NCERT Solutions for Class 12 Chemistry Chapter 1: Solutions

These Class 12 Chemistry Chapter 1 solutions cover the complete NCERT chapter Solutions for session 2026–27. Every Intext Question and every numbered Exercise (1.1–1.41) is reproduced verbatim and solved step by step — concentration units, Henry’s law, Raoult’s law, ideal and non-ideal solutions, colligative properties and van’t Hoff factor — with every numerical worked out fully, units shown, and answers cross-checked against the NCERT answer key.

Class: 12 Subject: Chemistry Chapter: 1 Chapter Name: Solutions Exercises: Intext + 1.1–1.41 Session: 2026–27
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Class 12 Chemistry Chapter 1 Solutions – Overview

A solution is a homogeneous mixture of two or more substances whose composition is uniform throughout. This chapter studies mostly liquid solutions: how their concentration is expressed (mass %, volume %, ppm, mole fraction, molarity, molality), how gases dissolve in liquids (Henry’s law), how vapour pressure of liquid mixtures behaves (Raoult’s law), the difference between ideal and non-ideal solutions (positive/negative deviations and azeotropes), and the four colligative properties — relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure — which depend only on the number of solute particles. Finally, the van’t Hoff factor (i) accounts for solutes that dissociate or associate, explaining abnormal molar masses.

Key Concepts

Molarity (M): moles of solute per litre of solution; depends on temperature (volume changes with T).

Molality (m): moles of solute per kilogram of solvent; independent of temperature.

Mole fraction (x): moles of a component ÷ total moles; sum of all mole fractions = 1; temperature-independent.

Henry’s law: at constant temperature the partial pressure of a gas above a solution is proportional to its mole fraction in solution, p = KHx. Higher KH means lower solubility.

Raoult’s law: for a solution of volatile liquids, the partial vapour pressure of each component is proportional to its mole fraction, pi = pi0xi.

Ideal solution: obeys Raoult’s law over the whole range; ΔmixH = 0 and ΔmixV = 0 (A–B forces ≈ A–A and B–B).

Non-ideal solution: positive deviation (A–B forces weaker; e.g. ethanol + acetone) or negative deviation (A–B forces stronger; e.g. chloroform + acetone). Large deviations give azeotropes.

Colligative properties: depend only on the number of solute particles — relative lowering of vapour pressure, elevation of boiling point (ΔTb = Kbm), depression of freezing point (ΔTf = Kfm), osmotic pressure (Π = CRT).

van’t Hoff factor (i): ratio of observed to calculated colligative property; i > 1 for dissociation, i < 1 for association.

Important Formulas

Mass % = (mass of component / total mass of solution) × 100

Mole fraction xA = nA / (nA + nB)

Molarity M = moles of solute / volume of solution (L)

Molality m = moles of solute / mass of solvent (kg)

Henry’s law: p = KHx

Raoult’s law: ptotal = p10x1 + p20x2

Relative lowering of VP: (p10 − p1) / p10 = x2

Elevation of boiling point: ΔTb = Kbm ;  M2 = (1000 · Kb · w2) / (ΔTb · w1)

Depression of freezing point: ΔTf = Kfm ;  M2 = (1000 · Kf · w2) / (ΔTf · w1)

Osmotic pressure: Π = CRT = (n2/V)RT ;  M2 = w2RT / (ΠV)

van’t Hoff factor: i = normal molar mass / abnormal molar mass = observed colligative property / calculated colligative property

Intext Questions — Solutions

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.

SOLUTION Total mass of solution = 22 g + 122 g = 144 g. Mass % of benzene = (22 / 144) × 100 = 15.28%. Mass % of CCl4 = (122 / 144) × 100 = 84.72%.

1.2 Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.

SOLUTION Take 100 g of solution → 30 g C6H6 and 70 g CCl4. Molar mass C6H6 = 78 g mol−1; moles = 30 / 78 = 0.3846 mol. Molar mass CCl4 = 154 g mol−1; moles = 70 / 154 = 0.4545 mol. x(benzene) = 0.3846 / (0.3846 + 0.4545) = 0.3846 / 0.8391 = 0.458 (CCl4 ≈ 0.542).

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.

SOLUTION (a) Molar mass Co(NO3)2·6H2O = 59 + 2(14 + 48) + 6(18) = 59 + 124 + 108 = 291 g mol−1. Moles = 30 / 291 = 0.1031 mol. Molarity = 0.1031 / 4.3 = 0.024 M. (b) Use M1V1 = M2V2: 0.5 × 30 = M2 × 500. M2 = (0.5 × 30) / 500 = 0.03 M.

1.4 Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution.

