a solution containing 6.56 × 10⁻³ g of ethane is 1 bar. If the solution contains 5.00 × 10⁻² g of ethane, then what shall be the partial pressure of the gas?
Concept. Henry’s law: p = K_H x. For a very dilute solution of the gas, the mole fraction x is proportional to the mass (moles) of dissolved gas for a fixed amount of solvent. So p ∝ mass of ethane dissolved.
p₁ = K_H x₁ and p₂ = K_H x₂, so p₂ ÷ p₁ = x₂ ÷ x₁ = (5.00 × 10⁻² ÷ 6.56 × 10⁻³).
p₂ = 1 bar × (5.00 × 10⁻² ÷ 6.56 × 10⁻³) = 7.62 bar
(Equivalent route: moles of ethane (C₂H₆ = 30 g mol⁻¹): n₁ = 2.187 × 10⁻⁴, n₂ = 1.667 × 10⁻³; K_H = p/x; the ratio gives the same 7.62 bar.)
Extra depth. The linear relation p ∝ x ∝ mass of dissolved gas is valid because ethane is sparingly soluble, so the mole fraction is proportional to the number of moles of ethane. The increase from 6.56 × 10⁻³ g to 5.00 × 10⁻² g is about 7.6-fold, giving a pressure of 7.62 bar.
Conclusion: The partial pressure of ethane = 7.62 bar.
Exercise 1.14 — What is meant by positive and negative deviations from Raoult’s law and how is the sign of Δ_mix H related to positive and negative deviations from Raoult’s law?
Background. Raoult’s law for a binary solution: p₁ = x₁p₁°, p₂ = x₂p₂° and p_total = p₁ + p₂. Solutions that obey it over the whole range are ideal; most real solutions do not — they are non-ideal and show deviations.
Positive deviation. The observed vapour pressure (partial and total) is higher than that predicted by Raoult’s law. It occurs when the A–B molecular interactions are weaker than the A–A and B–B interactions, so molecules escape more easily from the solution. Examples: ethanol + acetone, CS₂ + acetone, ethanol + water. In such solutions Δ_mix H is positive (endothermic: energy is absorbed to break stronger A–A/B–B interactions; solution cools) and Δ_mix V is positive.
Negative deviation. The observed vapour pressure is lower than Raoult’s law prediction. It occurs when A–B interactions are stronger than A–A and B–B (e.g. hydrogen bond formation between unlike molecules), so the escaping tendency falls. Examples: phenol + aniline, chloroform + acetone, nitric acid + water. Here Δ_mix H is negative (exothermic; the solution warms up) and Δ_mix V is negative.
| Property | Ideal | Positive deviation | Negative deviation |
|---|---|---|---|
| Vapour pressure | = Raoult’s law | > Raoult’s law | < Raoult’s law |
| A–B vs A–A, B–B | A–B ≈ A–A ≈ B–B | A–B weaker | A–B stronger |
| Δ_mix H | 0 | > 0 (+) | < 0 (−) |
| Δ_mix V | 0 | > 0 (+) | < 0 (−) |
| Azeotrope formed | — | Minimum boiling (ethanol–water) | Maximum boiling (HNO₃–water) |
| Examples | Benzene + toluene; n-hexane + n-heptane | Ethanol + acetone | Chloroform + acetone |
Extra depth. Deviations from Raoult’s law give information about intermolecular forces: a positive deviation means that A–B attraction is weaker than A–A/B–B (examples: ethanol–acetone, CS₂–acetone), a negative deviation means it is stronger (chloroform–acetone via hydrogen bonding, HNO₃–water). Maximum-boiling azeotropes arise from large negative deviations and minimum-boiling ones from large positive deviations.
Conclusion: Positive deviation ↔ Δ_mix H > 0 (weaker A–B interactions); negative deviation ↔ Δ_mix H < 0 (stronger A–B interactions).
Exercise 1.15 — An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
Concept. At the normal boiling point the vapour pressure of pure water equals 1 atm = 1.013 bar (p₁°). The solution exerts only 1.004 bar, so the relative lowering of vapour pressure gives the mole fraction of the solute: (p₁° − p₁) ÷ p₁° = n₂ ÷ (n₁ + n₂) ≈ n₂ ÷ n₁ for a dilute solution.
Step 1 — composition. A 2 % (w/w) solution = 2 g solute + 98 g water. n₁ = 98 ÷ 18 = 5.444 mol.
Step 2 — relative lowering. (1.013 − 1.004) ÷ 1.013 = 0.009 ÷ 1.013 = 0.00888.
Step 3 — moles of solute. Using n₂ ÷ (n₁ + n₂) = 0.00888 ⇒ n₂ = 0.00888 × 5.444 ÷ (1 − 0.00888) = 0.0488 mol. (The approximate form n₂ = 0.00888 × 5.444 = 0.04836 mol gives 41.35 g mol⁻¹.)
