Solutions
Welcome to HSLC Guru, your trusted companion for ASSEB Class 12 Chemistry. In this chapter, Solutions, we explore homogeneous mixtures of two or more components, the various ways of expressing concentration, the laws governing vapour pressure of solutions, and the colligative properties that depend solely on the number of solute particles. This English-medium guide is structured strictly according to the ASSEB syllabus and provides clear definitions, derivations, numerical practice, and exam-ready Q&A so that every learner can master this important physical chemistry chapter with confidence.
Chapter Summary
A solution is a homogeneous mixture of two or more pure substances whose composition can be varied within certain limits. The component present in larger quantity is called the solvent, and the component(s) present in smaller quantity are called the solute(s). Depending on the physical state of solute and solvent, solutions can be classified into nine types — solid in solid (alloys), solid in liquid (sugar in water), gas in liquid (oxygen in water), liquid in liquid (alcohol in water), and so on. The concentration of a solution is expressed in many ways: mass percentage (mass of solute per 100 g of solution), volume percentage, parts per million (ppm) for very dilute solutions, mole fraction (ratio of moles of one component to total moles), molarity (M) (moles of solute per litre of solution), molality (m) (moles of solute per kilogram of solvent), and normality (N) (gram equivalents per litre).
The solubility of a substance is the maximum amount that can be dissolved in a given amount of solvent at a particular temperature. For solids in liquids, if dissolution is endothermic the solubility increases with temperature, while exothermic dissolution shows the opposite trend. Gases dissolve in liquids according to Henry’s law, which states that the partial pressure of a gas above a liquid is directly proportional to its mole fraction in the solution. Increasing temperature decreases gas solubility, which is why aquatic life suffers in warm water. Raoult’s law for a solution of two volatile liquids states that the partial vapour pressure of each component is equal to the product of its mole fraction and its pure-component vapour pressure. Solutions that obey Raoult’s law over the entire range of composition are called ideal solutions; deviations occur when solute–solvent interactions differ from solute–solute and solvent–solvent interactions, giving positive deviation (weaker A–B interactions) or negative deviation (stronger A–B interactions). Mixtures showing maximum or minimum boiling points and behaving like a single substance during distillation are called azeotropes.
Properties that depend only on the number of solute particles and not on their nature are called colligative properties. The four important colligative properties are relative lowering of vapour pressure, elevation of boiling point ($\Delta T_b$), depression of freezing point ($\Delta T_f$), and osmotic pressure ($\pi$). Osmotic pressure is the most accurate for determining the molar mass of biomolecules because it gives a measurable value at room temperature. When solutes dissociate (electrolytes) or associate (carboxylic acids in benzene), the experimental molar mass differs from the theoretical value — this is called abnormal molar mass. To correct for this, the van’t Hoff factor (i) is introduced, defined as the ratio of the observed colligative property to the calculated colligative property. For dissociation, $i > 1$; for association, $i < 1$; and for non-electrolytes, $i = 1$.
The modified colligative property formulas incorporating the van’t Hoff factor are extensively used to compute molecular masses of electrolytes and to predict the freezing/boiling points of saline or sugary solutions. Real-world applications include preparation of antifreeze for car radiators using ethylene glycol, desalination of sea water by reverse osmosis, intravenous saline being isotonic with blood (0.9% NaCl), and sprinkling of salt on icy roads to lower the freezing point of water. Understanding solutions is foundational for biochemistry, pharmaceuticals, and industrial chemistry. The chapter combines conceptual reasoning with quantitative problem solving, so learners must practise numerical questions involving molarity, molality, and the four colligative properties along with the van’t Hoff factor adjustments.
Key Formulas (KaTeX)
$p = K_H \cdot x$
$p_A = p_A^\circ \cdot x_A$
$p_{\text{total}} = p_A^\circ x_A + p_B^\circ x_B$
$\frac{p_A^\circ – p_A}{p_A^\circ} = x_B$
$\Delta T_b = K_b \cdot m$
$\Delta T_f = K_f \cdot m$
$\pi = \frac{n}{V}RT = MRT$
$i = \frac{\text{observed colligative property}}{\text{calculated colligative property}}$
$\Delta T_b = i \cdot K_b \cdot m$
$\pi = iCRT$
1-Mark Questions
Q1. Define a solution.
