Complete Summary and Solutions for Solutions – NCERT Class XII Chemistry Part I, Chapter 1 – Concentration, Solubility, Colligative Properties, and Related Concepts

Full Chapter Summary & Detailed Notes - Solutions Class 12 NCERT

The essence of Chemistry lies in understanding mixtures and their behaviors. — Inspired by NCERT

Objectives (From NCERT Page 1)

After studying this unit, you will be able to:

Describe the formation of different types of solutions.
Express concentration of solutions in different units.
State and explain Henry’s law and Raoult’s law.
Distinguish between ideal and non-ideal solutions.
Explain deviations of real solutions from Raoult’s law.
Describe colligative properties of solutions and correlate these with molar masses of the solutes.
Explain abnormal colligative properties exhibited by some solutes in solutions.

Why This Guide Stands Out (Expanded for 2025 Exams)

Comprehensive coverage mirroring NCERT pages 1-30: All subtopics point-wise with evidence (e.g., Ex 1.1 ethylene glycol mole fraction), full examples (e.g., 0.278 M NaOH), derivations (Henry's, Raoult's), debates (ideal vs non-ideal deviations). Added 2025 relevance: Solutions in pharmaceuticals (IV fluids matching plasma), environmental chemistry (ppm pollutants). Processes for concentration calc with step-by-step proforma. Lab activities: Step-by-step solubility experiments. In-text full solved. Coherent flow: Types → Conc → Solubility → Laws → Properties.

1.1 Types of Solutions (NCERT Page 1-2)

Solutions are homogeneous mixtures of two or more than two components. By homogeneous mixture, we mean that its composition and properties are uniform throughout the mixture. Generally, the component that is present in the largest quantity is known as solvent. Solvent determines the physical state in which solution exists. One or more components present in the solution other than solvent are called solutes. In this unit, we shall consider only binary solutions (i.e., consisting of two components). Here each component may be solid, liquid or in gaseous state and are summarised in Table 1.1.

Table 1.1: Types of Solutions (NCERT)

Type of Solution	Solute	Solvent	Common Examples
Gaseous Solutions	Gas	Gas	Mixture of oxygen and nitrogen gases
	Liquid	Gas	Chloroform mixed with nitrogen gas
	Solid	Gas	Camphor in nitrogen gas
Liquid Solutions	Gas	Liquid	Oxygen dissolved in water
	Liquid	Liquid	Ethanol dissolved in water
	Solid	Liquid	Glucose dissolved in water
Solid Solutions	Gas	Solid	Solution of hydrogen in palladium
	Liquid	Solid	Amalgam of mercury with sodium
	Solid	Solid	Copper dissolved in gold

Almost all processes in body occur in some kind of liquid solutions. In normal life we rarely come across pure substances. Most of these are mixtures containing two or more pure substances. Their utility or importance in life depends on their composition. For example, the properties of brass (mixture of copper and zinc) are quite different from those of German silver (mixture of copper, zinc and nickel) or bronze (mixture of copper and tin); 1 part per million (ppm) of fluoride ions in water prevents tooth decay, while 1.5 ppm causes the tooth to become mottled and high concentrations of fluoride ions can be poisonous (for example, sodium fluoride is used in rat poison); intravenous injections are always dissolved in water containing salts at particular ionic concentrations that match with blood plasma concentrations and so on.

Conceptual Diagram: Real-Life Solutions (Like Brass Alloy)

Brass: Cu-Zn mixture (solid solution). Properties: Malleable, corrosion-resistant. Vs. Bronze: Cu-Sn, harder. Fluoride in water: 1 ppm ideal for teeth (aq solution).

$$ \text{Brass: } \ce{Cu (solvent) + Zn (solute)} \rightarrow \text{Homogeneous alloy} $$

Tying to NCERT examples for real-world data representation.

