NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions + Important Board Questions 2026

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Class 12 Chemistry Chapter 2

Solutions Complete NCERT Resource Guide

Mastering Class 12 Chemistry can feel like a mountain to climb, but Chapter 2: Solutions is one of the most scoring areas if you grasp the core concepts. This comprehensive guide provides Updated NCERT Solutions along with Important Questions tailored for your CBSE Board Exam 2026. Whether you are prepping for your boards or targeting competitive exams like NEET and JEE, these step-by-step explanations will help you ace your chemistry paper with ease.

Chapter NameSolutions (Chapter 2)
SubjectChemistry
ClassClass 12
BoardCBSE (Target Year 2026-27)
Important TopicsConcentration terms, Henry's Law, Raoult's Law, Ideal/Non-ideal solutions, Colligative properties, van 't Hoff factor
Difficulty LevelModerate (Concept-heavy + Numerical-based)
Exam WeightageApprox 7 Marks
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Learning Objectives

After completing this chapter, students will be able to:

Key Concepts, Definitions & Formulas

Before jumping into the solutions, let's review the fundamental rules and equations you will need:

Full NCERT Solutions (Step-by-Step)

Below are the detailed, step-by-step solutions for the exercises in Class 12 Chemistry Chapter 2 (Solutions).

Question 2.1: Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

Step 1: Definition. A solution is defined as a homogeneous mixture of two or more chemically non-reacting substances whose composition can be varied within certain limits.
Step 2: Gaseous Solutions. The solvent is a gas. Gas in Gas: Mixture of oxygen and nitrogen. Liquid in Gas: Chloroform mixed with nitrogen. Solid in Gas: Camphor in nitrogen.
Step 3: Liquid Solutions. The solvent is a liquid. Gas in Liquid: Oxygen dissolved in water. Liquid in Liquid: Ethanol dissolved in water. Solid in Liquid: Glucose or salt dissolved in water.
Step 4: Solid Solutions. The solvent is a solid. Gas in Solid: Solution of hydrogen in palladium. Liquid in Solid: Amalgam of mercury with sodium. Solid in Solid: Copper dissolved in gold (Alloys).

Question 2.2: Give an example of a solid solution in which the solute is a gas.

Step 1: Identification. An example of a solid solution where the solute is a gas is Hydrogen gas absorbed in Palladium metal ($\text{H}_2 \text{ in Pd}$).
Step 2: Mechanism. Here, hydrogen molecules fill the interstitial spaces within the crystalline lattice structure of solid palladium.

Question 2.3: Define the following terms: (i) Mole fraction (ii) Molarity (iii) Molality (iv) Mass percentage.

Step 1: Mole Fraction ($x$). It is the ratio of the number of moles of one component to the total number of moles of all the components present in the solution. It has no units.
Step 2: Molarity ($M$). It is the number of moles of solute dissolved per litre of the solution. Its unit is $\text{mol L}^{-1}$ or M. It changes with temperature.
Step 3: Molality ($m$). It is defined as the number of moles of solute dissolved per kilogram (1000 g) of the solvent. Its unit is $\text{mol kg}^{-1}$ or m. It is temperature-independent.
Step 4: Mass percentage (% W/W). It is the mass of a specific component per 100 g of the total solution.

Question 2.4: 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?

Step 1: Understand the given data. Mass percent of $\text{HNO}_3 = 68\%$. This means 68 g of $\text{HNO}_3$ is present in 100 g of solution. Density of solution ($\rho$) = 1.504 g mL⁻¹. Molar mass of $\text{HNO}_3 = 1 + 14 + (3 \times 16) = 63 \text{ g mol}^{-1}$.
Step 2: Calculate the volume of the solution. $$ \text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{100 \text{ g}}{1.504 \text{ g mL}^{-1}} = 66.49 \text{ mL} = 0.06649 \text{ L} $$
Step 3: Calculate the moles of $\text{HNO}_3$. $$ \text{Moles of solute } (n) = \frac{\text{Given Mass}}{\text{Molar Mass}} = \frac{68 \text{ g}}{63 \text{ g mol}^{-1}} = 1.079 \text{ moles} $$
Step 4: Calculate Molarity ($M$). $$ M = \frac{\text{Moles of solute}}{\text{Volume of solution in L}} = \frac{1.079}{0.06649} = 16.23 \text{ M} $$
Final Answer: The molarity of the nitric acid solution is 16.23 M.

