Tap "Show Solution" under each question to reveal the step-by-step working. All answers paraphrased for clarity; values verified against NCERT data tables.
1.1 Define the term solution. How is a solution different from a compound? Give examples of each type of binary solution.
Solution: A homogeneous mixture of two or more non-reacting substances whose composition can be varied within limits. Compounds, by contrast, have a fixed composition and the constituent elements are chemically bonded. Examples of nine binary solutions (see Part 1 Table 1.1): air (gas-in-gas), humid air (liquid-in-gas), smog/I₂ in air (solid-in-gas), soda water (gas-in-liquid), ethanol in water (liquid-in-liquid), sugar in water (solid-in-liquid), H₂ in Pd (gas-in-solid), dental amalgam (liquid-in-solid), brass (solid-in-solid).
1.2 Give an example of a solid solution in which the solute is a gas.
Hydrogen adsorbed in palladium metal is a classic case — the solute (H₂) is a gas and the solvent (Pd) is a solid. Another example is helium trapped inside metals during cosmic-ray exposure.
1.3 Define the terms molarity and molality. Why is molality preferred over molarity while reporting concentrations at different temperatures?
Molarity = moles of solute per litre of solution. Molality = moles of solute per kilogram of solvent.
Volume of a solution expands or contracts with temperature, so molarity is T-dependent. Mass of solvent is invariant, so molality is T-independent — essential for boiling-point elevation, freezing-point depression and other T-dependent experiments.
1.4 Calculate the mass percent of a solution containing 22 g of benzene (C₆H₆) in 122 g of CCl₄.
Total mass = 22 + 122 = 144 g.
\[\text{Mass \%}_{C_6H_6}=\frac{22}{144}\times100=\textbf{15.28\,\%}\]
Mass % of CCl₄ = 100 − 15.28 = 84.72 %.
1.5 Calculate (i) molality, (ii) molarity, and (iii) mole fraction of KI, if the density of a 20 % (w/w) aqueous KI solution is 1.202 g mL⁻¹. (MKI = 166)
Take 100 g solution → 20 g KI + 80 g water.
nKI = 20/166 = 0.1205 mol; nwater = 80/18 = 4.444 mol. Molality m = 0.1205 / 0.080 kg = 1.506 mol kg⁻¹.
Volume of solution = 100 g / 1.202 g mL⁻¹ = 83.19 mL = 0.0832 L. Molarity M = 0.1205 / 0.0832 = 1.449 mol L⁻¹. Mole fraction xKI = 0.1205 / (0.1205 + 4.444) = 0.0264.
1.6 H₂S, a toxic gas with rotten-egg smell, is used in qualitative analysis. If the solubility of H₂S in water at STP is 0.195 m, calculate Henry's-law constant.
Take 1 kg water → 0.195 mol H₂S, 55.56 mol water.
xH₂S = 0.195/(0.195+55.56) = 3.50 × 10⁻³.
At STP pH₂S = 0.987 bar.
\[K_H = \frac{p}{x}=\frac{0.987}{3.50\times10^{-3}}=\textbf{282\,bar}\]
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 soda water bottled under 2.5 atm CO₂ pressure.
p = 2.5 atm = 2.533 × 10⁵ Pa.
xCO₂ = p/KH = 2.533 × 10⁵ / 1.67 × 10⁸ = 1.517 × 10⁻³.
500 mL water ≈ 500 g → nwater = 500/18 = 27.78 mol.
Since xCO₂ is tiny, nCO₂ ≈ x · nwater = 1.517 × 10⁻³ × 27.78 = 4.215 × 10⁻² mol.
Mass = 4.215 × 10⁻² × 44 = 1.85 g CO₂.
1.8 The vapour pressures of pure liquids A and B are 450 and 700 mm Hg at 350 K. Find the composition of the liquid mixture for which total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.
Raoult: ptotal = 450 xA + 700 (1 − xA) = 700 − 250 xA.
600 = 700 − 250 xA → xA = 0.40, xB = 0.60.
yA = pA/ptotal = (450 × 0.40)/600 = 0.30; yB = 0.70.
Liquid 40 % A, vapour 30 % A — the vapour is richer in the more volatile B (higher p°).
1.9 Vapour pressure of water at 293 K is 17.535 mm Hg. Find the vapour pressure at 293 K when 25 g of glucose is dissolved in 450 g of water.
1.10 How much sucrose (M = 342) is to be added to 500 g of water so that it boils at 100.52 °C? (Kb = 0.52 K kg mol⁻¹; assume i = 1.)