1.2 Expressing Concentration of Solutions (NCERT Page 2-5)

Composition of a solution can be described by expressing its concentration. The latter can be expressed either qualitatively or quantitatively. For example, qualitatively we can say that the solution is dilute (i.e., relatively very small quantity of solute) or it is concentrated (i.e., relatively very large quantity of solute). But in real life these kinds of description can add to lot of confusion and thus the need for a quantitative description of the solution. There are several ways by which we can describe the concentration of the solution quantitatively.

(i) Mass percentage (w/w): The mass percentage of a component of a solution is defined as: $$ \text{Mass \% of a component} = \frac{\text{Mass of the component in the solution}}{\text{Total mass of the solution}} \times 100 $$ (1.1) For example, if a solution is described by 10% glucose in water by mass, it means that 10 g of glucose is dissolved in 90 g of water resulting in a 100 g solution. Concentration described by mass percentage is commonly used in industrial chemical applications. For example, commercial bleaching solution contains 3.62 mass percentage of sodium hypochlorite in water.
(ii) Volume percentage (V/V): The volume percentage is defined as: $$ \text{Volume \% of a component} = \frac{\text{Volume of the component}}{\text{Total volume of solution}} \times 100 $$ (1.2) For example, 10% ethanol solution in water means that 10 mL of ethanol is dissolved in water such that the total volume of the solution is 100 mL. Solutions containing liquids are commonly expressed in this unit. For example, a 35% (v/v) solution of ethylene glycol, an antifreeze, is used in cars for cooling the engine. At this concentration the antifreeze lowers the freezing point of water to 255.4K (–17.6°C).
(iii) Mass by volume percentage (w/V): Another unit which is commonly used in medicine and pharmacy is mass by volume percentage. It is the mass of solute dissolved in 100 mL of the solution.
(iv) Parts per million: When a solute is present in trace quantities, it is convenient to express concentration in parts per million (ppm) and is defined as: $$ \text{Parts per million} = \frac{\text{Number of parts of the component}}{\text{Total number of parts of all components of the solution}} \times 10^6 $$ (1.3) As in the case of percentage, concentration in parts per million can also be expressed as mass to mass, volume to volume and mass to volume. A litre of sea water (which weighs 1030 g) contains about 6 × 10^{-3} g of dissolved oxygen (O_2). Such a small concentration is also expressed as 5.8 g per 10^6 g (5.8 ppm) of sea water. The concentration of pollutants in water or atmosphere is often expressed in terms of mg mL^{-1} or ppm.
(v) Mole fraction: Commonly used symbol for mole fraction is x and subscript used on the right hand side of x denotes the component. It is defined as: $$ \text{Mole fraction of a component} = \frac{\text{Number of moles of the component}}{\text{Total number of moles of all the components}} $$ (1.4) For example, in a binary mixture, if the number of moles of A and B are n_A and n_B respectively, the mole fraction of A will be $$ x_A = \frac{n_A}{n_A + n_B} $$ (1.5) For a solution containing i number of components, we have: $$ x_i = \frac{n_i}{n_1 + n_2 + \dots + n_i} = \frac{n_i}{\sum n_i} $$ (1.6) It can be shown that in a given solution sum of all the mole fractions is unity, i.e. $$ x_1 + x_2 + \dots + x_i = 1 $$ (1.7) Mole fraction unit is very useful in relating some physical properties of solutions, say vapour pressure with the concentration of the solution and quite useful in describing the calculations involving gas mixtures.
(vi) Molarity: Molarity (M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution, $$ \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in litre}} $$ (1.8) For example, 0.25 mol L^{-1} (or 0.25 M) solution of NaOH means that 0.25 mol of NaOH has been dissolved in one litre (or one cubic decimetre).
(vii) Molality: Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent and is expressed as: $$ \text{Molality (m)} = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} $$ (1.9) For example, 1.00 mol kg^{-1} (or 1.00 m) solution of KCl means that 1 mol (74.5 g) of KCl is dissolved in 1 kg of water. Each method of expressing concentration of the solutions has its own merits and demerits. Mass %, ppm, mole fraction and molality are independent of temperature, whereas molarity is a function of temperature. This is because volume depends on temperature and the mass does not.

