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WBB • Class XI • Chemistry • Ch 5

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States of Matter: Gases and Liquids

Chemistry is the scientific discipline dedicated to investigating the composition, structure, physical and chemical properties, and transformations of matter. As the central science, it connects physical principles with biological mechanisms and material engineering. Chapter 1 establishes the quantitative, rigorous vocabulary of modern chemistry. Moving beyond qualitative descriptions, students explore the nature of matter, standard international (SI) base units, scientific notation, precision, accuracy, and significant figures. The historical foundations of atomic theory are developed through the classical laws of chemical combination: mass conservation, definite proportions, multiple proportions, reciprocal proportions, Gay-Lussac law of gaseous volumes, and Avogadro hypothesis. The chapter introduces the mole concept—chemistry supreme counting unit—which directly bridges the microscopic realm of atoms, molecules, and ions with macroscopically measurable quantities of mass and volume. Students master molar mass, average atomic mass derived from isotopic abundances, and the determination of empirical and molecular formulae via percentage composition and vapor density. Finally, the principles of chemical stoichiometry are formalized, enabling students to balance chemical equations, identify limiting reagents, evaluate reaction percentage yields, and calculate solution concentrations across molarity, molality, mole fraction, and mass percent.

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Why This Chapter Matters

Quantitative stoichiometry and the mole concept constitute the universal operating system of all experimental chemistry, chemical engineering, pharmaceutical synthesis, environmental monitoring, and biochemistry. In the pharmaceutical industry, precise stoichiometric calculations and limiting reagent analyses are mandatory to synthesize lifesaving active pharmaceutical ingredients like cisplatin, taxol, and AZT with minimal toxic byproducts and optimal percent yield. In environmental engineering, calculating parts per million (ppm) and molarity allows scientists to measure atmospheric carbon dioxide levels, heavy metal water contamination, and acid rain precipitation. In clinical medicine and biochemistry, understanding solution concentrations governs intravenous fluid therapy, drug dosing, electrolyte balances, and renal clearance rates. For students preparing for WBCHSE Class 11 annual examinations, WBJEE, and NEET/JEE Main, this introductory chapter provides the indispensable computational bedrock for subsequent physical chemistry chapters, including chemical thermodynamics, equilibrium, redox reactions, electrochemistry, and chemical kinetics.

Before You Begin (Prerequisites)

Class 10 Physical Science: Basic chemical symbols, atomic structure, valency, balancing simple equations
Basic algebra: Ratios, proportions, percentages, and scientific notation

Chapter Roadmap & Progression

1 Module 1: Scope of Chemistry, Natur...

2 Module 2: Laws of Chemical Combinat...

3 Module 3: Atomic Mass, Molecular Ma...

4 Module 4: Elemental Percentage Comp...

5 Module 5: Chemical Equations, Stoic...

6 Module 6: Concentration Terms of So...

Complete Concept Guide (100% Curriculum Coverage)

Module 1: Scope of Chemistry, Nature of Matter & Precision in Measurement

Chemistry is often called the central science because it bridges the foundational principles of physics with the applied complexities of biology, geology, and materials engineering.

1. Nature and Classification of Matter

Matter is formally defined as anything that possesses rest mass and occupies physical space (volume). Matter can be classified along two distinct axes:

Physical Classification: Matter exists in three primary physical states—solids (definite shape and volume due to strong intermolecular forces), liquids (definite volume but indefinite shape, fluid), and gases (neither definite shape nor volume, completely filling their container). Plasma and Bose-Einstein condensates represent additional extreme states.
Chemical Classification:
- Pure Substances: Possess uniform chemical composition throughout and cannot be separated into simpler substances by physical methods. Divided into:
  - Elements: Consist of only one type of atom (e.g., $\text{Fe, Cu, } \text{O}_2$).
  - Compounds: Consist of atoms of two or more different elements chemically bound in a fixed mass ratio (e.g., $\text{H}_2\text{O, NaCl, CO}_2$).
- Mixtures: Contain two or more pure substances physically combined in arbitrary proportions. Divided into:
  - Homogeneous Mixtures (Solutions): Composition is completely uniform throughout at the molecular level (e.g., air, aqueous sugar solution, brass alloy).
  - Heterogeneous Mixtures: Composition is non-uniform; physical phase boundaries are distinct (e.g., sand and iron filings, oil and water, smoke).

2. The International System of Units (SI Units)

The Système International d'Unités (SI) specifies seven fundamental base units from which all other chemical units are derived:

Physical Quantity	SI Base Unit	Symbol	Definition Basis
Length	meter	m	Speed of light in vacuum ($c = 299792458\text{ m/s}$)
Mass	kilogram	kg	Planck constant ($h = 6.62607015 \times 10^{-34}\text{ J}\cdot\text{s}$)
Time	second	s	Caesium-133 hyperfine transition frequency
Electric Current	ampere	A	Elementary charge ($e = 1.602176634 \times 10^{-19}\text{ C}$)
Thermodynamic Temperature	kelvin	K	Boltzmann constant ($k = 1.380649 \times 10^{-23}\text{ J/K}$)
Amount of Substance	mole	mol	Avogadro constant ($N_A = 6.02214076 \times 10^{23}\text{ mol}^{-1}$)
Luminous Intensity	candela	cd	Luminous efficacy of monochromatic radiation

3. Uncertainty, Precision, Accuracy & Significant Figures

Every experimental measurement contains intrinsic uncertainty arising from the limitations of the measuring instrument and human observer skill.

Precision: Refers to the closeness of agreement between successive measurements of the same quantity under identical conditions.
Accuracy: Refers to the closeness of agreement between an experimental value and the true or accepted reference value.