M₂ = w₂ ÷ n₂ = 2 g ÷ 0.0488 mol ≈ 41 g mol⁻¹
Extra depth. The exact solution from p₁ = x₁p₁° gives M ≈ 41 g mol⁻¹, in contrast to 41.35 by the approximation n₁ + n₂ ≈ n₁. Using the normal boiling point as the reference ensures p° = 1.013 bar, a useful trick when the vapour pressure of the pure solvent is not otherwise given.
Conclusion: Molar mass of the solute ≈ 41 g mol⁻¹ (41.35 g mol⁻¹ by the dilute-solution approximation).
Exercise 1.16 — Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?
Concept. An ideal solution obeys Raoult’s law for both components: p_total = x_h p_h° + x_o p_o°.
Moles. Heptane C₇H₁₆: M = 100 g mol⁻¹, n = 26.0 ÷ 100 = 0.26 mol. Octane C₈H₁₈: M = 114 g mol⁻¹, n = 35 ÷ 114 = 0.307 mol. Total = 0.567 mol.
Mole fractions. x(heptane) = 0.26 ÷ 0.567 = 0.4586; x(octane) = 0.307 ÷ 0.567 = 0.5414.
p(heptane) = 0.4586 × 105.2 = 48.24 kPa; p(octane) = 0.5414 × 46.8 = 25.34 kPa
p_total = 48.24 + 25.34 = 73.6 kPa
Extra depth. An ideal solution of n-alkanes (heptane and octane) obeys Raoult’s law because the intermolecular forces are almost identical; the total pressure 73.6 kPa lies between the pure pressures (46.8 and 105.2 kPa) in proportion to the composition. Such mixtures are the basis of the octane rating of petrol (isooctane/n-heptane).
Conclusion: The vapour pressure of the mixture is 73.6 kPa.
Exercise 1.17 — The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.
Concept. A 1 molal solution contains 1 mol of solute in 1000 g of water. n(water) = 1000 ÷ 18 = 55.5 mol.
x(water) = 55.5 ÷ (55.5 + 1) = 0.9823; x(solute) = 1 ÷ 56.5 = 0.0177
p = x₁p₁° = 0.9823 × 12.3 kPa = 12.08 kPa
Check through the lowering: Δp = x₂p₁° = 0.0177 × 12.3 = 0.218 kPa; 12.3 − 0.218 = 12.08 kPa.
Extra depth. The depressed vapour pressure for a 1 molal solution (12.08 kPa against 12.3 kPa) is a tiny change (1.8 %), illustrating why vapour-pressure lowering needs sensitive manometers. For dilute solutions Δp ≈ x₂p° ≈ m × M₁/1000 × p°, which gives the same answer to a good approximation.
Conclusion: Vapour pressure of the solution = 12.08 kPa.
Exercise 1.18 — Calculate the mass of a non-volatile solute (molar mass 40 g mol⁻¹) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
Concept. The vapour pressure is reduced to 80 % of p°, so p = 0.8 p°, i.e. the mole fraction of octane x₁ = 0.8 and that of the solute x₂ = 0.2.
Octane C₈H₁₈: M = 114 g mol⁻¹; n₁ = 114 ÷ 114 = 1 mol.
x₂ = n₂ ÷ (n₁ + n₂) = 0.2 ⇒ n₂ = 0.2 (1 + n₂) ⇒ 0.8 n₂ = 0.2 ⇒ n₂ = 0.25 mol
Mass of solute = 0.25 mol × 40 g mol⁻¹ = 10 g
Extra depth. For octane M = 114 g mol⁻¹, so 114 g is exactly 1 mol; this convenient number makes the calculation straightforward. Because x₂ = 0.2 here, the solution is not dilute and the exact formula must be used rather than the dilute-solution approximation, which would give 8 g (0.2 mol) instead of 10 g.
Conclusion: 10 g of the solute must be dissolved.
Exercise 1.19 — A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate: (i) molar mass of the solute (ii) vapour pressure of water at 298 K.
Concept. Raoult’s law for a non-volatile solute: (p° − p) ÷ p = n₂ ÷ n₁ (where n₂ ≪ n₁ is not assumed; written exactly as p° ÷ p = (n₁ + n₂) ÷ n₁). Let M be the molar mass of the solute and p° the vapour pressure of water.
Case 1: n₂ = 30 ÷ M; n₁ = 90 ÷ 18 = 5 mol. (p° − 2.8) ÷ 2.8 = n₂ ÷ 5 … (1)
Case 2: water = 90 + 18 = 108 g, n₁ = 6 mol. (p° − 2.9) ÷ 2.9 = n₂ ÷ 6 … (2)
From (1): n₂ = 5(p°/2.8 − 1) = 1.7857p° − 5. From (2): n₂ = 6(p°/2.9 − 1) = 2.0690p° − 6. Equating: 1.7857p° − 5 = 2.0690p° − 6 ⇒ 0.2833p° = 1 ⇒ p° = 3.53 kPa.
Then n₂ = 1.7857 × 3.53 − 5 = 1.304 mol, and M = 30 ÷ 1.304 = 23 g mol⁻¹.