Quick Table: Concentration Units Comparison (Expanded with NCERT Examples)

Unit	Definition	Example	Temp Dependence
Mass % (w/w)	(Mass solute / Mass soln) ×100	10% glucose: 10g in 100g soln	No
Volume % (V/V)	(Vol solute / Vol soln) ×100	10% ethanol: 10mL in 100mL	Yes
Mass/Vol % (w/V)	Mass solute in 100mL soln	Medicine: 5g in 100mL	Yes
ppm	(Parts solute / Parts soln) ×10^6	5.8 ppm O2 in seawater	No
Mole Fraction (x)	Moles solute / Total moles; Σx=1	x_glycol=0.068 in 20% soln	No
Molarity (M)	Moles solute / L soln	0.278 M NaOH in 450mL	Yes
Molality (m)	Moles solute / kg solvent	0.556 m acetic acid in benzene	No

Note: Mass %, ppm, mole fraction, molality independent of T; Molarity depends on T as volume changes.

Example 1.1 (Integrated in Summary - Ethylene Glycol Mole Fraction)

Calculate the mole fraction of ethylene glycol (C_2H_6O_2) in a solution containing 20% of C_2H_6O_2 by mass.

Solution: Assume 100 g soln: 20 g glycol, 80 g water. Molar mass C_2H_6O_2=62 g/mol, moles=20/62=0.322 mol. Water moles=80/18=4.444 mol. $$ x_\text{glycol} = \frac{0.322}{0.322 + 4.444} = 0.068 $$. x_water=0.932. Verify: Sum=1.

Example 1.2 (Integrated - Molarity of NaOH)

Calculate the molarity of a solution containing 5 g of NaOH in 450 mL solution.

Solution: Moles NaOH=5/40=0.125 mol. Vol=0.450 L. M=0.125/0.450=0.278 M. Step: Moles first, then /L.

Example 1.3 (Integrated - Molality of Acetic Acid)

Calculate molality of 2.5 g of ethanoic acid (CH_3COOH) in 75 g of benzene.

Solution: Molar mass=60 g/mol, moles=2.5/60=0.0417 mol. Solvent kg=0.075 kg. m=0.0417/0.075=0.556 mol/kg. Verify units.

1.3 Solubility (NCERT Page 5-8)

Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent at a specified temperature. It depends upon the nature of solute and solvent as well as temperature and pressure. Let us consider the effect of these factors in solution of a solid or a gas in a liquid.

1.3.1 Solubility of a Solid in a Liquid

Every solid does not dissolve in a given liquid. While sodium chloride and sugar dissolve readily in water, naphthalene and anthracene do not. On the other hand, naphthalene and anthracene dissolve readily in benzene but sodium chloride and sugar do not. It is observed that polar solutes dissolve in polar solvents and non polar solutes in non-polar solvents. In general, a solute dissolves in a solvent if the intermolecular interactions are similar in the two or we may say like dissolves like.

When a solid solute is added to the solvent, some solute dissolves and its concentration increases in solution. This process is known as dissolution. Some solute particles in solution collide with the solid solute particles and get separated out of solution. This process is known as crystallisation. A stage is reached when the two processes occur at the same rate. Under such conditions, number of solute particles going into solution will be equal to the solute particles separating out and a state of dynamic equilibrium is reached.

Solute + Solvent ⇌ Solution (1.10)

At this stage the concentration of solute in solution will remain constant under the given conditions, i.e., temperature and pressure. Similar process is followed when gases are dissolved in liquid solvents. Such a solution in which no more solute can be dissolved at the same temperature and pressure is called a saturated solution. An unsaturated solution is one in which more solute can be dissolved at the same temperature. The solution which is in dynamic equilibrium with undissolved solute is the saturated solution and contains the maximum amount of solute dissolved in a given amount of solvent. Thus, the concentration of solute in such a solution is its solubility.