Rules for Determining Significant Figures:

All non-zero digits are significant (e.g., $285\text{ cm}$ has 3 significant figures).
Zeros preceding the first non-zero digit are non-significant; they merely indicate the position of the decimal point (e.g., $0.0052$ has 2 significant figures).
Zeros between non-zero digits (captive zeros) are always significant (e.g., $4.008$ has 4 significant figures).
Zeros at the end of a number to the right of a decimal point are significant (e.g., $0.200\text{ g}$ has 3 significant figures; $100.0$ has 4 significant figures).
Exact counts of objects possess an infinite number of significant figures (e.g., 20 apples = $20.0000\dots$).
Calculation Rules: In addition and subtraction, the final result cannot have more decimal places than the measurement with the fewest decimal places. In multiplication and division, the final result must retain the same number of significant figures as the factor having the least significant figures.

Module 2: Laws of Chemical Combinations & Dalton's Atomic Theory

Modern quantitative chemistry was born out of five fundamental empirical laws governing the mass and volumetric ratios of reacting substances.

1. The Five Laws of Chemical Combination

Law of Conservation of Mass (Antoine Lavoisier, 1789):
"In any physical change or chemical reaction, the total mass of the products formed is always exactly equal to the total mass of the reacting substances consumed." \[\sum m_{\text{reactants}} = \sum m_{\text{products}}\] Modern Limitation: Under Einstein mass-energy equivalence ($E = \Delta m \cdot c^2$), in highly energetic nuclear reactions, a tiny mass loss transforms into immense binding energy. For ordinary chemical reactions, mass change is on the order of $10^{-10}\text{ g}$, making conservation of mass practically exact.
Law of Definite (Constant) Proportions (Joseph Proust, 1799):
"A pure chemical compound, regardless of its source or method of preparation, always contains the exact same elements combined together in the same fixed proportion by mass." Example: Pure water ($\text{H}_2\text{O}$) obtained from rain, river, or synthesis always contains hydrogen and oxygen in a $2 : 16 = 1 : 8$ mass ratio. Modern Limitation: Compounds synthesized from different isotopes of elements (e.g., $\text{H}_2\text{O}$ vs $\text{D}_2\text{O}$ heavy water) and non-stoichiometric berthollide compounds (e.g., $\text{Fe}_{0.95}\text{O}$) show slight deviations.
Law of Multiple Proportions (John Dalton, 1803):
"When two elements combine to form two or more chemical compounds, the different masses of one element that combine with a fixed mass of the other element stand to each other in a ratio of small whole numbers."
Example: Carbon and Oxygen form Carbon Monoxide ($\text{CO}$) and Carbon Dioxide ($\text{CO}_2$). In $\text{CO}$, $12\text{ g}$ of carbon combines with $16\text{ g}$ of oxygen. In $\text{CO}_2$, $12\text{ g}$ of carbon combines with $32\text{ g}$ of oxygen. The masses of oxygen combining with a fixed mass of carbon ($12\text{ g}$) are in the ratio $16 : 32 = 1 : 2$, a simple whole-number ratio.
Law of Reciprocal Proportions (Jeremias Richter, 1792):
"When two different elements combine separately with a fixed mass of a third element, the ratio of masses in which they do so is either the same as, or a simple whole-number multiple of, the ratio of masses in which they combine directly with each other."
Gay-Lussac's Law of Gaseous Volumes (Joseph Louis Gay-Lussac, 1808):
"When gases react together, they do so in volumes which bear a simple whole-number ratio to one another and to the volumes of the gaseous products, provided all volumes are measured at the same temperature and pressure."
Example: $2\text{ H}_2\text{(g)} + 1\text{ O}_2\text{(g)} \to 2\text{ H}_2\text{O(g)}$; volume ratio is $2 : 1 : 2$.

2. Avogadro's Hypothesis (Amedeo Avogadro, 1811)

To resolve the conflict between Gay-Lussac volume ratios and Dalton indivisible atoms, Avogadro introduced the crucial distinction between atoms (the smallest constituent of an element participating in chemical reactions) and molecules (the smallest entity of a substance capable of independent stable existence):

Avogadro's Law:

"Equal volumes of all gases under identical conditions of temperature and pressure contain equal numbers of molecules."
Mathematically: $V \propto n$ (at constant $T$ and $P$). Avogadro deduced that common elemental gases like hydrogen, oxygen, nitrogen, and chlorine are diatomic molecules ($\text{H}_2, \text{O}_2, \text{N}_2, \text{Cl}_2$).

3. Dalton's Atomic Theory & Modern Atomic Reality

Dalton proposed (1808): (1) Matter consists of indivisible atoms, (2) All atoms of a given element have identical mass and properties, (3) Atoms of different elements have different masses, (4) Compounds form when atoms combine in fixed whole-number ratios, (5) Chemical reactions involve reorganization of atoms.

Modern Atomic Theory Modifications:

Atoms are divisible into subatomic particles: electrons, protons, and neutrons.
Atoms of the same element can possess different masses: Isotopes (e.g., $^{35}\text{Cl}$ and $^{37}\text{Cl}$).
Atoms of different elements can possess identical masses: Isobars (e.g., $^{40}\text{Ar}$ and $^{40}\text{Ca}$).
Mass can be interconverted into energy via nuclear transmutation ($E = mc^2$).

Module 3: Atomic Mass, Molecular Mass & The Universal Mole Concept

Because individual atoms have masses on the order of $10^{-24}\text{ g}$, chemists use relative atomic mass scales and the mole concept to handle macroscopic amounts.

1. Unified Atomic Mass Unit (u or amu)

By international agreement (IUPAC, 1961), the standard reference isotope is Carbon-12 ($^{12}\text{C}$), assigned an exact mass of $12.0000\text{ u}$. One unified atomic mass unit ($1\text{ u}$) is defined as exactly one-twelfth of the mass of a single carbon-12 atom:

\[1\text{ u} = \frac{1}{12} \times \text{mass of one }^{12}\text{C atom} = \frac{1}{12} \times \frac{12.000\text{ g}}{6.02214 \times 10^{23}} = 1.66056 \times 10^{-24}\text{ g} = 1.66056 \times 10^{-27}\text{ kg}\]

Notice that $1\text{ u} = \frac{1}{N_A}\text{ grams}$.