Question 2.5: A solution of glucose ($\text{C}_6\text{H}_{12}\text{O}_6$) in water is labeled 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?

Step 1: Breakdown the 10% w/w definition. Mass of Glucose ($W_2$) = 10 g. Mass of Water ($W_1$) = 100 g - 10 g = 90 g = 0.09 kg. Molar mass of Glucose ($M_2$) = 180 g mol⁻¹. Molar mass of Water ($M_1$) = 18 g mol⁻¹.
Step 2: Calculate Molality ($m$). Moles of Glucose ($n_2$) = 10 / 180 = 0.0556 mol. $$ m = \frac{n_2}{\text{Mass of solvent in kg}} = \frac{0.0556}{0.09} = 0.618 \text{ m} $$
Step 3: Calculate Mole Fractions ($x$). Moles of Water ($n_1$) = 90 / 18 = 5.0 mol. Total Moles = 5.0 + 0.0556 = 5.0556 mol.
$$ x_{\text{glucose}} = \frac{0.0556}{5.0556} = 0.011 $$
$$ x_{\text{water}} = 1 - 0.011 = 0.989 $$
Step 4: Calculate Molarity ($M$). $$ \text{Volume of solution} = \frac{100 \text{ g}}{1.2 \text{ g mL}^{-1}} = 83.33 \text{ mL} = 0.08333 \text{ L} $$
$$ M = \frac{n_2}{\text{Volume in L}} = \frac{0.0556}{0.08333} = 0.667 \text{ M} $$
Final Answer: Molality = 0.618 m, $x_{\text{glucose}}$ = 0.011, $x_{\text{water}}$ = 0.989, Molarity = 0.667 M.

Question 2.6: How many mL of 0.1 M HCl are required to react completely with 1 g mixture of $\text{Na}_2\text{CO}_3$ and $\text{NaHCO}_3$ containing equimolar amounts of both?

Step 1: Setup variables for masses. Let mass of $\text{Na}_2\text{CO}_3$ be $x$ g. Mass of $\text{NaHCO}_3$ will be $(1-x)$ g. Molar mass of $\text{Na}_2\text{CO}_3$ = 106 g mol⁻¹. Molar mass of $\text{NaHCO}_3$ = 84 g mol⁻¹.
Step 2: Equimolar condition. Moles of $\text{Na}_2\text{CO}_3$ = Moles of $\text{NaHCO}_3$. $$ \frac{x}{106} = \frac{1-x}{84} \implies 84x = 106 - 106x \implies 190x = 106 \implies x = 0.558 \text{ g} $$
Mass of $\text{Na}_2\text{CO}_3$ = 0.558 g $\implies$ Moles = 0.558 / 106 = 0.00526 mol.
Mass of $\text{NaHCO}_3$ = 1 - 0.558 = 0.442 g $\implies$ Moles = 0.00526 mol.
Step 3: Reaction Stoichiometry with HCl.
$$ \text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 $$
$$ \text{NaHCO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 $$
Total moles of HCl needed = $(2 \times 0.00526) + (1 \times 0.00526) = 0.01578 \text{ mol}$.
Step 4: Find volume of 0.1 M HCl. $$ \text{Volume (L)} = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.01578}{0.1} = 0.1578 \text{ L} = 157.8 \text{ mL} $$
Final Answer: 157.8 mL of HCl is required.

Question 2.7: 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.

Step 1: Calculate total mass of solute.
Solute from 1st solution = 25% of 300 g = (25/100) $\times$ 300 = 75 g.
Solute from 2nd solution = 40% of 400 g = (40/100) $\times$ 400 = 160 g.
Total solute mass = 75 + 160 = 235 g.
Step 2: Calculate total mass of solution.
Total solution mass = 300 g + 400 g = 700 g.
Step 3: Calculate final mass percentage.
$$ \text{Mass Percentage of Solute} = \frac{235}{700} \times 100 = 33.57\% $$
$$ \text{Mass Percentage of Solvent} = 100 - 33.57 = 66.43\% $$
Final Answer: Solute mass percentage is 33.57% and solvent mass percentage is 66.43%.