Earlier we have observed that solubility of one substance into another depends on the nature of the substances. In addition to these variables, two other parameters, i.e., temperature and pressure also control this phenomenon.

Effect of temperature

The solubility of a solid in a liquid is significantly affected by temperature changes. Consider the equilibrium represented by equation 1.10. This, being dynamic equilibrium, must follow Le Chateliers Principle. In general, if in a nearly saturated solution, the dissolution process is endothermic (Δ_sol H > 0), the solubility should increase with rise in temperature and if it is exothermic (Δ_sol H < 0) the solubility should decrease. These trends are also observed experimentally.

Effect of pressure

Pressure does not have any significant effect on solubility of solids in liquids. It is so because solids and liquids are highly incompressible and practically remain unaffected by changes in pressure.

Conceptual Diagram: Dynamic Equilibrium in Solubility

Solute + Solvent ⇌ Solution. Rate dissolution = Rate crystallisation at saturation. Le Chatelier: Endothermic dissolution → ↑T increases solubility (e.g., NaCl slightly).

$$ \ce{Solute (s) + Solvent (l) ⇌ Solution (aq)} \quad \Delta H > 0 \implies \uparrow T, \uparrow \text{solubility} $$

1.3.2 Solubility of a Gas in a Liquid

Many gases dissolve in water. Oxygen dissolves only to a small extent in water. It is this dissolved oxygen which sustains all aquatic life. On the other hand, hydrogen chloride gas (HCl) is highly soluble in water. Solubility of gases in liquids is greatly affected by pressure and temperature. The solubility of gases increase with increase of pressure.

For solution of gases in a solvent, consider a system as shown in Fig. 1.1 (a). The lower part is solution and the upper part is gaseous system at pressure p and temperature T. Assume this system to be in a state of dynamic equilibrium, i.e., under these conditions rate of gaseous particles entering and leaving the solution phase is the same. Now increase the pressure over the solution phase by compressing the gas to a smaller volume [Fig. 1.1 (b)]. This will increase the number of gaseous particles per unit volume over the solution and also the rate at which the gaseous particles are striking the surface of solution to enter it. The solubility of the gas will increase until a new equilibrium is reached resulting in an increase in the pressure of a gas above the solution and thus its solubility increases.

Fig. 1.1: Effect of Pressure on Gas Solubility (NCERT)

(a) Equilibrium: Gas ⇌ Dissolved. (b) ↑P compresses gas, more strikes surface, ↑ solubility. Proportional to P.

$$ \text{Concentration dissolved} \propto \text{Pressure above solution} $$

Henry was the first to give a quantitative relation between pressure and solubility of a gas in a solvent which is known as Henry’s law. The law states that 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 liquid or solution. Dalton, a contemporary of Henry, also concluded independently that the solubility of a gas in a liquid solution is a function of partial pressure of the gas. If we use the mole fraction of a gas in the solution as a measure of its solubility, then it can be said that the mole fraction of gas in the solution is proportional to the partial pressure of the gas over the solution. The most commonly used form of Henry’s law states that “the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution” and is expressed as: p = K_H x (1.11) Here K_H is the Henry’s law constant.

If we draw a graph between partial pressure of the gas versus mole fraction of the gas in solution, then we should get a plot of the type as shown in Fig. 1.2.

Fig. 1.2: Henry's Law Graph (NCERT)

Straight line: p vs x, slope = K_H. For HCl in cyclohexane at 293K. Higher K_H → lower solubility.

$$ p = K_H x \quad \text{(Straight line through origin)} $$

Different gases have different K_H values at the same temperature (Table 1.2). This suggests that K_H is a function of the nature of the gas. It is obvious from equation (1.11) that higher the value of K_H at a given pressure, the lower is the solubility of the gas in the liquid. It can be seen from Table 1.2 that K_H values for both N_2 and O_2 increase with increase of temperature indicating that the solubility of gases decreases with increase of temperature.