Answer: A solution is a homogeneous mixture of two or more pure substances whose composition can be varied within certain limits.
Q2. What is meant by molarity?
Answer: Molarity is the number of moles of solute dissolved per litre of solution. It is denoted by M and depends on temperature.
Q3. Define molality.
Answer: Molality is the number of moles of solute dissolved per kilogram of solvent. It is denoted by m and is independent of temperature.
Q4. State Henry’s law.
Answer: Henry’s law states that at constant temperature, the partial pressure of a gas above a liquid solution is directly proportional to the mole fraction of the gas in the solution: $p = K_H \cdot x$.
Q5. What is an azeotrope?
Answer: An azeotrope is a binary mixture of liquids that boils at a constant temperature and distils without change in composition, behaving like a single component.
Q6. Define osmotic pressure.
Answer: Osmotic pressure is the excess pressure that must be applied on the solution side to prevent osmosis of pure solvent into the solution through a semipermeable membrane.
Q7. What is the van’t Hoff factor for NaCl assuming complete dissociation?
Answer: i = 2, since NaCl dissociates into Na+ and Cl- ions.
Q8. Name the colligative property used to determine molar masses of biomolecules.
Answer: Osmotic pressure, because it gives measurable values at low concentrations and at room temperature.
Q9. What is meant by isotonic solutions?
Answer: Two solutions are isotonic if they exert the same osmotic pressure at the same temperature; they have equal molar concentrations of solute particles.
Q10. Why does a sealed soda bottle fizz when opened?
Answer: Sudden release of pressure decreases the solubility of dissolved CO2 (Henry’s law), and the gas escapes rapidly causing fizz.
2–3 Marks Questions
Q1. Differentiate between molarity and molality.
Answer: (i) Molarity is moles of solute per litre of solution; molality is moles of solute per kg of solvent. (ii) Molarity changes with temperature because volume expands; molality is independent of temperature because mass is invariant. (iii) Molarity is more convenient for volumetric analysis while molality is preferred for studies involving colligative properties.
Q2. State Raoult’s law for a solution of two volatile liquids and write its mathematical form.
Answer: Raoult’s law states that for a solution of two volatile liquids A and B, the partial vapour pressure of each component is directly proportional to its mole fraction in solution. Mathematically: $p_A = p_A^\circ x_A$ and $p_B = p_B^\circ x_B$. Total vapour pressure: $p_{\text{total}} = p_A^\circ x_A + p_B^\circ x_B$.
Q3. Explain positive and negative deviations from Raoult’s law with one example each.
Answer: Positive deviation: 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 interactions. Example: ethanol + acetone. Negative deviation: Observed vapour pressure is lower than predicted. Occurs when A–B interactions are stronger (e.g., hydrogen bonding). Example: chloroform + acetone.
Q4. Why is the freezing point of sea water lower than that of pure water?
Answer: Sea water contains dissolved salts (mainly NaCl). The presence of solute particles lowers the chemical potential of the liquid phase, requiring a lower temperature for solid–liquid equilibrium. This depression of freezing point ($\Delta T_f = i K_f m$) is why sea water remains liquid below 0 °C.
Q5. Define the van’t Hoff factor and write its values for KCl, glucose, and benzoic acid in benzene.
Answer: The van’t Hoff factor (i) is the ratio of the observed colligative property to the calculated colligative property assuming no association/dissociation. For KCl (complete dissociation): i = 2. For glucose (non-electrolyte): i = 1. For benzoic acid in benzene (dimerises): i ≈ 0.5.
Q6. Explain why the boiling point of a solution is higher than that of the pure solvent.
Answer: Addition of a non-volatile solute lowers the vapour pressure of the solvent. Since boiling occurs when vapour pressure equals atmospheric pressure, a higher temperature is needed for the solution to attain that vapour pressure. Hence the boiling point rises by $\Delta T_b = K_b \cdot m$.
5–7 Marks Questions (Numerical + Theory)
Q1. Calculate the molality and mole fraction of solute in a solution containing 18 g of glucose (C6H12O6, M = 180 g/mol) in 500 g of water.
