Some basic concepts of Chemistry Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 1 Notes - Download PDF

Some Basic Concepts of Chemistry: the real starting point of your chemistry journey! This chapter is like the ABCs of the subject, where you get to know what matter is made of, how atoms and molecules interact, and why every tiny calculation in chemistry counts. It will form the basis for other chapters that you will learn later on.

This chapter covers essential topics such as the mole concept, molar mass, chemical equations, stoichiometry, chemical calculations and the law of conservation of mass. The NCERT notes for Class 11 chemistry Chapter 1 will help you revise these concepts in the easiest way possible. These NCERT notes will not only help you in your exams but will also improve your understanding of the topics. The important formulas are well discussed in these notes. The tables and charts are also included to help you learn better.

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NCERT Notes for Class 11 Chemistry Chapter 1

If you want to make your learning easier and interesting, then you are at the right place. These notes contain everything from simple to complex topics that will help you a lot in your exams. Scroll down to know more!

1.1 Importance Of Chemistry

Chemistry is a study that investigates the composition, structure, and properties of matter as well as the changes that occur during chemical reactions. It is derived from the Egyptian word kme (chem), which means "earth." Because it connects physical sciences (including chemistry) with life sciences and applied sciences, chemistry is sometimes referred to as a core science (such as medicine and engineering).

The branches of chemistry are as follows

Physical chemistry

The branch of chemistry that studies macroscopic and physical processes in the universe. It deals with the effect of a substance's physical properties on its chemical properties as well as its structure.

Inorganic chemistry

Inorganic chemistry is the field of chemistry that investigates substances that do not contain carbon or hydrogen atoms. To put it another way, it is the reverse of organic chemistry. Metals, salts, and chemicals are examples of substances that do not have carbon-hydrogen bonds.

Organic chemistry

Organic chemistry is the branch of science that studies the structure, composition, and chemical characteristics of organic molecules. It entails the investigation of carbon and its compounds.

Biochemistry

Biochemistry is the discipline of chemistry that studies the chemical processes that occur in and around organisms. It's a science that combines biology and chemistry in a laboratory setting. Biochemists can comprehend and address biological problems by applying chemical knowledge and technology.

Analytical chemistry

It is a discipline of chemistry that employs equipment and analytical procedures to figure out a substance's structure, functionality, and qualities.

1.2 Nature of matter

Anything that has mass and occupies space is called matter.

1.2.1 States of matter

In general, matter is grouped into three categories

Solids

Solids have the least flexibility of movement and have a distinct shape and volume. E.g., sugar, stone.

Liquids

The term "liquid" refers to a substance that has the shape of a container but has a fixed volume. Liquids can also flow or be poured. Water, milk, oil, mercury, and alcohol are a few examples.

Gas

Gases are substances that have neither a defined volume nor a defined shape. Gases usually entirely fill the container in which they are stored. For example, hydrogen, oxygen, and so on.

1.2.2. Classification of Matter

At the bulk or macroscopic level, matter is classified into two

• Mixtures

• Pure substance

Mixtures

A mixture is a material that contains two or more chemicals in any proportion. It is mostly divided into two types: heterogeneous and homogeneous mixtures

Homogeneous mixtures

Homogeneous mixtures form when two substances are mixed to form a single uniform phase. In homogeneous mixtures, the composition of substances present is uniform. Sugar solution and air are two examples of homogeneous mixtures.

Heterogeneous mixtures

Heterogeneous Mixtures are those in which two or more substances are combined to produce a combination with a non-uniform composition. Suspensions, which are made up of two solids, such as salt and sugar, are an example.

Pure substances

A pure substance is a material that contains only one type of particle. The following are the divisions of pure substances

Elements

An element is a pure material that contains only one type of atom and can't be broken down any further. Silver, hydrogen, oxygen, etc are examples.

Compounds

A compound is a pure material made up of two or more elements combined in a certain mass proportion. Furthermore, the properties of a composite differ from those of its constituent parts. Additionally, physical methods cannot separate the parts of a complex into simpler compounds. Chemical procedures are the only way to separate them. Examples include water, ammonia, carbon dioxide, etc.

1.3 Properties of Matter and Their Measurements

Every substance has its own distinct or distinguishing qualities. The two categories of attributes that are observed are physical and chemical properties.

1.3.1 Physical and chemical properties

Physical properties

Physical attributes are those that may be measured or observed without changing the identity or composition of a substance. Color, aroma, melting point, boiling temperature, density, and other physical characteristics are examples.

Chemical properties

Chemical properties are the characteristics of a substance that can be observed during a chemical reaction. Flammability, toxicity, heat of combustion, pH, radioactive decay rate, and chemical stability are some of the most important chemical properties.