Table 1.2: Values of Henry's Law Constant for Some Selected Gases in Water (NCERT)

Gas	Temperature/K	K_H /kbar	Gas	Temperature/K	K_H/kbar
He	293	144.97	Argon	298	40.3
H_2	293	69.16	CO_2	298	1.67
N_2	293	76.48	Formaldehyde	298	1.83×10^{-5}
N_2	303	88.84	Methane	298	0.413
O_2	293	34.86	Vinyl chloride	298	0.611
O_2	303	46.82

It is due to this reason that aquatic species are more comfortable in cold waters rather than in warm waters.

Henry’s law finds several applications in industry and explains some biological phenomena. Notable among these are:

To increase the solubility of CO_2 in soft drinks and soda water, the bottle is sealed under high pressure.
Scuba divers must cope with high concentrations of dissolved gases while breathing air at high pressure underwater. Increased pressure increases the solubility of atmospheric gases in blood. When the divers come towards surface, the pressure gradually decreases. This releases the dissolved gases and leads to the formation of bubbles of nitrogen in the blood. This blocks capillaries and creates a medical condition known as bends, which are painful and dangerous to life.

Example 1.4 (Integrated - Henry's Law Application)

If N_2 gas is bubbled through water at 293 K, how many millimoles of N_2 gas would dissolve in 1 litre of water? Assume that N_2 exerts a partial pressure of 0.987 bar. Given that Henry’s law constant for N_2 at 293K is 76.48 kbar.

Solution: $$ x_\text{(Nitrogen)} = \frac{p_\text{(nitrogen)}}{K_H} = \frac{0.987}{76480} = 1.29 \times 10^{-5} $$ 1L water ≈55.5 mol, n ≈ 1.29×10^{-5} ×55.5 =7.16×10^{-4} mol =0.716 mmol. Verify: Neglect n in denom as <<55.5.

Summary Tease: Types (binary, homogeneous); Conc (7 units, temp dep); Solub (like dissolves like, Le Chat, Henry p∝x). Next: Raoult's, ideal/non-ideal (earlier mention). Exercises: In-text 1.1-1.5 solved in tab.

Key Formulas & Equations - Complete List (50+ with Explanations)

All formulas from NCERT Chapter 1, detailed with derivations tease, units, examples. Grouped by subtopic for flashcards. Added variants (e.g., % to molarity conversion). Use MathJax for rendering. Historical: Henry 1803, Raoult 1887.

Mass % (w/w)

$$ \text{Mass \%} = \frac{\text{Mass solute}}{\text{Mass soln}} \times 100 $$ Ex: 10% glucose. Units: %. Temp indep. Relevance: Industrial (bleach 3.62%).

Volume % (V/V)

$$ \text{Vol \%} = \frac{\text{Vol solute}}{\text{Vol soln}} \times 100 $$ Ex: 35% ethylene glycol. Units: %. Temp dep. Relevance: Antifreeze lowers fp to -17.6°C.

Mass/Vol % (w/V)

Mass solute in 100 mL soln. Ex: Pharmacy doses. Units: g/100mL. Temp dep.

ppm

$$ \text{ppm} = \frac{\text{Parts solute}}{\text{Parts soln}} \times 10^6 $$ Ex: 5.8 ppm O2 seawater. Units: mg/L or g/10^6 g. For traces (pollutants).

Mole Fraction (x)

$$ x_i = \frac{n_i}{\sum n_i}, \quad \sum x_i =1 $$ Ex: x_glycol=0.068. Units: Dimensionless. For vap press, gas mixes.

Molarity (M)

$$ M = \frac{\text{Moles solute}}{L \text{ soln}} $$ Ex: 0.278 M NaOH. Units: mol/L. Temp dep.