SOLUTION Molar mass of urea = 60 g mol−1. Let mass of urea = w g, so mass of water = (2500 − w) g. Molality = (w / 60) / [(2500 − w)/1000] = 0.25. 1000w / [60(2500 − w)] = 0.25 ⇒ 1000w = 15(2500 − w) = 37500 − 15w. 1015w = 37500 ⇒ w = 36.94 g of urea.

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.

SOLUTION Take 100 g solution → 20 g KI, 80 g water. Molar mass KI = 39 + 127 = 166 g mol−1. Moles KI = 20 / 166 = 0.1205 mol. (a) Molality = 0.1205 / (80/1000) = 0.1205 / 0.080 = 1.51 mol kg−1 (≈ 1.5 m). (b) Volume of solution = 100 / 1.202 = 83.19 mL = 0.08319 L. Molarity = 0.1205 / 0.08319 = 1.45 mol L−1. (c) Moles water = 80 / 18 = 4.444 mol. x(KI) = 0.1205 / (0.1205 + 4.444) = 0.1205 / 4.5645 = 0.0264.

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.

SOLUTION 0.195 m means 0.195 mol H2S in 1 kg (1000 g) water = 1000/18 = 55.55 mol water. x(H2S) = 0.195 / (0.195 + 55.55) = 0.195 / 55.74 = 3.50 × 10−3. At STP, p(H2S) = 0.987 bar. KH = p / x = 0.987 / (3.50 × 10−3) = 282 bar (≈ 0.282 kbar).

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.

SOLUTION p(CO2) = 2.5 atm = 2.5 × 1.013 × 105 Pa = 2.533 × 105 Pa. x(CO2) = p / KH = (2.533 × 105) / (1.67 × 108) = 1.517 × 10−3. 500 mL water ≈ 500 g = 500/18 = 27.78 mol. Since n(CO2) « n(water): n(CO2) = x × 27.78 = 1.517 × 10−3 × 27.78 = 0.04215 mol. Mass of CO2 = 0.04215 × 44 = 1.85 g of CO2.

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.

SOLUTION ptotal = pA0xA + pB0(1 − xA): 600 = 450xA + 700(1 − xA). 600 = 700 − 250xA ⇒ 250xA = 100 ⇒ xA = 0.4, xB = 0.6. pA = 450 × 0.4 = 180 mm Hg; pB = 700 × 0.6 = 420 mm Hg. yA = 180 / 600 = 0.30; yB = 420 / 600 = 0.70.

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.

SOLUTION Moles urea = 50 / 60 = 0.8333 mol; moles water = 850 / 18 = 47.22 mol. x(urea) = 0.8333 / (0.8333 + 47.22) = 0.8333 / 48.06 = 0.01734. Relative lowering = (p0 − p)/p0 = x(solute) = 0.0173. p = p0(1 − x2) = 23.8 × (1 − 0.01734) = 23.8 × 0.98266 = 23.4 mm Hg.

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.

SOLUTION ΔTb = 100 − 99.63 = 0.37°C = 0.37 K. Kb(water) = 0.52 K kg mol−1. M2(sucrose) = 342 g mol−1. Using ΔTb = (1000 Kb w2)/(M2 w1): w2 = (ΔTb × M2 × w1) / (1000 × Kb) = (0.37 × 342 × 500) / (1000 × 0.52). w2 = 63270 / 520 = 121.67 g of sucrose.

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. Kf = 3.9 K kg mol−1.

SOLUTION ΔTf = 1.5 K; M2(ascorbic acid) = 72 + 8 + 96 = 176 g mol−1. w2 = (ΔTf × M2 × w1) / (1000 × Kf) = (1.5 × 176 × 75) / (1000 × 3.9). w2 = 19800 / 3900 = 5.08 g of ascorbic acid.

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.

SOLUTION n2 = 1.0 / 185000 = 5.405 × 10−6 mol; V = 0.450 L; T = 310 K; R = 8.314 J K−1 mol−1. Π = (n2RT)/V = (5.405 × 10−6 × 8.314 × 310) / (0.450 × 10−3 m3). Π = (0.01393) / (4.5 × 10−4) = 30.96 Pa.

NCERT Exercises (1.1–1.41) — Solutions

1.1 Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

ANSWER A solution is a homogeneous mixture of two or more components whose composition and properties are uniform throughout. The component present in the largest amount is the solvent; the others are solutes. Depending on the physical state of solvent there are three broad types (nine sub-types), each with solute as gas, liquid or solid: Gaseous solutions (solvent = gas): e.g. mixture of O2 and N2 (gas in gas), chloroform vapour in N2 (liquid in gas), camphor in N2 (solid in gas). Liquid solutions (solvent = liquid): e.g. O2 in water (gas in liquid), ethanol in water (liquid in liquid), glucose in water (solid in liquid). Solid solutions (solvent = solid): e.g. H2 in palladium (gas in solid), amalgam of mercury with sodium (liquid in solid), copper in gold (solid in solid).