Answer: Moles of glucose = 18 / 180 = 0.1 mol. Mass of solvent = 500 g = 0.5 kg. Molality m = 0.1 / 0.5 = 0.2 mol/kg. Moles of water = 500 / 18 = 27.78. Mole fraction of glucose = 0.1 / (0.1 + 27.78) = 0.1 / 27.88 = 0.00359. Mole fraction of water = 1 − 0.00359 = 0.99641.
Q2. 2 g of a non-volatile non-electrolyte solute is dissolved in 100 g of water. The boiling point of the solution is found to be 100.052 °C. Calculate the molar mass of the solute. (Kb for water = 0.52 K kg mol⁻¹).
Answer: $\Delta T_b$ = 100.052 − 100 = 0.052 K. Using $\Delta T_b = K_b \cdot m$, m = 0.052 / 0.52 = 0.1 mol/kg. Moles of solute in 100 g water = 0.1 × 0.1 = 0.01 mol. Molar mass M = mass / moles = 2 / 0.01 = 200 g/mol.
Q3. A 5% (w/w) aqueous solution of cane sugar (M = 342) has a freezing point of 271 K. Calculate the freezing point of 5% (w/w) glucose solution (M = 180) in water. (Kf = 1.86 K kg mol⁻¹, freezing point of pure water = 273.15 K).
Answer: For glucose: 5 g in 95 g water. Moles = 5/180 = 0.0278. Molality = 0.0278 / 0.095 = 0.292 mol/kg. $\Delta T_f$ = 1.86 × 0.292 = 0.543 K. Freezing point of glucose solution = 273.15 − 0.543 = 272.61 K. (Cane sugar shows smaller depression because its molar mass is larger, hence lower molality.)
Q4. The osmotic pressure of a solution containing 6.0 g of urea (M = 60) in 1 litre of solution at 27 °C is to be calculated. Use R = 0.0821 L atm K⁻¹ mol⁻¹.
Answer: Moles of urea = 6/60 = 0.1 mol. Concentration C = 0.1 mol/L. T = 300 K. Using $\pi = CRT$: π = 0.1 × 0.0821 × 300 = 2.463 atm.
Q5. 0.561 g of KOH dissolved in 200 g of water gives a freezing point depression of 0.093 K. Calculate the van’t Hoff factor and degree of dissociation. (Kf for water = 1.86 K kg mol⁻¹, M of KOH = 56).
Answer: Moles of KOH = 0.561/56 = 0.01 mol. Molality = 0.01 / 0.2 = 0.05 mol/kg. Theoretical $\Delta T_f$ = 1.86 × 0.05 = 0.093 K. Observed = 0.093 K. i = 0.093 / 0.093 = 1.0. (Note: with this data set i appears 1, indicating apparent no dissociation; using the corrected experimental value i = 1.86 gives degree of dissociation α = (i − 1)/(n − 1) where n = 2; here for ideal dissociation i = 2, α = 1.) For full dissociation expected i = 2 → α = 100%.
Multiple Choice Questions (MCQ)
Q1. Which of the following is NOT a colligative property?
(a) Osmotic pressure (b) Boiling point elevation (c) Refractive index (d) Freezing point depression
Answer: (c) Refractive index.
Q2. Henry’s law constant for a gas depends on:
(a) Pressure only (b) Nature of gas and temperature (c) Volume only (d) Mole fraction
Answer: (b) Nature of gas and temperature.
Q3. An ideal solution obeys Raoult’s law:
(a) Only at low concentration (b) Over the entire range of composition (c) Only at high temperature (d) Never
Answer: (b) Over the entire range of composition.
Q4. The van’t Hoff factor for K4[Fe(CN)6] (complete dissociation) is:
(a) 2 (b) 3 (c) 4 (d) 5
Answer: (d) 5.
Q5. Mixture showing positive deviation from Raoult’s law:
(a) CHCl3 + (CH3)2CO (b) HNO3 + H2O (c) C2H5OH + (CH3)2CO (d) Phenol + Aniline
Answer: (c) C2H5OH + (CH3)2CO.
Q6. Molality is expressed in:
(a) mol/L (b) g/L (c) mol/kg (d) g/kg
Answer: (c) mol/kg.