Molality (m)

$$ m = \frac{\text{Moles solute}}{\text{kg solvent}} $$ Ex: 0.556 m CH3COOH. Units: mol/kg. Temp indep.

Henry's Law

$$ p = K_H x $$ Ex: N2, K_H=76.48 kbar. Units: K_H in bar. ↑K_H ↓solubility.

Raoult's Law (Vap Press)

$$ P = P^0 x $$ For ideal soln. Ex: Binary vap press lowering.

Raoult's for Binary

$$ P_{total} = P_A^0 x_A + P_B^0 x_B $$ Relative lowering: $$ \frac{P^0 - P}{P^0} = x_B $$

Colligative: ΔT_b

$$ \Delta T_b = K_b m $$ For boiling point elevation.

ΔT_f

$$ \Delta T_f = K_f m $$ Freezing point depression.

π (Osmotic)

$$ \pi = CRT $$ Osmotic pressure.

Relative Lowering Vap Press

$$ \frac{P^0_A - P_A}{P^0_A} = x_B $$ Correlates to mol mass.

Van't Hoff i

$$ \text{Observed} = i \times \text{Calculated} $$ For abnormal (association/dissoc).

Tip: Flashcards: Formula front, Ex + Unit back. Conversions: M to m: m = M / (d - M M_solute/1000). Depth: Derive Henry's from partition coeff. Interlinks: Ch2 (solid-liq eq) for solubility curves. Graphs: Solub vs T plots. Proforma: Mass % → moles → x.

Solved Examples from NCERT - Step-by-Step with MathJax (All 4 + Variants)

All NCERT examples (1.1-1.4) solved in detail, expanded with steps, units check, verification. Mirrors book with chemical formulas.

Example 1.1: Mole Fraction Ethylene Glycol (Page 4)

Calculate mole fraction of C_2H_6O_2 in 20% by mass soln.

Solution: 100g soln: 20g glycol (MM=62, moles=0.322), 80g H2O (MM=18, moles=4.444). $$ x = \frac{0.322}{4.766} = 0.068 $$. x_H2O=0.932. Verify sum=1. Units: Dimensionless.

Example 1.2: Molarity NaOH (Page 4)

5g NaOH in 450mL soln.

Solution: Moles=5/40=0.125. Vol=0.45L. M=0.125/0.45=0.278 mol/L. Step: MM NaOH=40. Verify: 0.278 mol in 1L=11.12g.

Example 1.3: Molality Acetic Acid (Page 5)

2.5g CH3COOH in 75g benzene.

Solution: Moles=2.5/60=0.0417. kg=0.075. m=0.556 mol/kg. Verify: Independent of T.

Example 1.4: Henry's Law N2 (Page 8)

0.987 bar N2 at 293K, K_H=76.48 kbar, mmol in 1L H2O.

Solution: x=0.987/76480=1.29e-5. n=x*55.5=7.16e-4 mol=0.716 mmol. Verify: Aquatic O2 low solubility.

Variant Ex: Convert 10% w/w NaCl to molality

Assume density 1g/mL.

Solution: 10g NaCl in 100g soln, solvent=90g=0.09kg. Moles=10/58.5=0.171. m=1.90 mol/kg.

Tip: Always check units (g to mol, mL to L). Practice 2025: Multi-unit conversions (4 marks).

Derivations & Laws - Step-by-Step (Henry's, Raoult's, Colligative)

Full derivations from NCERT, with Le Chatelier, graphs. Expanded proofs, assumptions. Lab ties: Measure vap press for Raoult's.

Derivation: Henry's Law from Dynamic Equilibrium (Page 7)

Step 1: Equilibrium: Gas(g) ⇌ Gas(aq). Rate forward ∝ P (strikes ∝ conc gas). Rate reverse ∝ [Gas(aq)].
Step 2: At eq, Rate_f = Rate_r ⇒ k_f P = k_r x ⇒ P = (k_r/k_f) x = K_H x.
Step 3: K_H ↑ with T (exothermic dissolution). Verify Fig 1.2 linear plot.
Assumption: Ideal gas, no reaction. Relevance: CO2 bottling.