1.2 Give an example of a solid solution in which the solute is a gas.

ANSWER A solution of hydrogen gas dissolved (occluded) in palladium metal. Other valid examples include dissolved gases trapped in metals such as hydrogen in platinum.

1.3 Define the following terms: (i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage.

ANSWER (i) Mole fraction of a component = (moles of that component) / (total moles of all components). The sum of all mole fractions equals 1. (ii) Molality (m) = number of moles of solute per kilogram of solvent (unit: mol kg−1); independent of temperature. (iii) Molarity (M) = number of moles of solute per litre of solution (unit: mol L−1); depends on temperature. (iv) Mass percentage of a component = (mass of component / total mass of solution) × 100.

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?

SOLUTION Take 100 g solution → 68 g HNO3. Molar mass HNO3 = 63 g mol−1; moles = 68 / 63 = 1.079 mol. Volume of solution = mass / density = 100 / 1.504 = 66.49 mL = 0.06649 L. Molarity = 1.079 / 0.06649 = 16.23 M.

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?

SOLUTION Take 100 g solution → 10 g glucose, 90 g water. M(glucose) = 180 g mol−1; moles = 10/180 = 0.0556 mol. Moles water = 90/18 = 5.0 mol. Molality = 0.0556 / (90/1000) = 0.0556 / 0.090 = 0.617 mol kg−1. x(glucose) = 0.0556 / (0.0556 + 5.0) = 0.0556 / 5.0556 = 0.011; x(water) = 0.989. Volume = 100 / 1.2 = 83.33 mL = 0.08333 L. Molarity = 0.0556 / 0.08333 = 0.67 mol L−1.

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?

SOLUTION Let each be x mol. Masses: Na2CO3 = 106x, NaHCO3 = 84x. Total = 190x = 1 g ⇒ x = 5.263 × 10−3 mol of each. Reactions: Na2CO3 + 2HCl → 2NaCl + H2O + CO2 (needs 2 mol HCl); NaHCO3 + HCl → NaCl + H2O + CO2 (needs 1 mol HCl). Moles HCl = 2x + x = 3x = 3 × 5.263 × 10−3 = 0.01579 mol. Volume = moles / M = 0.01579 / 0.1 = 0.1579 L = 157.9 mL.

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.

SOLUTION Solute from first = 300 × 0.25 = 75 g; from second = 400 × 0.40 = 160 g. Total solute = 235 g. Total mass of solution = 300 + 400 = 700 g. Mass % = (235 / 700) × 100 = 33.57% (solute); water ≈ 66.43%.

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?

SOLUTION M(ethylene glycol) = 62 g mol−1; moles = 222.6 / 62 = 3.59 mol. Molality = 3.59 / (200/1000) = 3.59 / 0.200 = 17.95 mol kg−1. Total mass = 222.6 + 200 = 422.6 g; volume = 422.6 / 1.072 = 394.2 mL = 0.3942 L. Molarity = 3.59 / 0.3942 = 9.11 mol L−1.

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.

SOLUTION (i) 15 ppm means 15 g of CHCl3 in 106 g solution. Percent by mass = (15 / 106) × 100 = 1.5 × 10−3 %. (ii) M(CHCl3) = 119.5 g mol−1; moles = 15 / 119.5 = 0.1255 mol in ≈ 106 g (≈ 1000 kg) water. Molality = 0.1255 / 1000 = 1.25 × 10−4 mol kg−1.

1.10 What role does the molecular interaction play in a solution of alcohol and water?

ANSWER In pure alcohol and pure water, molecules are strongly held by their own hydrogen bonds. On mixing, the alcohol–water hydrogen bonds formed are weaker than the original alcohol–alcohol and water–water interactions. Because the new A–B interactions are weaker, the escaping tendency (and hence vapour pressure) of each component increases. The mixture therefore shows a positive deviation from Raoult’s law, with a small increase in volume and absorption of heat on mixing.

1.11 Why do gases always tend to be less soluble in liquids as the temperature is raised?

ANSWER Dissolution of a gas in a liquid is similar to condensation and is an exothermic process (heat is evolved): gas + solvent ⇌ solution + heat. By Le Chatelier’s principle, raising the temperature shifts this equilibrium backward (the gas is driven out), so the solubility of the gas decreases as temperature is raised.

1.12 State Henry’s law and mention some important applications.

ANSWER Henry’s law: at a constant temperature, the partial pressure of a gas in the vapour phase (p) is directly proportional to the mole fraction of the gas (x) in solution: p = KHx. Applications: (i) Soft drinks and soda water are sealed under high CO2 pressure to increase dissolved CO2. (ii) Scuba divers’ tanks are filled with air diluted with helium to avoid bends. (iii) At high altitudes low oxygen partial pressure lowers blood oxygen, causing anoxia in climbers.

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?