ΔTb = 0.52 K → m = ΔTb/Kb = 0.52/0.52 = 1.0 mol kg⁻¹.
For 500 g solvent, nsucrose = 0.5 mol → mass = 0.5 × 342 = 171 g sucrose.
1.11 Calculate the mass of ascorbic acid (C₆H₈O₆, M = 176) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5 °C. Kf(acetic acid) = 3.9 K kg mol⁻¹.
m = ΔTf/Kf = 1.5/3.9 = 0.3846 mol kg⁻¹.
n = m × 0.075 kg = 0.0288 mol; mass = 0.0288 × 176 = 5.08 g ascorbic acid.
1.12 Calculate the osmotic pressure in pascal exerted by a solution of 1.0 g of a polymer of molar mass 185 000 in 450 mL of water at 37 °C.
n = 1/185000 = 5.405 × 10⁻⁶ mol; V = 0.450 L; T = 310 K; R = 8.314 J mol⁻¹ K⁻¹.
\[\pi V = nRT \;\Rightarrow\; \pi = \frac{5.405\times10^{-6}\times8.314\times310}{4.5\times10^{-4}\,\text{m}^3}\]
π = 30.96 Pa ≈ 31 Pa.
1.13 The partial pressure of ethane over water at 25 °C is 1 bar when it is in equilibrium with a solution containing 6.56 × 10⁻³ g of ethane per 100 g water. If the partial pressure is raised to 5 bar, what will be the mass of ethane dissolved per 100 g water?
Henry's law: mass ∝ p. New mass = 6.56 × 10⁻³ × 5 = 3.28 × 10⁻² g = 32.8 mg per 100 g water.
1.14 An aqueous solution of 2 % non-volatile exerts a vapour pressure of 1.004 bar at the normal boiling point (373.15 K, pure water p° = 1.013 bar). Find the molar mass of the solute.
In 100 g solution: 2 g solute (M = ?) + 98 g water.
\[\frac{\Delta p}{p°}=\frac{1.013-1.004}{1.013}=\frac{0.009}{1.013}=8.89\times10^{-3}=x_B\]
For dilute: xB ≈ (2/M)/(98/18); solving:
\[\frac{2/M}{98/18}=8.89\times10^{-3}\Rightarrow\frac{2\times18}{M\times98}=8.89\times10^{-3}\]
M = 36/(98 × 8.89 × 10⁻³) = 41.3 g mol⁻¹.
1.15 Heptane and octane form an ideal solution. At 373 K, the vapour pressures of pure heptane (M = 100) and octane (M = 114) are 105.2 kPa and 46.8 kPa. What is the vapour pressure of a mixture containing 26.0 g of heptane and 35.0 g of octane?
1.19 Depression in freezing point of 0.15 M aqueous solution of KCl is 0.34 °C. Kf(water) = 1.86. Calculate the van't Hoff factor and the degree of dissociation.
Assume dilute → molality ≈ 0.15.
ΔTf,calc = 1.86 × 0.15 = 0.279 °C. i = 0.34/0.279 = 1.22.
For KCl, n = 2, i = 1 + α → α = 0.22 (22 % dissociated)… Note: real KCl is nearly fully dissociated; the low value here reflects a concentration effect.
1.20 19.5 g of CH₂FCOOH is dissolved in 500 g of water. The depression in the freezing point is 1.00 K. Calculate the van't Hoff factor and dissociation constant Ka.
M = 78 → n = 19.5/78 = 0.25 mol; m = 0.25/0.500 = 0.50 mol kg⁻¹.
ΔTf,calc = 1.86 × 0.50 = 0.93 K. i = 1.00/0.93 = 1.0753.
For HA ⇌ H⁺ + A⁻ (n = 2), α = i − 1 = 0.0753.
Ka = α² C / (1 − α) = (0.0753)² × 0.50 / 0.925 = 3.07 × 10⁻³.
1.21 Calculate the molarity of each of the following solutions: (a) 30 g of Co(NO₃)₂·6H₂O in 4.3 L solution (b) 30 mL of 0.5 M H₂SO₄ diluted to 500 mL.
1.22 Determine the amount of CaCl₂ (i = 2.47) dissolved in 2.5 L of water such that its osmotic pressure is 0.75 atm at 27 °C.
π = iCRT → C = π / (iRT) = 0.75 / (2.47 × 0.0821 × 300) = 0.01234 mol L⁻¹.
n = 0.01234 × 2.5 = 0.0308 mol. M(CaCl₂) = 111 → mass = 3.42 g.