(i) M = 23 g mol⁻¹ (ii) p° (water, 298 K) = 3.53 kPa
Extra depth. Two vapour-pressure measurements at known compositions let us find both the molar mass of the solute and the vapour pressure of the pure solvent without separately measuring the latter. The value p° = 3.53 kPa differs somewhat from the tabulated vapour pressure of water at 298 K (≈ 3.17 kPa) because the data supplied in the problem are rounded; the method, not the exact value, is being tested.
Conclusion: Molar mass of solute = 23 g mol⁻¹; vapour pressure of water at 298 K = 3.53 kPa.
Exercise 1.20 — A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
Concept. ΔT_f = K_f · m. K_f is a property of the solvent (water), so it can be obtained from the cane sugar data and reused for glucose.
Cane sugar (sucrose, C₁₂H₂₂O₁₁, M = 342). ΔT_f = 273.15 − 271 = 2.15 K. 5 g sucrose in 95 g water: m = (5 ÷ 342) ÷ 0.095 = 0.1539 mol kg⁻¹. Effective K_f = 2.15 ÷ 0.1539 = 13.97 K kg mol⁻¹ (note: this is not the textbook 1.86 — the data given in the question are approximate, so the question expects the ratio method).
Glucose (C₆H₁₂O₆, M = 180). 5 g in 95 g water: m = (5 ÷ 180) ÷ 0.095 = 0.2924 mol kg⁻¹.
ΔT_f = 13.97 × 0.2924 = 4.085 K
T_f (solution) = 273.15 − 4.085 = 269.07 K
Equivalent ratio method: ΔT_f(glucose) = ΔT_f(sugar) × (m_glucose ÷ m_sugar) = 2.15 × (342 ÷ 180) = 4.085 K, because equal masses give mole ratio 342/180.
Extra depth. The result K_f ≈ 13.97 shows that the “5 % cane sugar freezes at 271 K” data of the question are not consistent with water’s true K_f of 1.86; such inconsistent problem data are common, so we use the ratio of molalities, which is independent of the numerical value of K_f. Equal mass percentages of two non-electrolytes give depressions in the inverse ratio of molar masses (342/180).
Conclusion: The freezing point of the 5 % glucose solution is ≈ 269.07 K.
Exercise 1.21 — Two elements A and B form compounds having formula AB₂ and AB₄. When dissolved in 20 g of benzene (C₆H₆), 1 g of AB₂ lowers the freezing point by 2.3 K whereas 1.0 g of AB₄ lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol⁻¹. Calculate atomic masses of A and B.
Concept. Molar mass from freezing point depression: M₂ = (1000 × K_f × w₂) ÷ (ΔT_f × w₁).
M(AB₂) = (1000 × 5.1 × 1) ÷ (2.3 × 20) = 110.87 g mol⁻¹
M(AB₄) = (1000 × 5.1 × 1.0) ÷ (1.3 × 20) = 196.15 g mol⁻¹
Let the atomic masses be A and B. Then A + 2B = 110.87 … (i) and A + 4B = 196.15 … (ii). Subtract (i) from (ii): 2B = 85.28 ⇒ B = 42.64. Substitute in (i): A = 110.87 − 85.28 = 25.59.
| Compound | Molar mass (g mol⁻¹) |
|---|---|
| AB₂ | 110.87 |
| AB₄ | 196.15 |
Extra depth. This is a classic application of freezing-point depression to find atomic masses of elements from the molar masses of two compounds. The consistency check A + 2B = 110.87 and A + 4B = 196.15 gives B ≈ 42.6 and A ≈ 25.6, close to the values for Ca (40) and Mg (24).
Conclusion: Atomic mass of A = 25.59 u and of B = 42.64 u.
Exercise 1.22 — At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
Concept. π = CRT; at constant T, π is directly proportional to concentration (π₁ ÷ π₂ = C₁ ÷ C₂).
C₁ = 36 g L⁻¹ ÷ 180 g mol⁻¹ = 0.2 mol L⁻¹ at π₁ = 4.98 bar.
C₂ = C₁ × (π₂ ÷ π₁) = 0.2 × (1.52 ÷ 4.98) = 0.061 mol L⁻¹
(In mass terms 0.061 × 180 = 11 g glucose per litre.)
Extra depth. Isotonic solutions have the same osmotic pressure; a 0.2 M glucose solution (4.98 bar at 300 K) is hypertonic to a 0.061 M solution (1.52 bar). The ratio method works because R and T are the same, which is typical for osmotic-pressure comparisons in biology (intravenous fluids must be isotonic with blood).
Conclusion: The concentration of the more dilute solution is 0.061 M.
Exercise 1.23 — Suggest the most important type of intermolecular attractive interaction in the following pairs. (i) n-hexane and n-octane (ii) I₂ and CCl₄ (iii) NaClO₄ and water (iv) methanol and acetone (v) acetonitrile (CH₃CN) and acetone (C₃H₆O).