SOLUTION By Henry’s law, at fixed temperature solubility (mass dissolved, m) ∝ partial pressure p, so p1/m1 = p2/m2. p2 = p1 × (m2/m1) = 1 bar × (5.00 × 10−2) / (6.56 × 10−3). p2 = 1 × 7.62 = 7.62 bar.

1.14 What is meant by positive and negative deviations from Raoult’s law and how is the sign of ΔmixH related to positive and negative deviations from Raoult’s law?

ANSWER Positive deviation: the observed vapour pressure is higher than predicted by Raoult’s law. It occurs when A–B interactions are weaker than A–A and B–B; molecules escape more easily. Here mixing is endothermic, ΔmixH > 0 (and ΔmixV > 0). Example: ethanol + acetone. Negative deviation: the observed vapour pressure is lower than predicted. It occurs when A–B interactions are stronger than A–A and B–B; molecules escape less easily. Here mixing is exothermic, ΔmixH < 0 (and ΔmixV < 0). Example: chloroform + acetone.

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?

SOLUTION At the normal boiling point of water, p0 = 1.013 bar; p = 1.004 bar. 2% solution → 2 g solute, 98 g water. (p0 − p)/p0 = (w2 M1)/(M2 w1): (1.013 − 1.004)/1.013 = (2 × 18)/(M2 × 98). 0.009/1.013 = 36/(98 M2) ⇒ 8.884 × 10−3 = 36/(98 M2). M2 = 36 / (98 × 8.884 × 10−3) = 36 / 0.8707 = 41.35 g mol−1 (≈ 40.9 g mol−1).

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?

SOLUTION M(heptane, C7H16) = 100 g mol−1; moles = 26/100 = 0.26 mol. M(octane, C8H18) = 114 g mol−1; moles = 35/114 = 0.307 mol. x(heptane) = 0.26 / (0.26 + 0.307) = 0.26 / 0.567 = 0.4586; x(octane) = 0.5414. ptotal = 105.2 × 0.4586 + 46.8 × 0.5414 = 48.25 + 25.34 = 73.59 kPa.

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.

SOLUTION 1 molal = 1 mol solute in 1000 g water = 1000/18 = 55.55 mol water. x(solute) = 1 / (1 + 55.55) = 1 / 56.55 = 0.0177. (p0 − p)/p0 = x2 ⇒ p = p0(1 − x2) = 12.3 × (1 − 0.0177) = 12.3 × 0.9823 = 12.08 kPa.

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%.

SOLUTION Vapour pressure reduced to 80% means p = 0.8 p0, so (p0 − p)/p0 = 0.2 = x(solute). M(octane, C8H18) = 114 g mol−1; moles octane = 114/114 = 1 mol. Let solute moles = n2. x2 = n2 / (n2 + 1) = 0.2 ⇒ n2 = 0.2n2 + 0.2 ⇒ 0.8n2 = 0.2 ⇒ n2 = 0.25 mol. Mass of solute = 0.25 × 40 = 10 g.

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.

SOLUTION Let M = molar mass of solute and p0 = vapour pressure of pure water. Moles solute n2 = 30/M. Case 1: water = 90 g = 5 mol. (p0 − 2.8)/p0 = (30/M) / (30/M + 5). …(1) Case 2: water = 108 g = 6 mol. (p0 − 2.9)/p0 = (30/M) / (30/M + 6). …(2) Rearranging: p0/2.8 = (30/M + 5)/5 and p0/2.9 = (30/M + 6)/6. Dividing (1) by (2): 2.9/2.8 = [(30/M + 5)/5] ÷ [(30/M + 6)/6] = 6(30/M + 5) / [5(30/M + 6)]. Let a = 30/M. 1.0357 = 6(a + 5)/[5(a + 6)] ⇒ 1.0357 × 5(a + 6) = 6(a + 5) ⇒ 5.179(a + 6) = 6a + 30. 5.179a + 31.07 = 6a + 30 ⇒ 1.07 = 0.821a ⇒ a = 1.304 ⇒ M = 30/1.304 = 23.0 g mol−1. From Case 1: p0/2.8 = (1.304 + 5)/5 = 1.2608 ⇒ p0 = 3.53 kPa.

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.

SOLUTION For cane sugar: ΔTf = 273.15 − 271 = 2.15 K. 5% → 5 g sugar in 95 g water; M(sugar) = 342 g mol−1. ΔTf = (1000 Kf w2)/(M2 w1) ⇒ 2.15 = (1000 × Kf × 5)/(342 × 95) ⇒ Kf = (2.15 × 342 × 95)/5000 = 13.97 K kg mol−1. For glucose (M = 180): ΔTf = (1000 × 13.97 × 5)/(180 × 95) = 69850 / 17100 = 4.08 K. Freezing point of glucose solution = 273.15 − 4.08 = 269.07 K.

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.