2. Average Atomic Mass of Elements

Most elements exist naturally as a mixture of isotopes with distinct masses and percentage abundances. The average atomic mass $\bar{A}$ is the weighted arithmetic mean of these isotopic masses:

\[\bar{A} = \sum_{i=1}^k \left(\frac{\%_i}{100} \times A_i\right) = \frac{\sum f_i A_i}{\sum f_i}\]

Example: Chlorine consists of $75.77\%$ $^{35}\text{Cl}$ ($34.9689\text{ u}$) and $24.23\%$ $^{37}\text{Cl}$ ($36.9659\text{ u}$):

\[\bar{A}_{\text{Cl}} = (0.7577 \times 34.9689) + (0.2423 \times 36.9659) = 26.496 + 8.957 = 35.453\text{ u} \approx 35.5\text{ u}\]

3. The Mole Concept & Avogadro's Number

The mole (symbol: mol) is the SI unit for the amount of substance. Under the 2019 SI redefinition:

Definition of the Mole:

One mole contains exactly $6.02214076 \times 10^{23}$ elementary entities. This number is the fixed numerical value of the Avogadro constant, $N_A$, when expressed in $\text{mol}^{-1}$.
Elementary entities may be atoms, molecules, ions, electrons, or any specified group of particles.

4. Molar Mass and Molar Volume at STP

Molar Mass ($M$): The mass of one mole of a substance expressed in grams per mole ($\text{g/mol}$). Numerically, the molar mass in $\text{g/mol}$ equals the atomic or molecular mass in $\text{u}$. E.g., for $\text{H}_2\text{O}$, molecular mass = $18.015\text{ u}$, molar mass = $18.015\text{ g/mol}$.
Molar Gas Volume at STP: At Standard Temperature and Pressure (STP: $T = 273.15\text{ K} = 0^\circ\text{C}$, $P = 1\text{ atm}$), one mole of any ideal gas occupies: \[V_{\text{molar}} = 22.414\text{ Liters} \approx 22.4\text{ L/mol} = 22400\text{ mL}\] Note on IUPAC standard: At standard pressure of $1\text{ bar}$ ($10^5\text{ Pa}$) and $273.15\text{ K}$, $V_m = 22.71\text{ L/mol}$. In standard school board problems, unless specified as 1 bar, $22.4\text{ L}$ is universally applied.

5. The Master Mole Equations (The Mole Wheel)

To convert seamlessly between mass ($w$), number of particles ($N$), and gaseous volume at STP ($V$):

\[n = \frac{w}{M} = \frac{N}{N_A} = \frac{V_{\text{STP (L)}}}{22.4\text{ L}}\]

Module 4: Elemental Percentage Composition, Empirical & Molecular Formulae

Analytical chemists determine unknown molecular structures by measuring the elemental mass composition and relating it to empirical and molecular formulas.

1. Mass Percentage Composition

The mass percentage of each constituent element in a chemical compound is given by:

\[\text{Mass \% of an element} = \frac{\text{Total mass of that element in 1 mole of compound}}{\text{Molar mass of the compound}} \times 100\%\]

Example: In water ($\text{H}_2\text{O}$, $M = 18.016\text{ g/mol}$):

\[\%\text{ H} = \frac{2 \times 1.008}{18.016} \times 100 = 11.19\%, \quad \%\text{ O} = \frac{16.00}{18.016} \times 100 = 88.79\%\]

2. Empirical Formula vs. Molecular Formula

Empirical Formula: The chemical formula that indicates the simplest whole-number ratio of atoms of each element present in a compound. (E.g., for glucose, empirical formula is $\text{CH}_2\text{O}$).
Molecular Formula: The exact chemical formula showing the actual number of atoms of each element present in one molecule of the compound. (E.g., for glucose, molecular formula is $\text{C}_6\text{H}_{12}\text{O}_6$).

3. The Mathematical Relationship & Vapor Density

\[\text{Molecular Formula} = (\text{Empirical Formula})_n\] \[n = \frac{\text{Molar Mass (Molecular Weight)}}{\text{Empirical Formula Mass}}\]

For volatile gaseous or vaporized compounds, the molar mass is frequently determined using Vapor Density (V.D.):

\[\text{Vapor Density (V.D.)} = \frac{\text{Mass of a certain volume of gas at given } T, P}{\text{Mass of the same volume of hydrogen gas at same } T, P} = \frac{M_{\text{gas}}}{M_{\text{H}_2}} = \frac{M}{2}\] \[\text{Molar Mass } (M) = 2 \times \text{Vapor Density (V.D.)}\]

4. Systematic Algorithm for Empirical Formula Determination

Step	Operation	Mathematical Formula
1	Assume 100 g sample	Mass of element in grams = Mass percentage
2	Calculate relative number of moles	$n_i = \frac{\text{Mass of element } i}{\text{Atomic mass } A_i}$
3	Determine simplest atomic ratio	Divide each $n_i$ by the smallest $n_{\min}$ obtained
4	Convert to simplest whole numbers	If fractions occur (e.g., 1.5, 1.33), multiply all ratios by an integer (2, 3)
5	Find integer multiplier $n$	$n = \frac{\text{Molar mass}}{\text{Empirical formula mass}}$

Module 5: Chemical Equations, Stoichiometry & Limiting Reagent

Stoichiometry (from Greek stoicheion, element, and metron, measure) deals with the quantitative relationships among the amounts of reactants and products in a balanced chemical equation.