Concept. The type of attraction depends on whether the molecules are non-polar, polar, ionic or capable of hydrogen bonding.
| Pair | Nature of species | Most important interaction |
|---|---|---|
| (i) n-hexane + n-octane | Both non-polar hydrocarbons | London dispersion forces (instantaneous dipole – induced dipole) |
| (ii) I₂ + CCl₄ | Both non-polar | London dispersion forces (induced dipole – induced dipole) |
| (iii) NaClO₄ + water | Ionic solid + polar H₂O | Ion–dipole interaction (Na⁺ and ClO₄⁻ solvated by water) |
| (iv) CH₃OH + CH₃COCH₃ | Both polar; methanol O–H donor, acetone C=O acceptor | Dipole–dipole; hydrogen bonding between O–H of methanol and C=O of acetone |
| (v) CH₃CN + CH₃COCH₃ | Both polar aprotic (no O–H/N–H) | Dipole–dipole interaction |
Extra depth. Intermolecular forces in solutions are ranked: ion–dipole (strongest, for ionic solutes in water) > hydrogen bonding > dipole–dipole > London forces (weakest). The same ranking explains why NaClO₄ dissolves in water (ion–dipole) while I₂ dissolves in CCl₄ (induced dipole–induced dipole).
Conclusion: (i) and (ii) London forces; (iii) ion–dipole; (iv) and (v) dipole–dipole (H-bonding possible in (iv)).
Exercise 1.24 — Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH₃OH, CH₃CN.
Concept. n-Octane is a non-polar solvent held together by London dispersion forces. By “like dissolves like”, non-polar solutes dissolve best in it, and polar or ionic solutes are poorly soluble because their strong solute–solute interactions are not compensated by weak solute–octane interactions.
KCl < CH₃OH < CH₃CN < Cyclohexane
- KCl — an ionic solid with very high lattice energy; octane cannot solvate the ions (no ion–dipole interaction), so it is practically insoluble.
- CH₃OH — strongly hydrogen-bonded; octane cannot break these H-bonds, so very sparingly soluble (the small non-polar CH₃ group helps slightly).
- CH₃CN — polar with only dipole–dipole forces, which are weaker than H-bonds, and the nitrile has a non-polar CH₃ end; so it is more soluble than methanol.
- Cyclohexane — non-polar, interacts with octane through London dispersion forces of similar magnitude; fully miscible.
Extra depth. Solubility in a non-polar solvent follows the polarity of the solute — the less polar the solute, the more soluble. KCl (ionic, with a very large lattice enthalpy) is practically insoluble in octane because there is no ion–dipole stabilisation to pay back the lattice energy, whereas cyclohexane mixes in all proportions.
Conclusion: Increasing order of solubility in n-octane: KCl < CH₃OH < CH₃CN < cyclohexane.
Exercise 1.25 — Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water? (i) phenol (ii) toluene (iii) formic acid (iv) ethylene glycol (v) chloroform (vi) pentanol.
Concept. Solubility in water depends on the ability of the compound to form hydrogen bonds with water and on the size of its non-polar (hydrocarbon) part.
| Compound | Structure feature | Solubility in water |
|---|---|---|
| (i) Phenol | –OH H-bonds, but large non-polar benzene ring | Partially soluble (≈ 8 g per 100 mL) |
| (ii) Toluene | Non-polar hydrocarbon; no H-bonding | Insoluble |
| (iii) Formic acid | Small, –COOH forms strong H-bonds | Highly soluble (miscible) |
| (iv) Ethylene glycol | Two –OH groups, small molecule | Highly soluble (miscible) |
| (v) Chloroform | Weakly polar; hardly any H-bonding with water | Insoluble |
| (vi) Pentanol | –OH group but five-carbon hydrocarbon chain | Partially soluble |
Extra depth. Phenol is partly soluble because its hydrophobic benzene ring offsets the –OH H-bonding (about 8 g per 100 mL at 298 K); pentanol is partly soluble because its five-carbon chain is large (a 5-carbon alcohol dissolves only to ≈ 2.7 g per 100 mL), while methanol and ethanol are fully miscible. Chloroform and toluene are insoluble because they cannot form strong H-bonds with water.
Conclusion: Insoluble: toluene, chloroform; partially soluble: phenol, pentanol; highly soluble: formic acid, ethylene glycol.
Exercise 1.26 — If the density of some lake water is 1.25 g mL⁻¹ and contains 92 g of Na⁺ ions per kg of water, calculate the molarity of Na⁺ ions in the lake.
Step 1 — moles of Na⁺. n = 92 g ÷ 23 g mol⁻¹ = 4 mol per kg of water.
Step 2 — volume of solution. Mass of solution = 1000 g water + 92 g Na⁺ = 1092 g. V = 1092 ÷ 1.25 = 873.6 mL = 0.8736 L.