SOLUTION M2 = (1000 Kf w2)/(ΔTf w1). For AB2: M = (1000 × 5.1 × 1)/(2.3 × 20) = 5100/46 = 110.87 g mol−1. For AB4: M = (1000 × 5.1 × 1)/(1.3 × 20) = 5100/26 = 196.15 g mol−1. Let A = a, B = b. Then a + 2b = 110.87 and a + 4b = 196.15. Subtracting: 2b = 85.28 ⇒ b = 42.64 u. a = 110.87 − 2(42.64) = 110.87 − 85.28 = 25.59 u. So A ≈ 25.58 u, B ≈ 42.64 u.

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?

SOLUTION Π = CRT. At constant T, Π ∝ C, so C1/Π1 = C2/Π2. First solution: C1 = (36/180) mol / 1 L = 0.2 M for Π1 = 4.98 bar. C2 = C1 × (Π2/Π1) = 0.2 × (1.52/4.98) = 0.2 × 0.3052 = 0.061 M.

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).

ANSWER (i) n-hexane & n-octane: both non-polar → London (dispersion) forces. (ii) I2 & CCl4: both non-polar → London (dispersion) forces. (iii) NaClO4 & water: ionic solute in polar solvent → ion–dipole interaction. (iv) methanol & acetone: methanol is polar with –OH → dipole–dipole and hydrogen bonding. (v) acetonitrile & acetone: both polar → dipole–dipole interaction.

1.24 Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH3OH, CH3CN.

ANSWER n-octane is a non-polar solvent, so “like dissolves like”: non-polar solutes dissolve best; ionic solutes least. Increasing solubility: KCl < CH3OH < CH3CN < Cyclohexane. KCl is ionic and almost insoluble; CH3OH is strongly hydrogen-bonded and polar (low solubility); CH3CN is polar but less so; cyclohexane is non-polar like octane and is most soluble.

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.

ANSWER Highly soluble (form strong H-bonds / fully miscible with water): formic acid (iii) and ethylene glycol (iv). Partially soluble (have both polar and non-polar parts): phenol (i) and pentanol (vi). Insoluble (non-polar / cannot H-bond effectively): toluene (ii) and chloroform (v).

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.

SOLUTION Moles of Na+ = 92 / 23 = 4 mol (in 1 kg = 1000 g of water). Taking the solution mass ≈ 1000 g (dilute, ions per kg of water), volume = mass/density = 1000 / 1.25 = 800 mL = 0.8 L. Molarity of Na+ = 4 / 0.8 = 5 mol L−1 (5 M).

1.27 If the solubility product of CuS is 6 × 10−16, calculate the maximum molarity of CuS in aqueous solution.

SOLUTION CuS ⇌ Cu2+ + S2−. Let solubility = s mol L−1, so [Cu2+] = [S2−] = s. Ksp = s2 = 6 × 10−16 ⇒ s = √(6 × 10−16) = √6 × 10−8. s = 2.45 × 10−8 mol L−1. Maximum molarity of CuS = 2.45 × 10−8 M.

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.

SOLUTION Total mass of solution = 6.5 + 450 = 456.5 g. Mass % of aspirin = (6.5 / 456.5) × 100 = 1.424%.

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.

SOLUTION M(nalorphene) = 19(12) + 21(1) + 14 + 3(16) = 228 + 21 + 14 + 48 = 311 g mol−1. Moles in dose = 1.5 × 10−3 g / 311 g mol−1 = 4.823 × 10−6 mol. Molality 1.5 × 10−3 m means 1.5 × 10−3 mol per 1 kg (1000 g) water. Mass of water for our moles = (4.823 × 10−6 / 1.5 × 10−3) × 1000 = 3.215 g. Mass of solution = water + nalorphene = 3.215 + 0.0015 ≈ 3.22 g.

1.30 Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol.

SOLUTION Moles required = M × V = 0.15 mol L−1 × 0.250 L = 0.0375 mol. M(benzoic acid, C6H5COOH = C7H6O2) = 122 g mol−1. Mass = 0.0375 × 122 = 4.575 g of benzoic acid.

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.

ANSWER Depression of freezing point depends on the number of ions produced, i.e. on the acid’s degree of dissociation, which increases with the acidic strength. The order of acid strength is acetic acid < trichloroacetic acid < trifluoroacetic acid, because the strongly electron-withdrawing halogens stabilise the carboxylate ion; fluorine (most electronegative) does this most strongly. Greater dissociation gives more ions and a higher van’t Hoff factor (i), so ΔTf increases in the same order: acetic < trichloroacetic < trifluoroacetic acid.

1.32 Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water. Ka = 1.4 × 10−3, Kf = 1.86 K kg mol−1.