Question 2.8: An antifreeze solution is prepared from 222.6 g of ethylene glycol ($\text{C}_2\text{H}_6\text{O}_2$) 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: Molarity and Molality foundations. Mass of solute ($W_2$) = 222.6 g. Molar mass of ethylene glycol ($M_2$) = (2 $\times$ 12) + (6 $\times$ 1) + (2 $\times$ 16) = 62 g mol⁻¹. Moles of solute ($n_2$) = 222.6 / 62 = 3.59 mol. Mass of solvent ($W_1$) = 200 g = 0.2 kg.
Step 2: Calculate Molality ($m$). $$ m = \frac{3.59 \text{ mol}}{0.2 \text{ kg}} = 17.95 \text{ m} $$
Step 3: Calculate Molarity ($M$). Total Mass of Solution = 222.6 + 200 = 422.6 g.
Volume of Solution = 422.6 g / 1.072 g mL⁻¹ = 394.2 mL = 0.3942 L.
$$ M = \frac{3.59 \text{ mol}}{0.3942 \text{ L}} = 9.11 \text{ M} $$
Final Answer: Molality = 17.95 m and Molarity = 9.11 M.

Question 2.9: A sample of drinking water was found to be severely contaminated with chloroform ($\text{CHCl}_3$) 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.

Step 1: Deconstruct Parts Per Million (ppm). 15 ppm means 15 g of chloroform is present in $10^6$ g of solution.
(i) Percent by mass: $$ \text{Mass \%} = \frac{15}{10^6} \times 100 = 1.5 \times 10^{-3}\% $$
Step 2: Molality calculation. Since the solute mass (15 g) is negligible relative to $10^6$ g of solution, the mass of solvent (water) $\approx 10^6 \text{ g} = 1000 \text{ kg}$.
Molar mass of $\text{CHCl}_3$ = 12 + 1 + (3 $\times$ 35.5) = 119.5 g mol⁻¹.
Moles of Chloroform = 15 / 119.5 = 0.1255 mol.
$$ \text{Molality } (m) = \frac{0.1255 \text{ mol}}{1000 \text{ kg}} = 1.25 \times 10^{-4} \text{ m} $$
Final Answer: (i) 1.5 $\times 10^{-3}$% (ii) 1.25 $\times 10^{-4}$ m.

Question 2.10: What role does the molecular interaction play in a solution of alcohol and water?

Step 1: Initial States. Both alcohol (ethanol) and water molecules exhibit strong intermolecular hydrogen bonding in their pure states.
Step 2: Solution State Interactions. When alcohol and water are mixed, the alcohol-water molecules form weaker hydrogen bonds than the native water-water hydrogen bonds.
Step 3: Outcome. Because the intermolecular attractive forces become weaker in the solution, molecules can escape into the vapor phase more easily. This causes the solution to show a positive deviation from Raoult's Law, leading to a higher vapor pressure and a lower boiling point than predicted ideally.

Question 2.11: Why do gases always tend to be less soluble in liquids as the temperature is raised?

Step 1: Reaction Nature. Dissolution of a gas in a liquid is an exothermic process (heat is evolved): $$ \text{Gas} + \text{Liquid Solvent} \rightleftharpoons \text{Solution} + \text{Heat } (\Delta H = \text{-ve}) $$
Step 2: Le Chatelier's Principle. According to Le Chatelier's Principle, when the temperature is raised, the equilibrium shifts in the reverse direction to consume the added heat. Consequently, the gas escapes from the liquid, making it less soluble at higher temperatures.

Question 2.12: State Henry's law and mention some important applications.

Step 1: Statement. At a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of the liquid. Formally: $p = K_{\text{H}} \cdot x$.
Step 2: Key Applications.
1. To increase solubility of $\text{CO}_2$ in soft drinks: Bottles are sealed under high pressure.
2. Scuba Divers: High pressure deep down increases solubility of atmospheric gases in blood. To avoid "bends", tanks are diluted with less soluble Helium.
3. High Altitudes (Anoxia): Low partial pressure of oxygen leads to low concentrations of $\text{O}_2$ in blood, causing weakness.