Q7. The osmotic pressure of 0.1 M solution at 27 °C (R = 0.0821) is:
(a) 1.23 atm (b) 2.46 atm (c) 24.6 atm (d) 0.246 atm
Answer: (b) 2.46 atm.
Q8. Reverse osmosis is used in:
(a) Production of insulin (b) Desalination of sea water (c) Combustion (d) Distillation
Answer: (b) Desalination of sea water.
Q9. Which solution will have the highest boiling point (at same molality)?
(a) Glucose (b) Urea (c) NaCl (d) BaCl2
Answer: (d) BaCl2 (i ≈ 3, highest particle count).
Q10. Blood is isotonic with:
(a) 0.9% NaCl (b) 9% NaCl (c) Pure water (d) 1 M glucose
Answer: (a) 0.9% NaCl.
Fill in the Blanks
Q1. A solution that obeys Raoult’s law over the entire composition range is called an __________ solution.
Answer: ideal.
Q2. The molality of a solution is independent of __________.
Answer: temperature.
Q3. Henry’s law constant for CO2 in water is __________ than that of O2 at the same temperature.
Answer: smaller (CO2 more soluble).
Q4. The van’t Hoff factor for a non-electrolyte is equal to __________.
Answer: 1.
Q5. The relative lowering of vapour pressure equals the __________ of the solute in the solution.
Answer: mole fraction.
True / False
Q1. Solubility of gases in liquids increases with rise in temperature.
Answer: False. Gas solubility decreases as temperature rises.
Q2. Osmotic pressure is a colligative property.
Answer: True.
Q3. Azeotropic mixtures can be separated by simple distillation.
Answer: False. Azeotropes cannot be separated by simple distillation.
Q4. The van’t Hoff factor of acetic acid in benzene is greater than 1.
Answer: False. Acetic acid dimerises in benzene, so i < 1.
Q5. Molarity changes with temperature but molality does not.
Answer: True.
Glossary
| Term | Meaning |
|---|---|
| Solution | Homogeneous mixture of two or more components. |
| Solute | Component present in smaller amount. |
| Solvent | Component present in larger amount. |
| Molarity | Moles of solute per litre of solution. |
| Molality | Moles of solute per kg of solvent. |
| Mole fraction | Ratio of moles of one component to total moles. |
| Henry’s law | p = KH·x for gas dissolved in liquid. |
| Raoult’s law | Partial vapour pressure proportional to mole fraction. |
| Ideal solution | Obeys Raoult’s law throughout composition. |
| Azeotrope | Mixture distilling at constant composition. |
| Colligative property | Depends only on number of solute particles. |
| Osmosis | Flow of solvent through semipermeable membrane. |
| Osmotic pressure (π) | Pressure required to stop osmosis. |
| van’t Hoff factor (i) | Ratio of observed to calculated colligative property. |
| Isotonic | Solutions with equal osmotic pressure. |
Formula Table
| Quantity | Formula |
|---|---|
| Mass percentage | (mass of solute / mass of solution) × 100 |
| ppm | (mass of solute / mass of solution) × 10⁶ |
| Mole fraction (x) | n_component / n_total |
| Molarity (M) | moles of solute / volume of solution (L) |
| Molality (m) | moles of solute / mass of solvent (kg) |
| Henry’s law | p = KH·x |
| Raoult’s law (single) | p_A = p_A° · x_A |
| Total VP | p_total = p_A° x_A + p_B° x_B |
| Relative lowering | (p_A° − p_A)/p_A° = x_B |
| Boiling point elevation | ΔT_b = K_b · m |
| Freezing point depression | ΔT_f = K_f · m |
| Osmotic pressure | π = (n/V)RT = MRT |
| van’t Hoff factor | i = observed CP / calculated CP |
| Modified ΔT_b | ΔT_b = i · K_b · m |
| Modified osmotic pressure | π = iCRT |
This concludes Chapter 2 — Solutions. Practise the numericals on molarity/molality and all four colligative properties (with and without the van’t Hoff factor). Memorise Henry’s law, Raoult’s law, and the conditions for ideal vs. non-ideal solutions. Stay tuned to HSLC Guru for the next ASSEB Class 12 Chemistry chapter — Electrochemistry.