1.3.2 Measurement of physical properties

Physical quantities are those that we come upon during our scientific research. A physical quantity includes a number followed by a unit. The reference standard used to measure any physical quantity is referred to as a unit.

1.3.3 The International System of Units (SI)

The eleventh General Conference on Weights and Measures developed the International System of Units. The CGPM is an intergovernmental body established by the Meter Convention, a diplomatic convention signed in Paris in 1875. The SI system has seven base units. The seven essential scientific quantities are represented by these units. These quantities can be used to generate other physical quantities such as speed, volume, density, and so on.

1.3.4 Mass and Weight

A substance's mass is the amount of substance it contains, whereas its weight is the force of gravity acting on the object. The mass of matter is constant, but its weight varies from place to place due to changes in gravity. The kilogram is the SI unit of mass (kg). Newton is the SI derived unit of weight (derived unit of the SI base unit).

1.3.5 Volume

The amount of three-dimensional space surrounded by specific closed boundaries, such as the space occupied or contained by a material (solid, liquid, gas, or plasma), or the space occupied or contained by a shape. The most common way to measure volume is to use SI-derived units (cubic meters).

1.3.6 Density

A material's mass density, sometimes known as density, is defined as its mass per unit volume. The density is symbolized by the symbol ρ (the lowercase Greek letter rho). The SI unit for density is kilograms per cubic metre.

1.3.7 Temperature

The term "temperature" refers to a physical attribute of matter that quantifies the concepts of heat and cold. There are three popular temperature scales: Celsius, Fahrenheit, and (Fahrenheit) (Kelvin.

°F = 9/5(°C) + 32

K= °C +273.15

There are some prefixes that are used in the SI system. The values in the table below will help you with unit conversions.

1.4 Uncertainty in Measurements

In the study of chemistry, one must frequently deal with both experimental data and theoretical computations. There are practical methods for dealing with numbers and presenting facts in a realistic manner, to the extent possible.

1.4.1 Scientific Notation

While performing mathematical operations on numbers expressed in scientific notation, the following points are to be kept in mind.

Multiplication & Division
Follow exponent rules
-Multiply the coefficients
- Add or subtract the exponents

For example,

a) Multiplication-

$(5.6\times {10}^5)\times (6.9\times {10}^8)=38.64\times {10}^{13}$

$=3.864\times {10}^{14}$

b) Division-

$\frac{9.8\times {10}^{-2}}{2.5\times {10}^{-6}}=3.92\times {10}^4$

Addition & Subtraction
Make the exponents the same, then add/subtract the coefficients

For example,

a) Addition

$(6.65\times {10}^4)+(8.95\times {10}^3)$

$=(6.65+0.895)\times {10}^4=7.545\times {10}^4$

b) Subtraction

$(2.5\times {10}^{-2})-(4.8\times {10}^{-3})$

$=(2.5-0.48)\times {10}^{-2}=2.02\times {10}^{-2}$

1.4.2 Significant Figures

All experimental measurements are subject to some degree of ambiguity. When we talk about measurement, precision and accuracy are frequently mentioned. Significant figures are digits that are certain to be relevant.

Precision

Precision refers to the consistency of different measurements for the same amount. Accuracy, on the other hand, is the agreement of a given value with the true value of the result.

1.4.3 Dimensional analysis

We frequently have to convert units from one system to another in calculations. The factor label approach, also known as the unit factor method or dimensional analysis, is used to do this.

1.5 Laws of Chemical Combination

1.5.1 law of conservation of mass

“The mass in an isolated system can neither be created nor destroyed but can be transformed from one form to another.” According to the law of conservation of mass, in a low-energy thermodynamic process, the mass of the reactants must be equal to the mass of the products. This law was proposed by Antoine Lavoisier in 1789.

1.5.2 Law of Definite Proportions

The mass proportions of the constituents in a composite sample are always the same, according to the law of definite proportions. It was given by Joseph Proust, a French chemist.

1.5.3 law of multiple proportions

When two elements are mixed to make more than one compound, the weight of one element is proportionate to the fixed weight of the other element as a whole number, according to the law of multiple proportions. Dalton proposed this law in 1803.

1.5.4 Gay Lussac’s Law of Gaseous Volumes

"The link between the volume of a gaseous reactant and a product can be represented by a simple whole number," according to Gay Lussac's law, developed in 1808

1.5.5 Avogadro Law

Avogadro postulated in 1811 that the number of molecules in equal quantities of gases at the same temperature and pressure should be the same.

Here, in the figure below, two volumes of hydrogen gas combine with one volume of oxygen gas to form two volumes of water vapour that shows hydrogen and oxygen exist as diatomic molecules.