M = 4 mol ÷ 0.8736 L = 4.58 mol L⁻¹
Extra depth. Real-water samples such as lake water contain dissolved sodium salts in concentrations far below 4 M; the value here (4.58 M) is an exaggerated illustration of the interconversion between mass of solute per kg of water and molarity. Remember that the density refers to the solution (1.25 g mL⁻¹), not to the solvent.
Conclusion: Molarity of Na⁺ in the lake water = 4.58 M.
Exercise 1.27 — If the solubility product of CuS is 6 × 10⁻¹⁶, calculate the maximum molarity of CuS in aqueous solution.
Concept. CuS(s) ⇌ Cu²⁺(aq) + S²⁻(aq). If s is the solubility (maximum molarity), [Cu²⁺] = [S²⁻] = s, so K_sp = s × s = s².
s = √K_sp = √(6 × 10⁻¹⁶) = 2.45 × 10⁻⁸ mol L⁻¹
This extremely low value shows CuS is practically insoluble — that is why Cu²⁺ is precipitated as black CuS in group II qualitative analysis.
Extra depth. CuS is the least soluble of the common sulphides, which is why Cu²⁺ is precipitated from acidified solution in Group II by H₂S; in such a solution [S²⁻] is kept low by the common-ion effect of H⁺, and yet the ionic product still exceeds the tiny K_sp. A solubility of 2.45 × 10⁻⁸ M corresponds to about 2.3 μg per litre.
Conclusion: Maximum molarity of CuS = 2.45 × 10⁻⁸ M.
Exercise 1.28 — Calculate the mass percentage of aspirin (C₉H₈O₄) in acetonitrile (CH₃CN) when 6.5 g of C₉H₈O₄ is dissolved in 450 g of CH₃CN.
Concept. Mass % = (mass of solute ÷ mass of solution) × 100. Mass of solution = 6.5 g + 450 g = 456.5 g.
Mass % of aspirin = (6.5 ÷ 456.5) × 100 = 1.42 %
The mass of the solvent only (450 g) must not be used in the denominator — that would give 1.44 %.
Extra depth. Aspirin (acetylsalicylic acid, C₉H₈O₄, M = 180 g mol⁻¹) is soluble in acetonitrile; the solution here is dilute (1.42 %) so the density is close to that of acetonitrile. Mass percentage is the usual way pharmacists express drug content in preparations.
Conclusion: The solution is 1.42 % aspirin by mass.
Exercise 1.29 — Nalorphene (C₁₉H₂₁NO₃), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 × 10⁻³ m aqueous solution required for the above dose.
Concept. 1.5 × 10⁻³ m means 1.5 × 10⁻³ mol of nalorphene per 1 kg (1000 g) of water.
Molar mass of C₁₉H₂₁NO₃ = 19 × 12 + 21 × 1 + 14 + 3 × 16 = 228 + 21 + 14 + 48 = 311 g mol⁻¹. Mass of nalorphene in 1 kg of water = 1.5 × 10⁻³ × 311 = 0.4665 g = 466.5 mg.
Mass of that solution = 1000 g + 0.4665 g = 1000.4665 g, which contains 466.5 mg solute. So the required mass of solution for 1.5 mg of drug:
Mass = (1.5 mg ÷ 466.5 mg) × 1000.4665 g = 3.217 g ≈ 3.22 g
Extra depth. A very dilute solution (1.5 × 10⁻³ m) of nalorphene has a density nearly identical to water, so the mass of solution is almost equal to the water mass. This problem shows how drug dosages are converted into volumes/masses of a standard solution in clinical practice.
Conclusion: 3.22 g of the 1.5 × 10⁻³ m solution supplies the 1.5 mg dose.
Exercise 1.30 — Calculate the amount of benzoic acid (C₆H₅COOH) required for preparing 250 mL of 0.15 M solution in methanol.
Step 1 — moles needed. n = M × V = 0.15 mol L⁻¹ × 0.250 L = 0.0375 mol.
Step 2 — mass. Molar mass of C₆H₅COOH (C₇H₆O₂) = 84 + 6 + 32 = 122 g mol⁻¹.
Mass = 0.0375 × 122 = 4.575 g
Procedure: weigh 4.575 g benzoic acid, dissolve it in some methanol in a 250 mL volumetric flask, and make up to the mark with methanol.
Extra depth. Preparing a solution of a given molarity requires a volumetric flask: weigh 4.575 g of benzoic acid (M = 122 g mol⁻¹), dissolve it, transfer quantitatively and make up to the mark. Methanol is used as the solvent because benzoic acid is much more soluble in it than in cold water.
Conclusion: 4.575 g of benzoic acid is needed.
Exercise 1.31 — The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.
Concept. Depression in freezing point is a colligative property — it depends on the total number of particles in solution: ΔT_f = i K_f m. For a weak acid HA ⇌ H⁺ + A⁻, a larger degree of dissociation gives a larger van’t Hoff factor i and hence a larger ΔT_f for the same amount of acid.