Question 2.13: The partial pressure of ethane over a solution containing $6.56 \times 10^{-3} \text{ g}$ of ethane is 1 bar. If the solution contains $5.00 \times 10^{-2} \text{ g}$ of ethane, then what shall be the partial pressure of the gas?

Step 1: Henry's Law Application. The mass of gas dissolved ($m$) is directly proportional to its partial pressure ($p$): $m = K \cdot p$.
Step 2: Find Constant $K$. Case 1: $m_1 = 6.56 \times 10^{-3} \text{ g}$, $p_1 = 1 \text{ bar}$.
$$ 6.56 \times 10^{-3} = K \times 1 \implies K = 6.56 \times 10^{-3} \text{ g bar}^{-1} $$
Step 3: Solve for new pressure. Case 2: $m_2 = 5.00 \times 10^{-2} \text{ g}$, $p_2 = ?$
$$ 5.00 \times 10^{-2} = (6.56 \times 10^{-3}) \times p_2 $$
$$ p_2 = \frac{5.00 \times 10^{-2}}{6.56 \times 10^{-3}} = 7.62 \text{ bar} $$
Final Answer: The partial pressure will be 7.62 bar.

Question 2.14: What is meant by positive and negative deviations from Raoult's law and how is the sign of $\Delta_{\text{mix}}H$ related to positive and negative deviations?

Step 1: Positive Deviation. Occurs when the total vapor pressure of the solution is higher than that calculated from Raoult's law. Solute-solvent molecular interactions (A-B) are weaker than solute-solute (A-A) and solvent-solvent (B-B) interactions.
$$ \Delta_{\text{mix}}H = \text{+ve (Endothermic reaction, heat absorbed)} $$
Step 2: Negative Deviation. Occurs when the total vapor pressure of the solution is lower than that calculated from Raoult's law. Solute-solvent molecular interactions (A-B) are stronger than pure component interactions.
$$ \Delta_{\text{mix}}H = \text{-ve (Exothermic reaction, heat released)} $$

Question 2.15: 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?

Step 1: Identify conditions at normal boiling point. At the normal boiling point of water, pure vapor pressure ($p_1^{\circ}$) = 1 atm = 1.013 bar. Vapor pressure of solution ($p_1$) = 1.004 bar. Solute mass percentage = 2% $\implies$ $W_2$ = 2 g, $W_1$ (water) = 98 g. Molar mass of solvent ($M_1$) = 18 g mol⁻¹.
Step 2: Apply Relative Lowering of Vapor Pressure formula. $$ \frac{p_1^{\circ}-p_1}{p_1^{\circ}} = \frac{W_2 \times M_1}{M_2 \times W_1} $$
$$ \frac{1.013-1.004}{1.013} = \frac{2 \times 18}{M_2 \times 98} $$
$$ \frac{0.009}{1.013} = \frac{36}{98 \cdot M_2} $$
$$ M_2 = \frac{36 \times 1.013}{98 \times 0.009} = 41.35 \text{ g mol}^{-1} $$
Final Answer: The molar mass of the solute is 41.35 g mol⁻¹.

Question 2.16: Heptane and octane form an ideal solution. At 373 K, the vapor pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapor pressure of a mixture of 26.0 g of heptane and 35.0 g of octane?

Step 1: Calculate moles of components. Molar mass of heptane ($\text{C}_7\text{H}_{16}$) = 100 g mol⁻¹. Molar mass of octane ($\text{C}_8\text{H}_{18}$) = 114 g mol⁻¹.
$n_{\text{heptane}}$ = 26.0 / 100 = 0.260 mol.
$n_{\text{octane}}$ = 35.0 / 114 = 0.307 mol.
Step 2: Calculate mole fractions. Total Moles = 0.260 + 0.307 = 0.567 mol.
$x_{\text{heptane}}$ = 0.260 / 0.567 = 0.459.
$x_{\text{octane}}$ = 1 - 0.459 = 0.541.
Step 3: Apply Raoult's law for total pressure. $$ P_{\text{total}} = (p^{\circ} \cdot x)_{\text{heptane}} + (p^{\circ} \cdot x)_{\text{octane}} $$
$$ P_{\text{total}} = (105.2 \times 0.459) + (46.8 \times 0.541) = 48.29 + 25.32 = 73.61 \text{ kPa} $$
Final Answer: The total vapor pressure of the mixture is 73.61 kPa.