SOLUTION M(CH3CH2CHClCOOH, C4H7ClO2) = 48 + 7 + 35.5 + 32 = 122.5 g mol−1. Moles = 10 / 122.5 = 0.0816 mol. Molality = 0.0816 / 0.250 = 0.3265 mol kg−1. The acid dissociates: HA ⇌ H+ + A−. Degree α = √(Ka/C) = √(1.4 × 10−3 / 0.3265) = √(4.288 × 10−3) = 0.0655. van’t Hoff factor i = 1 + α = 1.0655. ΔTf = i Kf m = 1.0655 × 1.86 × 0.3265 = 0.65 K.

1.33 19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.00°C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.

SOLUTION M(CH2FCOOH, C2H3FO2) = 24 + 3 + 19 + 32 = 78 g mol−1. Moles = 19.5 / 78 = 0.25 mol; molality = 0.25 / 0.5 = 0.5 mol kg−1. Calculated ΔTf (no dissociation) = Kf m = 1.86 × 0.5 = 0.93 K. i = observed/calculated = 1.00 / 0.93 = 1.0753. Degree α = i − 1 = 0.0753. Ka = Cα2/(1 − α) = (0.5 × 0.07532)/(1 − 0.0753) = (0.5 × 5.67 × 10−3)/0.9247 = 3.07 × 10−3.

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.

SOLUTION Moles glucose = 25/180 = 0.1389 mol; moles water = 450/18 = 25 mol. x(glucose) = 0.1389 / (0.1389 + 25) = 0.1389 / 25.139 = 5.525 × 10−3. p = p0(1 − x2) = 17.535 × (1 − 5.525 × 10−3) = 17.535 × 0.99448 = 17.44 mm Hg.

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.

SOLUTION By Henry’s law p = KHx ⇒ x(methane) = p / KH = 760 / (4.27 × 105). x(methane) = 1.78 × 10−3. Thus the solubility (mole fraction) of methane = 1.78 × 10−3.

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.

SOLUTION Moles A = 100/140 = 0.7143 mol; moles B = 1000/180 = 5.556 mol. Total = 6.270 mol. xA = 0.7143/6.270 = 0.1139; xB = 0.8861. pB = pB0xB = 500 × 0.8861 = 443.1 torr. ptotal = pA + pB ⇒ pA = 475 − 443.1 = 31.9 torr (≈ 32 torr, vapour pressure of A in solution). pA0 = pA/xA = 31.9 / 0.1139 = 280.5 torr (≈ 280.7 torr) for pure liquid A.

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 ptotal, pchloroform, and pacetone as a function of xacetone. 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.

ANSWER For an ideal solution, pacetone = 741.8 · xacetone, pchloroform = 632.8 · (1 − xacetone), and ptotal = pacetone + pchloroform; each is a straight line. The table below gives the ideal values for the same compositions:
100·xacetone011.823.436.050.858.264.572.1
pacetone (ideal)087.5173.6267.0376.8431.7478.5534.9
pchloroform (ideal)632.8558.1484.7405.0311.4264.5224.6176.6
ptotal (ideal)632.8645.6658.3672.0688.2696.2703.1711.5
ptotal (experimental)632.8603.0579.5562.1580.4599.5615.3641.8
The experimental partial and total vapour pressures lie below the straight ideal-solution lines, and ptotal shows a minimum. Therefore the acetone–chloroform mixture shows a negative deviation from Raoult’s law (acetone and chloroform form a hydrogen bond, strengthening A–B interactions). It forms a maximum-boiling azeotrope.

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.

SOLUTION Moles benzene (78) = 80/78 = 1.026 mol; moles toluene (92) = 100/92 = 1.087 mol. Total = 2.113 mol. x(benzene) = 1.026/2.113 = 0.486; x(toluene) = 0.514. p(benzene) = 50.71 × 0.486 = 24.64 mm Hg; p(toluene) = 32.06 × 0.514 = 16.48 mm Hg; ptotal = 41.12 mm Hg. y(benzene) = p(benzene)/ptotal = 24.64/41.12 = 0.60 (toluene ≈ 0.40).

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.

SOLUTION Total pressure = 10 atm = 7600 mm Hg. Partial pressures: p(O2) = 0.20 × 7600 = 1520 mm Hg; p(N2) = 0.79 × 7600 = 6004 mm Hg. x(O2) = p(O2)/KH = 1520 / (3.30 × 107) = 4.61 × 10−5. x(N2) = p(N2)/KH = 6004 / (6.51 × 107) = 9.22 × 10−5.

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.

SOLUTION Π = i (n2/V) RT ⇒ n2 = ΠV / (i RT). T = 300 K, R = 0.0821 L atm K−1 mol−1. n2 = (0.75 × 2.5) / (2.47 × 0.0821 × 300) = 1.875 / 60.84 = 0.0308 mol (≈ 0.03 mol CaCl2). M(CaCl2) = 40 + 71 = 111 g mol−1. Mass = 0.0308 × 111 = 3.42 g of CaCl2 (≈ 0.03 mol).