Acid strength. The electron-withdrawing halogen atoms stabilise the carboxylate anion through the –I (inductive) effect and increase the acidity: F is more electronegative than Cl, and three halogens act more strongly than none.
| Acid | Inductive effect | Dissociation | i | ΔT_f |
|---|---|---|---|---|
| CH₃COOH | +I of CH₃ (destabilises anion) | Least | smallest | smallest |
| CCl₃COOH | –I of three Cl | Higher | larger | larger |
| CF₃COOH | –I of three F (strongest) | Highest | largest | largest |
Extra depth. The acid strength follows the –I effect of the substituents: CF₃ (strongest withdrawal) > CCl₃ > CH₃. Acetic acid has pK_a ≈ 4.76, whereas trichloroacetic and trifluoroacetic acids are several orders of magnitude stronger; CF₃COOH is almost completely dissociated, so i approaches 2 — roughly twice the depression of an undissociated solute.
Conclusion: Acidity (and so dissociation, i and ΔT_f) increases in the order CH₃COOH < CCl₃COOH < CF₃COOH, so the freezing point depression increases in that order.
Exercise 1.32 — Calculate the depression in the freezing point of water when 10 g of CH₃CH₂CHClCOOH is added to 250 g of water. K_a = 1.4 × 10⁻³, K_f = 1.86 K kg mol⁻¹.
Concept. The acid is weak and partly dissociates (HA ⇌ H⁺ + A⁻), so the van’t Hoff factor i = 1 + α is required: ΔT_f = i K_f m.
Step 1 — molality. Molar mass of C₄H₇ClO₂ = 48 + 7 + 35.5 + 32 = 122.5 g mol⁻¹. n = 10 ÷ 122.5 = 0.0816 mol; m = 0.0816 ÷ 0.250 = 0.3265 mol kg⁻¹ (taken ≈ molarity C).
Step 2 — degree of dissociation. K_a = Cα² ÷ (1 − α): 1.4 × 10⁻³ = 0.3265 α² ÷ (1 − α). Solving the quadratic 0.3265α² + 1.4 × 10⁻³α − 1.4 × 10⁻³ = 0 gives α = 0.0634 (the approximation α = √(K_a ÷ C) = 0.0655 gives nearly the same).
Step 3 — i. i = 1 + α = 1.0634.
ΔT_f = 1.0634 × 1.86 × 0.3265 = 0.646 K ≈ 0.65 K
Extra depth. Because the acid is weak, only about 6.3 % dissociates at this concentration (α = 0.063); the exact quadratic is used rather than α = √(K_a/C) = 0.0655, which is slightly higher. The resulting i = 1.063 raises the depression from 0.607 K (undissociated) to 0.65 K — a small but real correction.
Conclusion: Depression in freezing point ≈ 0.65 K.
Exercise 1.33 — 19.5 g of CH₂FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0 °C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.
Step 1 — molality. Molar mass of CH₂FCOOH (C₂H₃FO₂) = 24 + 3 + 19 + 32 = 78 g mol⁻¹. n = 19.5 ÷ 78 = 0.25 mol; m = 0.25 ÷ 0.5 = 0.5 mol kg⁻¹.
Step 2 — calculated depression (no dissociation) = K_f m = 1.86 × 0.5 = 0.93 K. Observed = 1.0 K.
i = observed ÷ calculated = 1.0 ÷ 0.93 = 1.0753
Step 3 — degree of dissociation. For HA ⇌ H⁺ + A⁻: i = 1 + α ⇒ α = 0.0753.
Step 4 — K_a. Concentration of H⁺ = A⁻ = Cα = 0.5 × 0.0753 = 0.03765 M; HA = C(1 − α) = 0.5 × 0.9247 = 0.4624 M.
K_a = Cα² ÷ (1 − α) = 0.5 × (0.0753)² ÷ 0.9247 = 3.07 × 10⁻³
Extra depth. Fluoroacetic acid dissociates to about 7.5 % at this concentration, giving i = 1.0753; the value of K_a obtained (3.07 × 10⁻³) lies between that of chloroacetic acid (1.4 × 10⁻³) and dichloroacetic acid (5 × 10⁻²), in line with the –I effect of fluorine. Colligative-property measurements are a standard way to measure the degree of dissociation of weak electrolytes.
Conclusion: van’t Hoff factor i = 1.0753; K_a = 3.07 × 10⁻³.
Exercise 1.34 — Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
Moles. n(glucose) = 25 ÷ 180 = 0.1389 mol; n(water) = 450 ÷ 18 = 25 mol.
x(water) = 25 ÷ (25 + 0.1389) = 0.99447
p = x₁ p₁° = 0.99447 × 17.535 = 17.44 mm Hg
Lowering = 17.535 − 17.44 = 0.097 mm Hg (relative lowering = x₂ = 0.00553).
Extra depth. The vapour-pressure lowering of 0.097 mm Hg is very small for a 5.3 % glucose solution — the reason why a sensitive differential manometer is needed. Raoult’s law is exact only for ideal solutions, but dilute solutions of non-electrolytes behave almost ideally.
Conclusion: Vapour pressure of the solution = 17.44 mm Hg.