Question 2.17: The vapor pressure of water is 12.3 kPa at 300 K. Calculate vapor pressure of 1 molal solution of a non-volatile solute in it.

Step 1: Extract information from "1 molal solution". A 1 molal solution contains 1 mole of solute in 1000 g of water (solvent).
Moles of solute ($n_2$) = 1 mol.
Moles of water ($n_1$) = 1000 / 18 = 55.56 mol.
Step 2: Calculate mole fraction of solute ($x_2$). $$ x_2 = \frac{1}{1+55.56} = \frac{1}{56.56} = 0.0177 $$
Step 3: Relate with relative lowering formula. $$ \frac{p^{\circ} - p}{p^{\circ}} = x_2 \implies \frac{12.3 - p}{12.3} = 0.0177 $$
$$ 12.3 - p = 12.3 \times 0.0177 = 0.218 \text{ kPa} $$
$$ p = 12.3 - 0.218 = 12.08 \text{ kPa} $$
Final Answer: The vapor pressure of the solution is 12.08 kPa.

Question 2.18: 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 vapor pressure to 80%.

Step 1: Parse the phrase "reduce its vapor pressure to 80%". If pure vapor pressure ($p^{\circ}$) = 100 units, then solution vapor pressure ($p$) = 80 units. $$ \frac{p^{\circ} - p}{p^{\circ}} = \frac{100-80}{100} = 0.20 $$
Step 2: Setup composition math. Molar mass of solute ($M_2$) = 40 g mol⁻¹. Mass of octane ($W_1$) = 114 g. Molar mass of octane ($M_1$, $\text{C}_8\text{H}_{18}$) = 114 g mol⁻¹. Moles of solvent octane ($n_1$) = 114 / 114 = 1 mol.
Step 3: Formulate and solve. $$ \frac{p^{\circ}-p}{p^{\circ}} = \frac{n_2}{n_1+n_2} \implies 0.20 = \frac{n_2}{1 + n_2} $$
$$ 0.20(1 + n_2) = n_2 \implies 0.20 + 0.20n_2 = n_2 \implies 0.80n_2 = 0.20 \implies n_2 = 0.25 \text{ moles} $$
$$ \text{Mass of solute } (W_2) = \text{Moles} \times \text{Molar mass} = 0.25 \times 40 = 10 \text{ g} $$
Final Answer: 10 g of non-volatile solute should be added.

Question 2.19: A solution containing 30 g of a non-volatile solute exactly in 90 g of water has a vapor pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution, the new vapor pressure becomes 2.9 kPa at 298 K. Calculate: (i) molar mass of the solute, (ii) vapor pressure of water at 298 K.

Step 1: Setup Equation 1 (Initial State). Let molar mass of solute be $M$. Moles of solute ($n_2$) = 30 / $M$. Moles of water ($n_1$) = 90 / 18 = 5 mol. $$ \frac{p^{\circ} - 2.8}{p^{\circ}} = \frac{30/M}{5 + 30/M} = \frac{30}{5M + 30} = \frac{6}{M + 6} $$
Hence: $\frac{2.8}{p^{\circ}} = 1 - \frac{6}{M+6} = \frac{M}{M+6}$ --- (Eq i)
Step 2: Setup Equation 2 (After adding 18g water). Moles of water now ($n_1'$) = (90+18)/18 = 6 mol. $$ \frac{p^{\circ} - 2.9}{p^{\circ}} = \frac{30/M}{6 + 30/M} = \frac{30}{6M + 30} = \frac{5}{M + 5} $$
Hence: $\frac{2.9}{p^{\circ}} = 1 - \frac{5}{M+5} = \frac{M}{M+5}$ --- (Eq ii)
Step 3: Solve for $M$ by dividing (Eq i) by (Eq ii). $$ \frac{2.8}{2.9} = \frac{M/(M+6)}{M/(M+5)} = \frac{M+5}{M+6} $$
$$ 2.8(M+6) = 2.9(M+5) \implies 2.8M + 16.8 = 2.9M + 14.5 \implies 0.1M = 2.3 \implies M = 23 \text{ g mol}^{-1} $$
Step 4: Solve for $p^{\circ}$. Substitute $M=23$ back into Eq i: $$ \frac{2.8}{p^{\circ}} = \frac{23}{23+6} = \frac{23}{29} \implies p^{\circ} = \frac{2.8 \times 29}{23} = 3.53 \text{ kPa} $$
Final Answer: (i) Molar mass of solute = 23 g mol⁻¹ (ii) Pure vapor pressure of water = 3.53 kPa.