1. Balanced Chemical Equations & Stoichiometric Coefficients

Consider the general balanced reaction:

\[a\text{A} + b\text{B} \to c\text{C} + d\text{D}\]

The stoichiometric coefficients $a, b, c, d$ indicate the exact mole ratios:

\[\frac{n_A}{a} = \frac{n_B}{b} = \frac{n_C}{c} = \frac{n_D}{d}\]

From this fundamental equivalence, four quantitative calculations follow:

Mole-Mole Relationships: $n_C = n_A \times \frac{c}{a}$.
Mass-Mass Relationships: Convert mass of reactant to moles ($n = w/M$), apply mole ratio, and multiply by product molar mass.
Mass-Volume Relationships: Relate grams of solid/liquid reactant to liters of gas produced at STP using $22.4\text{ L/mol}$.
Volume-Volume Relationships: For purely gaseous reactions at constant $T, P$, volume ratios equal mole ratios directly (Gay-Lussac law).

2. Limiting Reagent (Limiting Reactant) Concept

In practical chemical reactions, reactants are rarely present in exact stoichiometric proportions. One reactant will be exhausted first.

Limiting Reagent Definition & Identification Algorithm:

The Limiting Reagent (L.R.) is the reactant that is completely consumed first in a reaction. It limits the maximum amount of product that can be formed.
Identification Rule: For each reactant, calculate the ratio: \[\text{Ratio} = \frac{\text{Actual initial moles of reactant }(n_i)}{\text{Stoichiometric coefficient in balanced equation }(\nu_i)}\] The reactant with the smallest ratio is the Limiting Reagent!

All calculations of product quantities and amounts of excess reactants consumed MUST be calculated strictly based on the Limiting Reagent.

3. Theoretical Yield, Actual Yield & Percentage Yield

Theoretical Yield: The maximum calculated amount of product obtained from the limiting reagent assuming 100% reaction completion.
Actual Yield: The real amount of product isolated experimentally in the laboratory (often less due to side reactions, incomplete reaction, or loss during isolation).
Percentage Yield: \[\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%\]

Module 6: Concentration Terms of Solutions & Interconversions

In chemistry, most reactions occur in homogeneous liquid solutions. The amount of solute dissolved in a specified amount of solvent or solution is its concentration.

1. Standard Concentration Expressions

Concentration Unit	Definition & Mathematical Formula	Units	Temperature Effect
Mass Percent (% w/w)	$\frac{w_{\text{solute}}}{w_{\text{solution}}} \times 100 = \frac{w_B}{w_A + w_B} \times 100$	Dimensionless (%)	Temperature Independent
Volume Percent (% v/v)	$\frac{V_{\text{solute}}}{V_{\text{solution}}} \times 100$	Dimensionless (%)	Temperature Dependent
Mass by Volume (% w/v)	$\frac{w_{\text{solute (g)}}}{V_{\text{solution (mL)}}} \times 100$	$\text{g/100 mL}$	Temperature Dependent
Mole Fraction ($x$)	$x_B = \frac{n_B}{n_A + n_B}, \quad x_A + x_B = 1$	Unitless	Temperature Independent
Molarity ($M$)	$M = \frac{n_B}{V_{\text{L}}} = \frac{w_B \times 1000}{M_B \times V_{\text{mL}}}$	$\text{mol/L}$ or M	Temperature Dependent (liquid expands)
Molality ($m$)	$m = \frac{n_B}{w_{\text{solvent (kg)}}} = \frac{w_B \times 1000}{M_B \times w_A\text{ (g)}}$	$\text{mol/kg}$ or m	Temperature Independent

2. Dilution & Mixing Laws

Dilution Equation: When solvent is added, moles of solute remain constant: \[M_1 V_1 = M_2 V_2\]
Mixing Equation: When two solutions of the same solute are mixed: \[M_R = \frac{M_1 V_1 + M_2 V_2}{V_1 + V_2}\]

3. Master Interconversion: Molarity vs. Molality

If a solution has molarity $M$, solute molar mass $M_B$, and solution density $\rho$ (in $\text{g/mL}$ or $\text{g/cm}^3$):

\[m = \frac{1000 \times M}{1000 \times \rho - M \times M_B}\]

Key Formulas, Reactions & Definitions

The Master Mole Equations (The Mole Wheel)

$$n = \frac{w}{M} = \frac{N}{N_A} = \frac{V_{\text{STP (L)}}}{22.4\text{ L}}$$

Average Atomic Mass from Isotopic Abundance

$$\bar{A} = \sum_{i=1}^k \left(\frac{\%_i}{100} \times A_i\right) = \frac{\sum f_i A_i}{\sum f_i}$$

Empirical & Molecular Formula with Vapor Density

$$\text{Molecular Formula} = (\text{Empirical Formula})_n, \quad n = \frac{M}{\text{Empirical Mass}}, \quad M = 2 \times \text{V.D.}$$

Molarity of a Solution & Dilution Law

$$M = \frac{w_B \times 1000}{M_B \times V_{\text{mL}}}, \quad M_1 V_1 = M_2 V_2$$

Molality of a Solution

$$m = \frac{w_B \times 1000}{M_B \times w_A\text{ (g)}}$$

Molarity-Molality Density Interconversion

$$m = \frac{1000 \times M}{1000 \times \rho - M \times M_B}, \quad x_B = \frac{n_B}{n_A + n_B}$$

Conceptual Solved Examples & Case Studies

Example 1

State the number of significant figures in: (a) 0.0048, (b) 205.08, (c) 500.0, and (d) 2.00 x 10^4. Also convert the density of mercury, 13.6 g/cm^3, into SI units (kg/m^3).

Step-by-Step Solution:

Step 1: Determine Significant Figures

(a) 0.0048: The leading zeros before 4 are non-significant; only 4 and 8 are significant. 2 significant figures.
(b) 205.08: All non-zero digits and captive zeros between them are significant. 5 significant figures.
(c) 500.0: The trailing zero after the decimal point makes all preceding zeros significant. 4 significant figures.
(d) 2.00 x 10^4: In scientific notation, only the mantissa determines significance (2, 0, 0). 3 significant figures.