1.6 Dalton’s Atomic Theory

Dalton wrote 'A New System of Chemical Philosophy' in 1808, in which he proposed:

• Atoms are the building blocks of matter.

• A given element's atoms all share the same properties, including the same mass. The mass of atoms in different elements differs.

• When atoms of different elements mix in a specific ratio, compounds are produced.

• Reorganization of atoms occurs during chemical processes. In a chemical reaction, they are neither generated nor destroyed.

1.7 Atomic and Molecular Masses

1.7.1 atomic mass

Atomic mass is defined as the mass of a single atom, expressed in atomic mass units (amu) or unified mass units (u)

1 a.m.u = 1/12 th mass of 1 carbon-12 atom

$1\mathrm{amu}=1.66056\times {10}^{-24}\mathrm{g}$

1.7.2 Average atomic mass

The average atomic mass of an element can be estimated from distinct isotopes of an element based on their relative abundance.

For example, there are three isotopes of C – (12C, 13C, 14C) with their relative abundance (%), 98.892, 1.108, and 2 × 10^-10 with their atomic mass (AMU) as 12, 13.00335, and 14.00317, respectively.
The average atomic mass of carbon will be
$=(0.98892)(12u)+(0.01108)(13.00335u)+(2\times {10}^{-12})(14.00317u)$
= 12.011 u.

1.7.3 Molecular mass

The masses of the atoms that make up a molecule are added together to determine its mass. As a result, the two methods for measuring the mass of an atom, amu and g / mol, can also be used to represent the mass of a molecule.

1.7.4 Formula mass

Formula mass is used for compounds like NaCl, in which positive and negative entities are arranged in a three-dimensional structure.

Formula mass of NaCl = Atomic mass of sodium + Atomic mass of chlorine

= 23.00 u + 35.5 u = 56.5 u.

1.8 Mole Concept

Mole is a unit of amount of substance. A mole is defined as the amount of something that includes the same number of elementary particles (atoms, molecules, or ions) as 12 g of carbon dioxide (C-12).

1 mol of any substance contains 6.023 x 10²³ particles. 6.023 x 10²³ is known as the Avogadro constant or Avogadro number.

Eg : 1 mole water = 6.023 x 10²³ water molecules

The mass of 1 mole of a substance is called the molar mass.

1.9 Percentage Composition

Mass% of an element = Mass of that element in the compound* 100/Molar mass of the compound

Eg : Molar mass of water = 18.02 g/mol

Mass % of hydrogen = 2*1.00810018

= 11.18

Mass % of oxygen = 16*10018

= 88.79

1.9.1 Empirical formula for molecular formula

The compound's formula that gives the simplest whole-number ratio of the atoms of various elements present in a single molecule is called the empirical formula.

The actual ratio of the atoms of various elements present in one molecule of a compound is given by the substance's molecular formula.

1.10 Stoichiometry and Stoichiometric Calculations

Stoichiometry is the study of chemical processes and calculations. The stoichiometric coefficient is the coefficient that is utilized to balance the reaction. The stoichiometric coefficient, rather than the mass, is the ratio of moles of molecules of atoms that react. The stoichiometric ratio can only be used to forecast the number of moles of product generated when all reactants are present in a stoichiometric ratio. The real quantity of product created is always smaller than the theoretical calculation predicts.

1.10.1 Limiting reagent

If the reagents are not employed in a stoichiometric ratio, the limiting reagent, which is less than the needed amount, dictates how much product is created, and the excess reagent is called the excess reagent. When we burn carbon in the air (which has an infinite supply of oxygen), the amount of CO₂ created is proportional to the amount of carbon burned. Carbon is the limiting reactant in this situation, whereas O₂ is the excess reactant.

1.10.2 Reactions in Solutions

Because the two solids cannot react in the solid state, they must be dissolved in a liquid. When solutes dissolve in a solvent, they form a solution, which is a single phase in which they coexist. The strength of the solution is measured using several factors. The amount of solute in a solution is indicated by the strength of the solution.

Mass percentage

It is the mass of the solute in grams per 100 grams of solution. A 10 % solution of sodium chloride, for example, contains 10 g of NaCl per 100 g of solution.

Mass $\mathrm{\%}$ of the solute $=\frac{\text{ Mass of solute }}{\text{ Mass of solution }}\times 100$

Mole fraction

The mole fraction of any component of a solution is the ratio of that component's number of moles to the total number of moles of all the solution's components.