Question 2.20: A 5% solution (by mass) of cane sugar (molar mass = 342 g mol⁻¹) in water has freezing point of 271 K. Calculate the freezing point of 5% glucose (molar mass = 180 g mol⁻¹) in water if freezing point of pure water is 273.15 K.

Step 1: Analyze Cane Sugar Data to discover $K_f$. Mass of solute ($W_2$) = 5 g. Mass of water ($W_1$) = 95 g = 0.095 kg.
$\Delta T_f = T_f^{\circ} - T_f = 273.15 - 271 = 2.15 \text{ K}$.
$$ 2.15 = K_f \times \frac{5/342}{0.095} \text{ --- (Eq 1)} $$
Step 2: Analyze Glucose Data. Mass of solute ($W_2$) = 5 g. Mass of water ($W_1$) = 95 g = 0.095 kg.
$$ \Delta T_f = K_f \times \frac{5/180}{0.095} \text{ --- (Eq 2)} $$
Step 3: Divide Eq 2 by Eq 1. $$ \frac{\Delta T_f}{2.15} = \frac{5/180}{5/342} = \frac{342}{180} = 1.9 $$
$$ \Delta T_f = 1.9 \times 2.15 = 4.085 \text{ K} $$
Step 4: Determine Final Freezing Point. $$ T_f = T_f^{\circ} - \Delta T_f = 273.15 - 4.085 = 269.06 \text{ K} $$
Final Answer: The freezing point of the glucose solution is 269.06 K.

Extra Important Questions (Board Style)

Target verification tasks matching high-yield board testing frameworks engineered to lock down full marks in your 2026 exams.

Section A: Multiple Choice Questions (MCQs)

Q1. Which of the following concentration units is independent of temperature changes?

A) Molarity
B) Normality
C) Molality
D) Formality

Correct Answer Choice: C
Step 1: Metric Verification. Molality maps the number of moles of solute per kg of solvent. Since mass does not expand or contract with temperature variations (unlike solution volume), molality remains constant.

Q2. If an ideal solution is prepared by mixing two pure liquid elements, which thermodynamic condition is true?

A) $\Delta_{\text{mix}}V > 0$
B) $\Delta_{\text{mix}}H = 0$
C) $\Delta_{\text{mix}}S = 0$
D) $\Delta_{\text{mix}}G = 0$

Correct Answer Choice: B
Step 1: Structural Verification. For completely ideal solutions, there is no structural net heat released or absorbed when breaking and remaking interactions, meaning enthalpy change $\Delta H = 0$. However, entropy $\Delta S$ is always positive when mixing.

Section B: Assertion-Reason Questions

Directions: Choose option (A) if both statements are true and the reason provides the exact explanation. Choose (B) if both are true but independent. Choose (C) if the assertion is true but the reason is false. Choose (D) if assertion is false.

Q3. Assertion (A): Adding NaCl to water elevates its boiling point temperature.
Reason (R): The vapor pressure of an aqueous solution decreases when a non-volatile solute is dissolved.

Correct Answer Choice: A
Step 1: Phenomenon Validation. Dissolving a non-volatile solute like salt introduces particles that occupy surface area, suppressing the escape capability of water molecules, dropping the overall vapor pressure. Consequently, the liquid requires extra heating to bring its vapor pressure up to standard atmospheric conditions, elevating the boiling point.

Section C: Short Answer Questions (SAQs)

Q4. What are azeotropes? Distinguish briefly between minimum boiling and maximum boiling azeotropes.