Step 2: Convert Density of Mercury to SI Units

\[\rho = 13.6\text{ g/cm}^3\] \[1\text{ kg} = 10^3\text{ g} \implies 1\text{ g} = 10^{-3}\text{ kg}\] \[1\text{ m} = 100\text{ cm} = 10^2\text{ cm} \implies 1\text{ cm} = 10^{-2}\text{ m} \implies 1\text{ cm}^3 = 10^{-6}\text{ m}^3\] \[\rho = 13.6 \times \frac{10^{-3}\text{ kg}}{10^{-6}\text{ m}^3} = 13.6 \times 10^3\text{ kg/m}^3 = 1.36 \times 10^4\text{ kg/m}^3\]

Final Answer: (a) 2, (b) 5, (c) 4, (d) 3; Density in SI = $1.36 \times 10^4\text{ kg/m}^3$ (or $13600\text{ kg/m}^3$).

Example 2

In a 4.4 g sample of carbon dioxide gas (CO2), calculate: (i) the number of moles of CO2, (ii) the total number of CO2 molecules, (iii) the number of oxygen atoms present, and (iv) the volume occupied by the gas at STP (0°C, 1 atm). (Atomic masses: C = 12 u, O = 16 u).

Step-by-Step Solution:

Step 1: Calculate Molar Mass of CO2

\[M_{\text{CO}_2} = 12.0 + (2 \times 16.0) = 44.0\text{ g/mol}\]

Step 2: Sub-problem (i) — Number of Moles of CO2

\[n = \frac{w}{M} = \frac{4.4\text{ g}}{44.0\text{ g/mol}} = 0.10\text{ mol}\]

Step 3: Sub-problem (ii) — Number of CO2 Molecules

\[N_{\text{molecules}} = n \times N_A = 0.10\text{ mol} \times 6.022 \times 10^{23}\text{ molecules/mol} = 6.022 \times 10^{22}\text{ molecules}\]

Step 4: Sub-problem (iii) — Number of Oxygen Atoms

Each $\text{CO}_2$ molecule contains exactly 2 oxygen atoms:

\[N_{\text{O atoms}} = 2 \times N_{\text{molecules}} = 2 \times 6.022 \times 10^{22} = 1.2044 \times 10^{23}\text{ atoms}\]

Step 5: Sub-problem (iv) — Volume at STP

\[V = n \times 22.4\text{ L} = 0.10\text{ mol} \times 22.4\text{ L/mol} = 2.24\text{ L} = 2240\text{ mL}\]

Final Answer: (i) $0.10\text{ mol}$, (ii) $6.022 \times 10^{22}\text{ molecules}$, (iii) $1.2044 \times 10^{23}\text{ atoms}$, (iv) $2.24\text{ L}$.

Example 3

An organic compound contains 40.0% carbon, 6.67% hydrogen, and the remainder oxygen by mass. Its vapor density is measured to be 30. Determine: (i) the empirical formula of the compound, and (ii) its molecular formula. (Atomic masses: C = 12 u, H = 1 u, O = 16 u).

Step-by-Step Solution:

Step 1: Calculate Percentage of Oxygen

\[\%\text{ O} = 100 - (40.0 + 6.67) = 100 - 46.67 = 53.33\%\]

Step 2: Determine Relative Moles in 100 g Sample

$\text{Moles of C} = \frac{40.0}{12.0} = 3.333\text{ mol}$
$\text{Moles of H} = \frac{6.67}{1.0} = 6.670\text{ mol}$
$\text{Moles of O} = \frac{53.33}{16.0} = 3.333\text{ mol}$

Step 3: Determine Simplest Molar Ratio

Divide each mole value by the smallest value ($3.333$):

$\text{C} = \frac{3.333}{3.333} = 1.0$
$\text{H} = \frac{6.670}{3.333} = 2.0$
$\text{O} = \frac{3.333}{3.333} = 1.0$

The simplest whole number ratio is $\text{C} : \text{H} : \text{O} = 1 : 2 : 1$.

\[\text{Empirical Formula} = \text{CH}_2\text{O}\]

Step 4: Determine Empirical Formula Mass & Molar Mass

\[\text{Empirical Formula Mass} = 12.0 + (2 \times 1.0) + 16.0 = 30.0\text{ g/mol}\] \[\text{Molar Mass } (M) = 2 \times \text{Vapor Density} = 2 \times 30 = 60.0\text{ g/mol}\]

Step 5: Determine Multiplier n and Molecular Formula

\[n = \frac{\text{Molar Mass}}{\text{Empirical Formula Mass}} = \frac{60.0}{30.0} = 2\] \[\text{Molecular Formula} = (\text{CH}_2\text{O})_2 = \text{C}_2\text{H}_4\text{O}_2\text{ (Acetic Acid / Ethanoic Acid)}\]

Final Answer: (i) Empirical Formula = $\text{CH}_2\text{O}$; (ii) Molecular Formula = $\text{C}_2\text{H}_4\text{O}_2$.

Example 4

In the Haber process, 50.0 kg of N2(g) and 10.0 kg of H2(g) are reacted according to: N2(g) + 3 H2(g) -> 2 NH3(g). (i) Identify the limiting reagent, (ii) Calculate the theoretical mass of NH3(g) produced in kilograms, and (iii) Determine the mass of the excess reactant remaining unreacted.

Step-by-Step Solution:

Step 1: Convert Initial Masses into Moles

\[M_{\text{N}_2} = 28.02\text{ g/mol}, \quad M_{\text{H}_2} = 2.016\text{ g/mol}\] \[n_{\text{N}_2} = \frac{50.0 \times 10^3\text{ g}}{28.02\text{ g/mol}} = 1.784 \times 10^3\text{ mol} = 1784\text{ mol}\] \[n_{\text{H}_2} = \frac{10.0 \times 10^3\text{ g}}{2.016\text{ g/mol}} = 4.960 \times 10^3\text{ mol} = 4960\text{ mol}\]

Step 2: Sub-problem (i) — Identify the Limiting Reagent

Balanced Equation: $1\text{ N}_2 + 3\text{ H}_2 \to 2\text{ NH}_3$.

\[\text{For } \text{N}_2: \frac{n_{\text{N}_2}}{1} = \frac{1784}{1} = 1784\] \[\text{For } \text{H}_2: \frac{n_{\text{H}_2}}{3} = \frac{4960}{3} = 1653.3\]

Since $1653.3 < 1784$, Hydrogen ($\text{H}_2$) is the Limiting Reagent (L.R.), and Nitrogen ($\text{N}_2$) is in excess.