Exercise 1.35 — Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 × 10⁵ mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.
Concept. Here Henry’s law is written for molality: p = K_H · m. Hence molality of dissolved methane = p ÷ K_H.
m = 760 mm Hg ÷ 4.27 × 10⁵ mm Hg mol⁻¹ kg = 1.78 × 10⁻³ mol kg⁻¹
The solubility is 1.78 × 10⁻³ mol of CH₄ per kg of benzene (≈ 0.0285 g CH₄ per kg). A K_H of this size means methane is only slightly soluble, as expected for a non-polar gas at 1 atm.
Extra depth. Methane is non-polar and dissolves in benzene (also non-polar) much more readily than in water; the solubility 1.78 × 10⁻³ mol kg⁻¹ at 760 mm Hg is typical of a sparingly soluble gas. Because K_H here is given for molality, the law reads p = K_H m — different forms of Henry’s law exist for mole fraction, molality and molarity.
Conclusion: Solubility of methane = 1.78 × 10⁻³ mol kg⁻¹.
Exercise 1.36 — 100 g of liquid A (molar mass 140 g mol⁻¹) was dissolved in 1000 g of liquid B (molar mass 180 g mol⁻¹). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.
Moles. n_A = 100 ÷ 140 = 0.714 mol; n_B = 1000 ÷ 180 = 5.556 mol; total = 6.270 mol.
x_A = 0.714 ÷ 6.270 = 0.1139; x_B = 0.8861.
Partial pressure of B = x_B p_B° = 0.8861 × 500 = 443.05 torr. Partial pressure of A in solution = 475 − 443.05 = 31.95 torr.
p_A° = p_A ÷ x_A = 31.95 ÷ 0.1139 = 280.5 torr
| Quantity | Value |
|---|---|
| Vapour pressure of A in solution | 31.9 torr |
| Vapour pressure of pure A (p_A°) | 280.4 torr |
Extra depth. Liquid A is only about 9 % of the solution by mass and 11 mole per cent, so its vapour pressure in the solution (31.9 torr) is a small fraction of p_A° (280.4 torr), while the total pressure (475 torr) is only slightly below p_B° = 500 torr because B dominates the solution.
Conclusion: p_A° ≈ 280.4 torr; vapour pressure of A in solution ≈ 31.9 torr.
Exercise 1.37 — Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot p_total, p_chloroform and p_acetone as a function of x_acetone. The experimental data observed for different compositions of mixture is given. Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.
Step 1 — ideal (Raoult’s law) lines. p_acetone = x·741.8; p_chloroform = (1 − x)·632.8; p_total = p_chloroform° + (p_acetone° − p_chloroform°) x = 632.8 + 109 x. Each is a straight line: p_acetone from 0 to 741.8, p_chloroform from 632.8 to 0, p_total from 632.8 to 741.8.
| 100 x(acetone) | p(acetone) ideal | p(acetone) observed | p(CHCl₃) ideal | p(CHCl₃) observed | p total ideal | p total observed |
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 632.8 | 632.8 | 632.8 | 632.8 |
| 11.8 | 87.5 | 54.9 | 558.1 | 548.1 | 645.6 | 603.0 |
| 23.4 | 173.6 | 110.1 | 484.7 | 469.4 | 658.3 | 579.5 |
| 36.0 | 267.0 | 202.4 | 405.0 | 359.7 | 672.0 | 562.1 |
| 50.8 | 376.8 | 322.7 | 311.3 | 257.7 | 688.1 | 580.4 |
| 58.2 | 431.7 | 405.9 | 264.5 | 193.6 | 696.2 | 599.5 |
| 64.5 | 478.5 | 454.1 | 224.7 | 161.2 | 703.2 | 615.3 |
| 72.1 | 534.8 | 521.1 | 176.5 | 120.7 | 711.3 | 641.8 |
Step 2 — interpretation. At every composition the observed partial pressures of acetone and of chloroform, and hence the total pressure, are lower than the Raoult’s law values. The observed curves sag below the ideal straight lines (maximum deviation near x ≈ 0.5, total 580 mm Hg against 688 mm Hg ideal).
Reason. Chloroform’s C–H is acidic enough to form a hydrogen bond with the carbonyl oxygen of acetone (Cl₃C–H···O=C(CH₃)₂). These A–B interactions are stronger than A–A and B–B interactions, reducing the escaping tendency of the molecules, so Δ_mix H < 0.
Extra depth. The observed pressures lie below the Raoult-law lines for all compositions, giving a negative deviation. The consequence of such a negative deviation is a negative Δ_mix H (heat is evolved when the liquids are mixed) and, if the deviation is large enough, a maximum-boiling azeotrope — acetone–chloroform is a standard example.
Conclusion: Acetone–chloroform shows negative deviation from Raoult’s law.
Exercise 1.38 — Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.