1.23 Determine the osmotic pressure at 27 °C of a solution prepared by dissolving 25 mg of K₂SO₄ in 2 L of water, assuming K₂SO₄ is completely dissociated.
1.24 Concentrated nitric acid is about 68 % HNO₃ by mass with density 1.504 g mL⁻¹. Calculate the molarity of the solution.
Take 1 L = 1504 g. Mass HNO₃ = 0.68 × 1504 = 1022.7 g → n = 1022.7/63 = 16.23 mol.
M = 16.23 mol L⁻¹.
1.25 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 the solution is 1.2 g mL⁻¹, then what shall be the molarity?
100 g solution → 10 g glucose (0.0556 mol) + 90 g water (5.0 mol). m = 0.0556/0.090 = 0.617 mol kg⁻¹.
xglucose = 0.0556/5.0556 = 0.011; xwater = 0.989.
V = 100/1.2 = 83.33 mL = 0.0833 L → M = 0.667 mol L⁻¹.
1.26 If the density of a lake water is 1.25 g mL⁻¹ and it contains 92 ppm of Na⁺ ions (by mass), calculate the molarity of Na⁺ ions in the water.
92 ppm = 92 g Na⁺ per 10⁶ g of lake water.
Take 1 L of water = 1250 g → Na⁺ mass = 92 × 1250 / 10⁶ = 0.115 g.
nNa⁺ = 0.115/23 = 5.0 × 10⁻³ mol → M = 5.0 × 10⁻³ mol L⁻¹.
Chapter Wrap-up
You have traversed the full landscape of solutions — from the nine physical-state types and the half-a-dozen ways to express concentration, through Henry's and Raoult's laws, onwards to ideal and non-ideal behaviour, azeotropes, colligative properties and finally abnormal molar masses corrected by the van't Hoff factor. The unifying thread: all four colligative properties count particles, not identities, which is why they form independent tools for determining molar mass and why a correct count (through i) is essential whenever a solute dissociates or associates.
Master the formulas, practise the conversions, and the chapter's reputation as a "numericals-heavy" unit becomes a source of easy marks in the board exam.
Frequently Asked Questions - NCERT Exercises and Solutions: Solutions
What are the key NCERT exercise types in Chapter 1 Solutions?
NCERT Class 12 Chemistry Chapter 1 Solutions exercises cover definitions, structure-property relationships, reaction mechanisms, numerical problems, and predict-the-product questions. The MyAiSchool solution set provides step-by-step worked solutions for every NCERT question, aligned with the CBSE board exam pattern. Students should focus on reaction mechanisms, IUPAC nomenclature, and stoichiometric reasoning to score full marks.
How should students approach reaction mechanism questions in Solutions?
For reaction mechanism questions in NCERT Class 12 Chemistry Chapter 1 Solutions: (1) identify the substrate, reagent, solvent, and conditions, (2) draw curly-arrow electron movement at each step, (3) label intermediates (carbocation, carbanion, free radical, transition state), (4) state stereochemistry where relevant. The MyAiSchool solutions show full curly-arrow mechanisms for every multi-step reaction.
What are the most-asked CBSE board questions from Chapter 1?
From NCERT Class 12 Chemistry Chapter 1 (Solutions), the most-asked CBSE board questions test conceptual understanding, structure-property logic, distinguishing tests, IUPAC naming, and short numerical problems. 5-mark questions usually combine structural reasoning + mechanism + application. The MyAiSchool exercise set tags each question by mark weight and Bloom level for prioritized prep.
How do I balance chemical equations in NCERT exercises?
For balancing equations in NCERT Class 12 Chemistry Chapter 1: (1) write skeletal equation with correct formulas, (2) balance atoms other than H, O first, (3) balance O, then H (using H2O for organic reactions; or in acidic/basic medium for redox), (4) balance charge using e- in redox, (5) cross-check atom count and charge on both sides. The MyAiSchool solutions include balanced redox half-reactions where applicable.
What are common mistakes students make in Chapter 1 exercises?
Common mistakes in NCERT Class 12 Chemistry Chapter 1 (Solutions) include: (1) incorrect IUPAC names (wrong locants or suffix), (2) skipping reaction conditions (catalyst, temperature, solvent), (3) wrong mechanism arrows, (4) sign errors in numerical (Ecell, Kc, ΔG), (5) forgetting stereochemistry (retention/inversion/racemisation). The MyAiSchool solutions flag these traps for each question.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 12 Chemistry Chapter 1 Solutions solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with chemical reasoning, curly-arrow mechanisms, IUPAC naming verifications, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Verified against NCERT and CBSE marking schemes.
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