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.

SOLUTION M(K2SO4) = 2(39) + 32 + 4(16) = 78 + 32 + 64 = 174 g mol−1. Moles = 25 × 10−3 / 174 = 1.437 × 10−4 mol. K2SO4 → 2K+ + SO42−, so complete dissociation gives i = 3. Π = i (n/V) RT = 3 × (1.437 × 10−4 / 2) × 0.0821 × 298. Π = 3 × 7.184 × 10−5 × 24.47 = 5.27 × 10−3 atm.

Extra Practice Questions

Short Answer Type Questions

Q1. Why is molality preferred over molarity in colligative-property calculations?

ANSWERMolality is based on mass of solvent, which does not change with temperature, whereas molarity is based on solution volume, which expands or contracts with temperature. Since colligative experiments may involve temperature changes, the temperature-independent molality gives reliable results.

Q2. Why does a 0.1 m NaCl solution have a higher boiling point than a 0.1 m glucose solution?

ANSWERNaCl dissociates into Na+ and Cl− giving nearly twice the number of particles (i ≈ 2), while glucose stays as single molecules (i = 1). Elevation of boiling point depends on the number of particles, so ΔTb for NaCl is about double that of glucose.

Q3. What is an azeotrope? Name its two types.

ANSWERAn azeotrope is a binary liquid mixture that has the same composition in liquid and vapour phase and boils at a constant temperature, so it cannot be separated by fractional distillation. The two types are minimum-boiling azeotrope (large positive deviation, e.g. ethanol–water) and maximum-boiling azeotrope (large negative deviation, e.g. nitric acid–water).

Q4. Why does the addition of a non-volatile solute lower the vapour pressure of a solvent?

ANSWERIn the solution, part of the liquid surface is occupied by non-volatile solute particles, so fewer solvent molecules occupy the surface and escape into the vapour. The reduced number of escaping solvent molecules lowers the vapour pressure relative to the pure solvent.

Q5. Define isotonic, hypertonic and hypotonic solutions.

ANSWERTwo solutions with the same osmotic pressure are isotonic (no net osmosis between them). A solution with osmotic pressure higher than that of a cell’s fluid is hypertonic (water flows out, cell shrinks); one with lower osmotic pressure is hypotonic (water flows in, cell swells).

Long Answer Type Questions

Q1. Explain, with molecular reasoning and examples, positive and negative deviations from Raoult’s law and how azeotropes arise from them.

ANSWERRaoult’s law assumes A–B interactions equal A–A and B–B. In positive deviation, A–B forces are weaker, so molecules escape more readily, raising the vapour pressure above the ideal value; mixing is endothermic (ΔmixH > 0) with volume expansion (e.g. ethanol + acetone, where acetone breaks ethanol’s hydrogen bonds). In negative deviation, A–B forces are stronger, lowering vapour pressure below the ideal value; mixing is exothermic (ΔmixH < 0) with volume contraction (e.g. chloroform + acetone, which form a new H-bond). When deviations become very large, the ptotal curve passes through a maximum or minimum where liquid and vapour have the same composition — this constant-boiling mixture is an azeotrope. Large positive deviation gives a minimum-boiling azeotrope (ethanol–water, ~95% ethanol); large negative deviation gives a maximum-boiling azeotrope (nitric acid–water, ~68% HNO3). Such mixtures cannot be separated by fractional distillation.

Q2. State all four colligative properties and explain how each can be used to determine the molar mass of a solute.

ANSWERThe four colligative properties are relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure; all depend only on the number of solute particles. (1) From relative lowering, (p0 − p)/p0 = w2M1/(M2w1), so M2 can be found. (2) From ΔTb = Kbm, M2 = 1000Kbw2/(ΔTbw1). (3) From ΔTf = Kfm, M2 = 1000Kfw2/(ΔTfw1). (4) From Π = (n2/V)RT, M2 = w2RT/(ΠV). Osmotic pressure is best for large biomolecules and polymers because it is measured near room temperature and gives a large, easily measured value even for dilute solutions.

Q3. What is the van’t Hoff factor? Discuss its values for dissociation and association with examples, and how it modifies colligative-property equations.

ANSWERThe van’t Hoff factor (i) accounts for solutes that dissociate or associate in solution. It is defined as i = (normal molar mass)/(abnormal molar mass) = (observed colligative property)/(calculated colligative property) = (moles of particles after dissociation/association)/(moles before). For dissociation the number of particles increases, so i > 1 (e.g. KCl → K+ + Cl−, i ≈ 2; K2SO4, i ≈ 3 in dilute solution). For association the number of particles decreases, so i < 1 (e.g. ethanoic acid dimerises in benzene, i ≈ 0.5). The corrected equations are: relative lowering = i·x2, ΔTb = i Kb m, ΔTf = i Kf m, and Π = i (n2/V)RT.