Step 3: Sub-problem (ii) — Calculate Theoretical Mass of NH3 Produced

According to stoichiometry: $3\text{ moles of H}_2$ produce $2\text{ moles of NH}_3$:

\[n_{\text{NH}_3} = n_{\text{H}_2} \times \frac{2}{3} = 4960 \times \frac{2}{3} = 3306.7\text{ mol}\] \[M_{\text{NH}_3} = 14.01 + 3(1.008) = 17.034\text{ g/mol}\] \[\text{Mass of NH}_3 = 3306.7\text{ mol} \times 17.034\text{ g/mol} = 56326\text{ g} = 56.33\text{ kg}\]

Step 4: Sub-problem (iii) — Determine Unreacted Excess N2

Moles of $\text{N}_2$ consumed by $4960\text{ mol of H}_2$:

\[n_{\text{N}_2\text{ consumed}} = \frac{1}{3} \times n_{\text{H}_2} = \frac{4960}{3} = 1653.3\text{ mol}\] \[n_{\text{N}_2\text{ unreacted}} = 1784 - 1653.3 = 130.7\text{ mol}\] \[\text{Mass of excess N}_2 = 130.7\text{ mol} \times 28.02\text{ g/mol} = 3662\text{ g} = 3.66\text{ kg}\]

Sanity Check (Mass Conservation): Initial mass = $50.0 + 10.0 = 60.0\text{ kg}$. Final mass = $56.33\text{ kg (NH}_3) + 3.66\text{ kg (N}_2) = 59.99 \approx 60.0\text{ kg}$.

Final Answer: (i) Limiting Reagent = $\text{H}_2$; (ii) Mass of $\text{NH}_3$ = $56.33\text{ kg}$; (iii) Excess $\text{N}_2$ left = $3.66\text{ kg}$.

Example 5

An aqueous solution of glucose (C6H12O6, molar mass = 180.0 g/mol) is 10.0% by mass (w/w) and has a density of 1.20 g/mL. Calculate: (i) the molality (m) of the solution, (ii) the molarity (M) of the solution, and (iii) the mole fraction of glucose (xB).

Step-by-Step Solution:

Step 1: Basis of Calculation (100 g Solution)

Mass of solute (Glucose, $w_B$) = $10.0\text{ g}$
Mass of solvent (Water, $w_A$) = $100.0 - 10.0 = 90.0\text{ g} = 0.090\text{ kg}$
Moles of glucose $n_B = \frac{10.0\text{ g}}{180.0\text{ g/mol}} = 0.0556\text{ mol}$
Moles of water $n_A = \frac{90.0\text{ g}}{18.02\text{ g/mol}} = 4.994\text{ mol}$

Step 2: Sub-problem (i) — Molality (m)

\[m = \frac{n_B}{w_{A\text{ (kg)}}} = \frac{0.0556\text{ mol}}{0.090\text{ kg}} = 0.618\text{ mol/kg} = 0.618\text{ m}\]

Step 3: Sub-problem (ii) — Molarity (M)

\[\text{Volume of solution } V = \frac{\text{Mass of solution}}{\text{Density}} = \frac{100.0\text{ g}}{1.20\text{ g/mL}} = 83.33\text{ mL} = 0.08333\text{ L}\] \[M = \frac{n_B}{V_{\text{L}}} = \frac{0.0556\text{ mol}}{0.08333\text{ L}} = 0.667\text{ mol/L} = 0.667\text{ M}\]

Step 4: Sub-problem (iii) — Mole Fraction of Glucose (xB)

\[x_B = \frac{n_B}{n_A + n_B} = \frac{0.0556}{4.994 + 0.0556} = \frac{0.0556}{5.0496} = 0.0110\]

Final Answer: (i) Molality = $0.618\text{ m}$; (ii) Molarity = $0.667\text{ M}$; (iii) Mole Fraction = $0.011$.

Example 6

A 2.50 g sample of chalk containing 80.0% pure calcium carbonate (CaCO3, M = 100.0 g/mol) is treated with 0.50 M hydrochloric acid (HCl). The reaction is: CaCO3(s) + 2 HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l). (i) Calculate the volume of 0.50 M HCl solution required for complete reaction, (ii) Calculate the volume of dry CO2 gas liberated at STP (273 K, 1 atm), and (iii) Find the mass of calcium chloride (CaCl2, M = 111.0 g/mol) produced.

Step-by-Step Solution:

Step 1: Calculate Mass and Moles of Pure CaCO3

\[\text{Mass of pure CaCO}_3 = 2.50\text{ g} \times \frac{80.0}{100} = 2.00\text{ g}\] \[n_{\text{CaCO}_3} = \frac{2.00\text{ g}}{100.0\text{ g/mol}} = 0.020\text{ mol}\]

Step 2: Sub-problem (i) — Volume of 0.50 M HCl Required

From the balanced equation, $1\text{ mole of CaCO}_3$ reacts with $2\text{ moles of HCl}$:

\[n_{\text{HCl}} = 2 \times n_{\text{CaCO}_3} = 2 \times 0.020\text{ mol} = 0.040\text{ mol}\]

Using the Molarity formula $M = \frac{n}{V_{\text{L}}}$:

\[V_{\text{HCl (L)}} = \frac{n_{\text{HCl}}}{M} = \frac{0.040\text{ mol}}{0.50\text{ mol/L}} = 0.080\text{ L} = 80.0\text{ mL}\]