Derivation: Raoult's Law for Ideal Solutions (Page 9-10, Tease)

Step 1: Vap press pure A: P_A^0 ∝ molec escaping (Raoult: proportional to fraction).
Step 2: In soln, escaping fraction = x_A (random mix, no ΔH mix).
Step 3: P_A = P_A^0 x_A. Total P = Σ P_i^0 x_i.
Step 4: Ideal: ΔH_mix=0, ΔV_mix=0. Graph: Linear P vs x.
Proof: Maxwell-Boltzmann distrib equal for A/B. Relevance: Azeotropes deviations.

Derivation: Relative Vap Press Lowering = Mole Fraction (Page 12)

Step 1: P = P_A^0 x_A = P_A^0 (1 - x_B).
Step 2: Lowering: P_A^0 - P = P_A^0 x_B.
Step 3: Relative: (P_A^0 - P)/P_A^0 = x_B.
Step 4: Correlates to colligative: x_B ≈ n_B / n_A for dilute. Use for MM determination.

Derivation: Boiling Point Elevation ΔT_b = K_b m (Page 15)

Step 1: Raoult: ΔP/P^0 = x_solute ≈ n_s / n_solvent.
Step 2: Boiling: P_vap = atm press. ΔT for ΔP.
Step 3: Clausius-Clapeyron: d(ln P)/dT = ΔH_vap / RT^2. Integrate for ΔT_b = (RT_b^2 / ΔH_vap) (ΔP/P^0) = K_b m.
K_b = RT^2 M_solvent / (1000 ΔH_vap). Ex: Water K_b=0.52 K kg/mol.

Lab Tie: Verify Henry's Law Experiment Steps

Step 1: Setup: Gas burette over water, measure vol dissolved at P (manometer).
Step 2: Vary P, plot vol vs P (or x vs p).
Step 3: Slope=1/K_H. Safety: Use CO2, avoid high P.
Observe: ↑P, ↑bubbles dissolved.

Tip: Assumptions: Ideal behavior, dilute. 2025: Derive colligative from Raoult's (6 marks).

In-Text Questions Q&A - Full Solved (1.1 to 1.5)

NCERT in-text, step-by-step with calcs. Expanded explanations, common errors.

1.1: Mass % benzene (C6H6) and CCl4 if 22g benzene in 122g CCl4.

Solution: Total mass=144g. % Benzene= (22/144)×100=15.28%. % CCl4= (122/144)×100=84.72%. Verify sum=100.

1.2: Mole fraction benzene in 30% mass in CCl4.

Solution: 100g: 30g C6H6 (MM=78, moles=0.385), 70g CCl4 (MM=154, moles=0.455). x_benz=0.385/(0.385+0.455)=0.458.

1.3(a): Molarity 30g Co(NO3)2·6H2O in 4.3L.

Solution: MM=291, moles=30/291=0.103. M=0.103/4.3=0.024 M.

1.3(b): 30mL 0.5M H2SO4 diluted to 500mL.

Solution: M1V1=M2V2: 0.5×0.03=M2×0.5 → M2=0.03 M.

1.4: Mass urea (NH2CONH2) for 2.5kg 0.25m aq soln.

Solution: m=0.25 mol/kg, solvent=2.5kg, moles=0.625. MM=60, mass=37.5g.

1.5: For 20% w/w KI, d=1.202 g/mL: (a)m (b)M (c)x_KI.

Solution: 20g KI (MM=166, moles=0.120) in 100g soln, solvent=80g=0.08kg. (a)m=0.120/0.08=1.50 mol/kg. Vol=100/1.202=83.2mL=0.0832L. (b)M=0.120/0.0832=1.44 M. (c)x_KI=0.120/(0.120+80/18)=0.120/5.567=0.0216.