Moles. Benzene C₆H₆ = 78 g mol⁻¹: n = 80 ÷ 78 = 1.0256 mol. Toluene C₇H₈ = 92 g mol⁻¹: n = 100 ÷ 92 = 1.0870 mol. Total = 2.1126 mol.
x(benzene) = 1.0256 ÷ 2.1126 = 0.4855; x(toluene) = 0.5145.
p(benzene) = 0.4855 × 50.71 = 24.62 mm Hg; p(toluene) = 0.5145 × 32.06 = 16.49 mm Hg
p_total = 41.11 mm Hg; y(benzene) = 24.62 ÷ 41.11 = 0.599 ≈ 0.60
The vapour is richer in benzene (0.60 against 0.49 in the liquid) because benzene is more volatile; y(toluene) = 0.40.
Extra depth. Benzene–toluene is the textbook ideal solution: both are non-polar aromatics with similar size and forces, so Δ_mix H = 0 and Δ_mix V = 0. Industrial distillation columns exploit the difference in volatility (y_B = 0.60 versus x_B = 0.49) to separate the two.
Conclusion: Mole fraction of benzene in the vapour phase ≈ 0.60.
Exercise 1.39 — The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry’s law constants for oxygen and nitrogen at 298 K are 3.30 × 10⁷ mm and 6.51 × 10⁷ mm respectively, calculate the composition of these gases in water.
Concept. Volume % of an ideal gas mixture = mole %, so partial pressure = fraction × total pressure. Total pressure = 10 atm = 10 × 760 = 7600 mm Hg.
p(O₂) = 0.20 × 7600 = 1520 mm Hg; p(N₂) = 0.79 × 7600 = 6004 mm Hg.
x(O₂) = p ÷ K_H = 1520 ÷ 3.30 × 10⁷ = 4.6 × 10⁻⁵
x(N₂) = 6004 ÷ 6.51 × 10⁷ = 9.22 × 10⁻⁵
So 100 mol of water at 10 atm dissolves about 4.6 × 10⁻³ mol O₂ and 9.2 × 10⁻³ mol N₂. Oxygen is relatively more soluble than its partial pressure suggests (K_H of O₂ is half that of N₂), which is why dissolved air is richer in oxygen than the atmosphere.
Extra depth. The dissolved gas content of water in equilibrium with air at 10 atm (diving depth ≈ 90 m) is about ten times that at 1 atm, which is why divers are at risk of nitrogen narcosis and the bends; the mole fraction of N₂ is about twice that of O₂ even though the air is only 4 times richer in N₂ — a result of the different K_H values (6.51 × 10⁷ against 3.30 × 10⁷ mm).
Conclusion: In water: mole fraction of O₂ = 4.6 × 10⁻⁵; of N₂ = 9.22 × 10⁻⁵.
Exercise 1.40 — Determine the amount of CaCl₂ (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27 °C.
Concept. π = i C R T = i (n ÷ V) R T = i (w ÷ M V) R T.
R = 0.0821 L atm K⁻¹ mol⁻¹; T = 27 + 273 = 300 K; V = 2.5 L; M(CaCl₂) = 40 + 71 = 111 g mol⁻¹.
w = π M V ÷ (i R T) = (0.75 × 111 × 2.5) ÷ (2.47 × 0.0821 × 300) = 208.1 ÷ 60.84 = 3.42 g
Equivalently, C = π ÷ (iRT) = 0.75 ÷ 60.84 = 0.01233 M; n = 0.01233 × 2.5 = 0.0308 mol; mass = 0.0308 × 111 = 3.42 g.
Extra depth. Calcium chloride is a strong electrolyte whose van’t Hoff factor i = 2.47 (close to the theoretical 3) reflects ion-pair formation at finite concentration. With the value of π = 0.75 atm the solution is dilute (about 0.0123 M), so i is close to its limiting value.
Conclusion: 3.42 g of CaCl₂ must be dissolved.
Exercise 1.41 — Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K₂SO₄ in 2 litre of water at 25 °C, assuming that it is completely dissociated.
Concept. Complete dissociation: K₂SO₄ → 2K⁺ + SO₄²⁻, so 3 ions are formed and i = 3.
Molar mass K₂SO₄ = 78 + 32 + 64 = 174 g mol⁻¹. n = 0.025 g ÷ 174 = 1.437 × 10⁻⁴ mol. C = 1.437 × 10⁻⁴ ÷ 2 L = 7.18 × 10⁻⁵ mol L⁻¹. T = 298 K.
π = i C R T = 3 × 7.18 × 10⁻⁵ × 0.0821 × 298 = 5.27 × 10⁻³ atm
(≈ 0.534 kPa or 534 Pa.)
Extra depth. For a 1:2 electrolyte at this very low concentration (7.2 × 10⁻⁵ M) the dissociation is essentially complete, so i = 3 is a good approximation. The small osmotic pressure (5.27 × 10⁻³ atm, ≈ 534 Pa) illustrates why osmotic pressure of very dilute solutions of simple salts is measurable only with sensitive instruments.
Conclusion: Osmotic pressure ≈ 5.27 × 10⁻³ atm.