Step 3: Sub-problem (ii) — Volume of CO2 Gas Liberated at STP

From the balanced equation, $1\text{ mole of CaCO}_3$ produces $1\text{ mole of CO}_2$:

\[n_{\text{CO}_2} = n_{\text{CaCO}_3} = 0.020\text{ mol}\] \[V_{\text{CO}_2\text{ at STP}} = 0.020\text{ mol} \times 22.4\text{ L/mol} = 0.448\text{ L} = 448\text{ mL}\]

Step 4: Sub-problem (iii) — Mass of CaCl2 Produced

From the balanced equation, $1\text{ mole of CaCO}_3$ yields $1\text{ mole of CaCl}_2$:

\[n_{\text{CaCl}_2} = 0.020\text{ mol}\] \[M_{\text{CaCl}_2} = 40.08 + (2 \times 35.45) = 110.98 \approx 111.0\text{ g/mol}\] \[\text{Mass of CaCl}_2 = 0.020\text{ mol} \times 111.0\text{ g/mol} = 2.22\text{ g}\]

Final Answer: (i) Volume of $0.50\text{ M HCl}$ = $80.0\text{ mL}$; (ii) Volume of $\text{CO}_2$ at STP = $448\text{ mL}$ (or $0.448\text{ L}$); (iii) Mass of $\text{CaCl}_2$ = $2.22\text{ g}$.

Common Misconceptions & Examiner Traps

Common Misconception

Confusing Molarity (M) with Molality (m) and Overlooking Temperature Effects

Scientific Reality & Correction

Molarity is defined as moles of solute per liter of SOLUTION, whereas Molality is moles of solute per kilogram of SOLVENT. Because liquid volume expands or contracts with temperature, Molarity changes with temperature. Molality depends purely on mass and remains completely invariant with temperature changes.

Common Misconception

Identifying the Limiting Reagent by Comparing Initial Masses Directly

Scientific Reality & Correction

Assuming that the reactant with smaller initial mass is automatically the limiting reagent. Chemical reactions occur on a mole-to-mole basis governed by stoichiometric coefficients, not grams. 10 grams of H2 contains 5 moles, whereas 50 grams of N2 contains only 1.78 moles.

Common Misconception

Applying the 22.4 L Molar Gas Volume to Liquids or Solids at STP

Scientific Reality & Correction

Applying the 22.4 L/mol conversion to liquid water, liquid bromine, or solid precipitates at STP. The 22.4 L/mol volume rule is derived from the ideal gas law (PV = nRT) and applies ONLY to substances existing in the gaseous state at STP.

Common Misconception

Equating Vapor Density to Absolute Density

Scientific Reality & Correction

Treating Vapor Density as an absolute density with units of g/L. Vapor Density is a dimensionless relative ratio comparing the density of a gas to that of hydrogen gas at identical temperature and pressure. Hence, Molar Mass = 2 x V.D.

Common Misconception

Applying Significant Figure Rules Inconsistently across Operations

Scientific Reality & Correction

Applying the least significant figures rule to addition/subtraction or the least decimal places rule to multiplication/division. In addition/subtraction, decimal places dictate precision; in multiplication/division, total significant figures dictate precision.

The Mole Wheel, Chemical Laws, Formulae & Solution Stoichiometry Architecture

Chapter Summary & 10 Key Takeaways

Takeaway 1

Matter is anything that possesses mass and occupies space; it is classified chemically into pure substances (elements and compounds) and mixtures (homogeneous and heterogeneous).

Takeaway 2

The SI system defines 7 base physical units: meter (m), kilogram (kg), second (s), ampere (A), kelvin (K), mole (mol), and candela (cd).

Takeaway 3

Significant figures communicate measurement precision: non-zero digits are always significant, leading zeros are not, captive zeros are significant, and trailing zeros with a decimal point are significant.

Takeaway 4

The Law of Conservation of Mass states that mass can neither be created nor destroyed in a chemical reaction; total mass of reactants equals total mass of products.

Takeaway 5

The Law of Multiple Proportions (Dalton) states that when two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers.

Takeaway 6

Avogadro Law establishes that equal volumes of all gases under identical temperature and pressure contain an equal number of molecules.

Takeaway 7

One mole is the amount of substance containing exactly 6.02214076 x 10^23 elementary entities (Avogadro constant NA); one mole of an ideal gas occupies 22.4 L at STP (0°C, 1 atm).

Takeaway 8

The Empirical Formula represents the simplest whole-number ratio of atoms in a compound, while the Molecular Formula shows the exact number: Molecular Formula = (Empirical Formula)n, where n = Molar Mass / Empirical Mass.

Takeaway 9

The Limiting Reagent is the reactant that is completely consumed first in a reaction, thereby dictating and limiting the maximum theoretical yield of products.

Takeaway 10

Molarity (M = moles of solute / liters of solution) depends on temperature due to liquid thermal expansion, whereas Molality (m = moles of solute / kilograms of solvent) and Mole Fraction are independent of temperature.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

How many moles and molecules of methane (CH4) are present in 3.2 g of the gas? (Atomic masses: C = 12 u, H = 1 u).

Reveal Answer & Explanation

Answer: Molar mass of CH4 = 12 + 4(1) = 16 g/mol. Number of moles n = 3.2 g / 16 g/mol = 0.20 mol. Number of molecules N = 0.20 mol * 6.022 x 10^23 molecules/mol = 1.2044 x 10^23 molecules.
Calculate the molar mass of CH4 (12 + 4 = 16 g/mol). Use n = w/M and N = n * NA.

Which law of chemical combination is illustrated by the pair of compounds carbon monoxide (CO) and carbon dioxide (CO2)? Explain briefly.

Reveal Answer & Explanation

Answer: It illustrates the Law of Multiple Proportions (John Dalton). In CO, 12 g of carbon combines with 16 g of oxygen. In CO2, 12 g of carbon combines with 32 g of oxygen. The masses of oxygen combining with a fixed mass of carbon are in the ratio 16 : 32 = 1 : 2, which is a simple whole-number ratio.
Consider the ratio of masses of oxygen combining with a fixed 12 g mass of carbon in both compounds.