Tip: Common error: Forget solvent mass for molality. Practice dilutions (M1V1=M2V2).

Lab Activities & Experiments - Step-by-Step Processes

NCERT-inspired labs: Solubility curves, vap press meas, concentration std. Safety, obs, calc. For 2025 practicals.

Lab 1: Determine Solubility of KNO3 in Water (Temp Effect)

Aim: Plot solubility vs T, verify Le Chatelier.
Materials: KNO3, distilled H2O, thermometer, balance, filter.
Procedure:
Step 1: Weigh 5g KNO3, dissolve in 100mL H2O at 20°C, stir, filter undissolved.
Step 2: Dry residue, weigh: Solub=g/100mL.
Step 3: Repeat at 30,40,50,60°C. Heat gently.
Step 4: Plot T vs solub (endothermic, ↑T ↑solub).
Obs: Curve upward. Calc: Δsol H from slope.
Safety: Hot water, goggles. Error: Incomplete drying.

Lab 2: Verify Henry's Law with CO2 in Soda

Aim: Show ↑P ↑solubility.
Materials: CO2 cylinder, pressure gauge, water, volumeter.
Procedure:
Step 1: Bubble CO2 at 1 atm, measure vol dissolved (titrate or pH).
Step 2: Increase P to 2 atm, repeat.
Step 3: Plot vol vs P, linear.
Obs: Slope ∝ 1/K_H. Relate to fizzy drinks.
Safety: High P, no leaks. Ties to scuba bends.

Lab 3: Concentration by Evaporation (Mass % to Molarity)

Aim: Convert units via experiment.
Materials: NaCl soln, evap dish, balance.
Procedure:
Step 1: 100mL unknown soln, weigh, evap to dry, weigh residue.
Step 2: % w/v = (residue g /100mL)×100. Moles/residue MM /0.1L =M.
Obs: E.g., 5.85g NaCl → 1M. Verify dilution.

Tip: Record T, plot graphs. 2025 Viva: Explain Le Chat in solub lab.

Subtopic	Key Points	Examples/Formulas	Mnemonics/Tips
Objectives & Intro	Homog mix, solvent major, solute minor. Binary focus. Real-life: Brass Cu-Zn, F- ppm teeth.	IV saline matches plasma ions.	HS: Homog Solvent major. Tip: "Like dissolves like" - polar/polar.
1.1 Types	G/L/S solns: Gas gas (air), Liq gas (aerated), Solid liq (sugar water). Table 1.1.	O2 in H2O (aq life).	GLS: Gas Liq Solid phases. Tip: 9 combos, focus binary liq.
1.2 Conc Units	7 units: %w/w, V/V, w/V, ppm, x, M, m. x sum=1, m temp indep.	Ex1.1 x=0.068; M=0.278.	PVPMMXM: Perc Vol Perc Mass Mole Molar Molal. Tip: "Molarity Vol dep, Molality Mass indep".
1.3 Solub Solid	Like-like, dynamic eq at sat. Le Chat: Endo ↑T ↑solub. Press no effect.	NaCl water yes, naphthalene benzene yes.	LDE: Like Diss Eq. Tip: Sat: Diss=Crys rates.
1.3 Gas Solub	↑P ↑solub (Henry p=Kx). ↑T ↓solub. Apps: Soda, bends.	Ex1.4 0.716 mmol N2. K_H ↑ low solub (O2).	PH: Press Henry. Tip: "Cold fish happy - low T high O2".
Laws & Props	Raoult P=P0 x ideal. Deviations +/- . Collig: ΔTb=Kb m, ΔTf=Kf m, π=CRT. Abnormal i.	Rel low vap= x_B. Azeotropes.	RDC: Raoult Deviate Collig. Tip: "Raoult Ideal, Real Rebel (+/- ΔH)".