MCQs & Assertion–Reason

1. Which concentration unit is independent of temperature?

(a) Molarity    (b) Normality    (c) Molality    (d) Formality

2. According to Henry’s law, the solubility of a gas in a liquid is directly proportional to:

(a) temperature    (b) partial pressure of the gas    (c) volume of the liquid    (d) molar mass of the gas

3. A solution showing positive deviation from Raoult’s law has:

(a) ΔmixH < 0    (b) ΔmixH = 0    (c) ΔmixH > 0    (d) ΔmixV < 0

4. Which pair forms a maximum-boiling azeotrope?

(a) ethanol + water    (b) nitric acid + water    (c) benzene + toluene    (d) n-hexane + n-heptane

5. The relative lowering of vapour pressure is equal to the:

(a) mole fraction of solvent    (b) mole fraction of solute    (c) molality    (d) molarity

6. For the same molality, which solution shows the greatest depression of freezing point?

(a) glucose    (b) urea    (c) NaCl    (d) K2SO4

7. The van’t Hoff factor for a solute that dimerises completely in solution is:

(a) 2    (b) 1    (c) 0.5    (d) 0

8. Osmotic pressure (Π) of a dilute solution is given by:

(a) Π = mRT    (b) Π = CRT    (c) Π = Kbm    (d) Π = KHx

9. Solubility of a gas in a liquid generally decreases with:

(a) increase in pressure    (b) increase in temperature    (c) decrease in temperature    (d) decrease in KH

10. Blood cells placed in a hypertonic solution will:

(a) swell    (b) burst    (c) shrink    (d) remain unchanged

Answer key: 1-(c), 2-(b), 3-(c), 4-(b), 5-(b), 6-(d), 7-(c), 8-(b), 9-(b), 10-(c).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: Molality does not change with temperature.

Reason: Molality is defined in terms of mass of solvent, which is independent of temperature.

A-R 2. Assertion: Aquatic species are more comfortable in cold water than in warm water.

Reason: The solubility of gases in water increases with decrease of temperature.

A-R 3. Assertion: A mixture of chloroform and acetone shows positive deviation from Raoult’s law.

Reason: Chloroform and acetone form intermolecular hydrogen bonds with each other.

A-R 4. Assertion: The van’t Hoff factor of ethanoic acid in benzene is less than one.

Reason: Ethanoic acid molecules associate to form dimers in benzene, reducing the number of particles.

A-R 5. Assertion: Osmotic pressure is preferred for measuring molar masses of polymers and proteins.

Reason: Osmotic pressure is measured near room temperature and gives a large, easily measured value even for dilute solutions.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).

Common Mistakes to Avoid

Watch out for these

  • Confusing molarity (per litre of solution) with molality (per kg of solvent) — read the data carefully.
  • Forgetting the van’t Hoff factor (i) for electrolytes (NaCl, CaCl2, K2SO4) in colligative-property numericals.
  • Mixing up positive and negative deviation with the sign of ΔmixH (positive → endothermic; negative → exothermic).
  • Using the wrong value of R: 0.0821 L atm K−1 mol−1 for atm, 8.314 J K−1 mol−1 for SI (Pa), 0.083 L bar K−1 mol−1 for bar.
  • Not converting temperature to kelvin or volume to litres before substituting in Π = CRT.
  • Adding solute moles to solvent moles incorrectly when finding mole fraction in dilute-solution shortcuts.

How to score full marks in this chapter

Always write the formula first, substitute with units, then compute. For colligative numericals decide early whether the solute is an electrolyte (use i). Memorise the standard molar masses (urea 60, glucose 180, sucrose 342, NaCl 58.5) and Kb/Kf for water (0.52 / 1.86 K kg mol−1). For theory questions, link molecular interactions (A–A, B–B, A–B) to deviations and to the sign of ΔmixH, and quote the textbook examples (ethanol–acetone, chloroform–acetone, ethanol–water, HNO3–water).

Frequently Asked Questions

What is Class 12 Chemistry Chapter 1 Solutions about?

Chapter 1 Solutions deals with types of solutions, ways of expressing concentration (molarity, molality, mole fraction, ppm), Henry’s law, Raoult’s law, ideal and non-ideal solutions, the four colligative properties, and the van’t Hoff factor used to explain abnormal molar masses.

What are colligative properties?

Colligative properties are properties that depend only on the number of solute particles, not on their nature — relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure. They are used to determine the molar mass of a solute.

Are all the NCERT exercises of Chapter 1 solved here?

Yes. Every Intext Question and all numbered Exercises (1.1 to 1.41) are reproduced verbatim and solved step by step, with each numerical answer cross-checked against the NCERT answer key for session 2026–27.

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