Step 1: Definition. Azeotropes are binary liquid mixtures having a constant composition that boil completely at a uniform temperature without undergoing fractional distillation change.
Step 2: Minimum Boiling. Formed by non-ideal solutions showing a significant positive deviation from Raoult's Law (e.g., 95% ethanol + 5% water). Their boiling point is lower than either component.
Step 3: Maximum Boiling. Formed by solutions displaying a heavy negative deviation from Raoult's Law (e.g., 68% nitric acid + 32% water). Their boiling point is higher than either component.

Q5. Why is osmotic pressure measurement preferred over other colligative properties to determine macromolecular masses like proteins or polymers?

Step 1: Temperature Safety. Pressure readings are taken systematically around standard room temperature, preventing natural protein structural denaturing that occurs at high temperatures (like boiling points).
Step 2: Scale Advantage. It yields high-magnitude, easily measurable values even for highly dilute solutions, whereas changes like $\Delta T_f$ are too small to gauge accurately.

Section D: Case-Based / Integrated Long Questions

Visual Note for Students: When practicing these biological fluid numericals, it helps to sketch out two containers: one depicting a normal red blood cell in balance, and another showing it shrinking or swelling depending on the salt concentration outside.

Q6. Case Study: Blood cells are dynamic osmotically active biological systems enclosed by semipermeable membranes. (i) What happens physically to a red blood cell when placed inside an aqueous solution containing 1.5% NaCl mass fraction? Why? (ii) Calculate the structural osmotic pressure of a human blood environment at a standard body climate of 37°C, assuming it matches an equivalent concentration of 0.15 M non-permeable solutes ($R = 0.0821 \text{ L atm mol}^{-1} \text{ K}^{-1}$).

Step 1: Cell Physical Reaction. Human blood fluid is osmotically isotonic with roughly 0.9% (w/v) NaCl solution. An external concentration of 1.5% creates a hypertonic environment. Water inside the cell will flow out via osmosis, causing the red blood cells to shrink (crenation).
Step 2: Osmotic Pressure Calculation.
$C = 0.15 \text{ mol L}^{-1}$.
$T = 37 + 273.15 = 310.15 \text{ K}$.
$$ \pi = C \cdot R \cdot T = 0.15 \times 0.0821 \times 310.15 = 3.82 \text{ atm} $$

Common Mistakes Students Make

Exam Preparation Tips

Frequently Asked Questions (FAQs)

1. Is Chapter 2 Solutions high scoring for CBSE Class 12 Chemistry?
Yes! It carries about 7 marks in the core board framework, split predictably between conceptual definitions (like Henry's Law or Azeotropes) and direct formula-based numerical problems.
2. Where can I quickly download verified NCERT textbook solutions?
You can access all official, curriculum-aligned text files directly from the NCERT official portal, or use this guide as your permanent verified reference copy.
3. What is the difference between an ideal and a non-ideal solution?
An ideal solution obeys Raoult's Law at all concentrations and temperatures, showing no heat change ($\Delta H=0$) or volume change ($\Delta V=0$) upon mixing. A non-ideal solution deviates from Raoult's Law due to differences in intermolecular forces.
4. Why does raw ice melt quickly when salt is sprinkled on it?
Salt acts as a non-volatile foreign solute that lowers the freezing point of water. This shifts the freezing threshold below the ambient freezing point, making ice melt quickly at regular winter conditions.
5. What exactly does an abnormal molar mass signify?
It indicates that the solute particles have either underwent dissociation (splitting into more ions, lowering apparent molecular mass) or association (dimerizing into clusters, raising apparent molecular mass) inside the solution phase.

Conclusion: Understanding the chemistry of Solutions comes down to practicing standard numerical calculations and clear conceptual definitions. Make sure you revise Henry's and Raoult's laws, pay attention to the van 't Hoff factor during exams, and solve past year papers to boost your confidence. Keep studying regularly and you'll do great in your Boards 2026!

More Class 12 Chemistry Chapters

Ch 1: The Solid State Ch 2: Solutions Ch 3: Electrochemistry Ch 4: Chemical Kinetics Ch 5: Surface Chemistry Ch 6: General Principles Ch 7: The p-Block Elements Ch 8: The d- & f-Block Elements Ch 9: Coordination Compounds Ch 10: Haloalkanes Ch 11: Alcohols & Phenols Ch 12: Aldehydes & Ketones Ch 13: Amines Ch 14: Biomolecules All Subjects Dashboard