Calculate the molarity of a solution prepared by dissolving 4.0 g of sodium hydroxide (NaOH, M = 40.0 g/mol) in enough distilled water to make exactly 250 mL of solution.

Reveal Answer & Explanation

Answer: Moles of NaOH = 4.0 g / 40.0 g/mol = 0.10 mol. Volume of solution = 250 mL = 0.250 L. Molarity M = 0.10 mol / 0.250 L = 0.40 mol/L (0.40 M).
Use M = (wB * 1000) / (MB * V_mL).

What mass of calcium oxide (CaO) is obtained by heating 10.0 g of pure calcium carbonate (CaCO3)? (Ca = 40, C = 12, O = 16).

Reveal Answer & Explanation

Answer: CaCO3 -> CaO + CO2. Molar mass of CaCO3 = 100 g/mol, Molar mass of CaO = 40 + 16 = 56 g/mol. Moles of CaCO3 = 10.0 / 100 = 0.10 mol. Since mole ratio is 1:1, moles of CaO formed = 0.10 mol. Mass of CaO = 0.10 mol * 56 g/mol = 5.60 g.
Reaction: CaCO3 -> CaO + CO2. 100 g of CaCO3 decomposes into 56 g of CaO and 44 g of CO2.

Why is molality preferred over molarity in experimental studies of colligative properties?

Reveal Answer & Explanation

Answer: Molality is defined per kilogram of solvent (mass), whereas molarity is defined per liter of solution (volume). Liquid volume expands or contracts with temperature, causing molarity to vary with temperature. Mass is strictly temperature-invariant, so molality remains constant throughout heating or cooling.
Consider how volume and mass behave when temperature changes.

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All Class 11 Chemistry Chapters

Ch 1: Some Basic Concepts of Chemistry Ch 2: Structure of an Atom Ch 3: Classification of Elements and Periodicity in Properties Ch 4: Chemical Bonding and Molecular Structure Ch 5: States of Matter: Gases and Liquids Ch 6: Thermodynamics Ch 7: Equilibrium Ch 8: Redox Reactions Ch 9: s-Block Elements (Group 1 and Group 2 elements) Ch 10: p-Block Elements (Group 13 and Group 14 elements) Ch 11: Organic Chemistry: Some Basic Principles and Techniques Ch 12: Hydrocarbons Ch 13: Environmental Chemistry

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States of Matter: Gases and Liquids

Why This Chapter Matters

Before You Begin (Prerequisites)

Chapter Roadmap & Progression

Complete Concept Guide (100% Curriculum Coverage)

Module 1: Scope of Chemistry, Nature of Matter & Precision in Measurement

Module 2: Laws of Chemical Combinations & Dalton's Atomic Theory

Module 3: Atomic Mass, Molecular Mass & The Universal Mole Concept

Module 4: Elemental Percentage Composition, Empirical & Molecular Formulae

Module 5: Chemical Equations, Stoichiometry & Limiting Reagent

Module 6: Concentration Terms of Solutions & Interconversions

Key Formulas, Reactions & Definitions

Conceptual Solved Examples & Case Studies

Common Misconceptions & Examiner Traps

Common Misconception

Scientific Reality & Correction

Common Misconception

Scientific Reality & Correction

Common Misconception

Scientific Reality & Correction

Common Misconception

Scientific Reality & Correction

Common Misconception

Scientific Reality & Correction

The Mole Wheel, Chemical Laws, Formulae & Solution Stoichiometry Architecture

Chapter Summary & 10 Key Takeaways

Check Your Understanding (Diagnostic Practice Questions)

Timed CBT Practice Tests (Exam Simulator)

All Class 11 Chemistry Chapters

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Physical Quantity	SI Base Unit	Symbol	Definition Basis
Length	meter	m	Speed of light in vacuum (\(c = 299792458\text{ m/s}\))
Mass	kilogram	kg	Planck constant (\(h = 6.62607015 \times 10^{-34}\text{ J}\cdot\text{s}\))
Time	second	s	Caesium-133 hyperfine transition frequency
Electric Current	ampere	A	Elementary charge (\(e = 1.602176634 \times 10^{-19}\text{ C}\))
Thermodynamic Temperature	kelvin	K	Boltzmann constant (\(k = 1.380649 \times 10^{-23}\text{ J/K}\))
Amount of Substance	mole	mol	Avogadro constant (\(N_A = 6.02214076 \times 10^{23}\text{ mol}^{-1}\))
Luminous Intensity	candela	cd	Luminous efficacy of monochromatic radiation

Concentration Unit	Definition & Mathematical Formula	Units	Temperature Effect
Mass Percent (% w/w)	\(\frac{w_{\text{solute}}}{w_{\text{solution}}} \times 100 = \frac{w_B}{w_A + w_B} \times 100\)	Dimensionless (%)	Temperature Independent
Volume Percent (% v/v)	\(\frac{V_{\text{solute}}}{V_{\text{solution}}} \times 100\)	Dimensionless (%)	Temperature Dependent
Mass by Volume (% w/v)	\(\frac{w_{\text{solute (g)}}}{V_{\text{solution (mL)}}} \times 100\)	\(\text{g/100 mL}\)	Temperature Dependent
Mole Fraction (\(x\))	\(x_B = \frac{n_B}{n_A + n_B}, \quad x_A + x_B = 1\)	Unitless	Temperature Independent
Molarity (\(M\))	\(M = \frac{n_B}{V_{\text{L}}} = \frac{w_B \times 1000}{M_B \times V_{\text{mL}}}\)	\(\text{mol/L}\) or M	Temperature Dependent (liquid expands)
Molality (\(m\))	\(m = \frac{n_B}{w_{\text{solvent (kg)}}} = \frac{w_B \times 1000}{M_B \times w_A\text{ (g)}}\)	\(\text{mol/kg}\) or m	Temperature Independent