Mole fraction of A,

$X_A=\frac{\text{ No. of moles of }A}{\text{ No. of moles of }A+\text{ No. of moles of }B}$

$=\frac{n_{\mathrm{A}}}{n_{\mathrm{A}}+n_{\mathrm{B}}}$

and the mole fraction of B

${\mathrm{X}}_{\mathrm{B}}$ = $\frac{\text{ No. of moles of B }}{\text{ No. of moles of A + No. of moles of B }}$

$\begin{array}{rl}& =\frac{n_{\mathrm{B}}}{n_{\mathrm{A}}+n_{\mathrm{B}}}\\ {\mathrm{X}}_{\mathrm{A}}+{\mathrm{X}}_{\mathrm{B}}& =\frac{n_{\mathrm{A}}}{n_{\mathrm{A}}+n_{\mathrm{B}}}+\frac{n_{\mathrm{B}}}{n_{\mathrm{A}}+n_{\mathrm{B}}}\\ & =\frac{n_{\mathrm{A}}+n_{\mathrm{B}}}{n_{\mathrm{A}}+n_{\mathrm{B}}}=1\\ {\mathrm{X}}_{\mathrm{A}}+{\mathrm{X}}_{\mathrm{B}}& =1\end{array}$

Molarity

The number of moles of solute dissolved per dm3 (or litre, L) of a solution is referred to as its molarity. Temperature affects the molarity of any solution. As a result, any solution's molarity is specified for a specific temperature.

Molarity of solution $=\frac{\text{ No. of moles of the solute }}{\text{ Volume of the solution in litres (or in }{\mathrm{dm}}^3\text{ ) }}=\frac{n}{\mathrm{V}}$

Molality

The number of moles of solute per kg of solvent is the molality of a solution. If n moles of a solute are dissolved in W kg of solvent to make a solution,

Molality, $(m)=\frac{\text{ No. of moles of solute }}{\text{ Mass of the solvent in kg. }}$

Important Questions of Class 11 Chemistry Chapter 1

Question: On combustion 0.210 g of an organic compound containing $\mathrm{C},\mathrm{H}$ and O gave $0.127\mathrm{g}{\mathrm{H}}_2\mathrm{O}$ and $0.307\mathrm{g}{\mathrm{CO}}_2$. The percentages of hydrogen and oxygen in the given organic compound are:

(1) $53.41,39.6$

(2) $6.72,53.41$

(3) $7.55,43.85$

(4) $6.72,39.87$

Answer:

Mass of organic compound $=0.210\mathrm{g}$
Mass of water formed $=0.127\mathrm{g}$
Mass of ${\mathrm{CO}}_2$ formed $=0.307\mathrm{g}$
Mass of hydrogen $=\frac{0.127\times 2}{18}=0.014\mathrm{g}$
Percentage of hydrogen $=\frac{0.014\times 100}{0.210}=6.72\mathrm{\%}$
Mass of carbon $=\frac{0.307\times 12}{44}=0.084\mathrm{g}$
Percentage of carbon $=\frac{0.084\times 100}{0.210}=39.87\mathrm{\%}$
Percentage of oxygen $=100-(39.87+6.72)=53.41\mathrm{\%}$

Hence, the correct answer is option (2).

Question: Among ${10}^{-9}\mathrm{g}$ (each) of the following elements, which one will have the highest number of atoms?

Element : $\mathrm{Pb},\mathrm{Po},\mathrm{Pr}$ and Pt

(1) Po

(2) Pr

(3) Pb

(4) Pt

Answer: $\text{ No. of atoms }=\frac{\text{ Mass}}{\text{ MolarMas }(\mathrm{g}/\mathrm{mol})}\times {\mathrm{N}}_{\mathrm{A}}$

From this formula, it's clear that for a given mass, the element with the smallest molar mass will have the greatest number of atoms, as it appears in the denominator.
${\mathrm{M}}_{{\mathrm{P}}_{\mathrm{o}}}=209$
${\mathrm{M}}_{\mathrm{pr}}=141$
${\mathrm{M}}_{\mathrm{Pb}}=207$
${\mathrm{M}}_{\mathrm{Pt}}=195$

Pr has the lowest molar mass ( $M_{\mathrm{Pr}}=141\mathrm{g}/\mathrm{mol}$ ), ${10}^{-9}$ grams of Pr will contain the highest number of atoms compared to the other elements listed.

Hence, the correct answer is option (2).

Question: 16 g of oxygen has the same number of molecules as
(i) 16 g of CO
(ii) 28 g of $N_2$
(iii) 14 g of $N_2$
(iv)) 1.0 g of ${\mathrm{H}}_2$

Choose the answer from the options below

1) (i) and (ii)

2) (ii) and (iv)

3) (iii) and (iv)

4) None of the above

Answer:

The number of molecules of oxygen in 16g of oxygen is

$\begin{array}{rl}& \frac{16}{32}\times (6.023\times {10}^{23})\\ & =0.5\times {(6.023\times {10}^{23})}_{}\end{array}$

Number of molecules of ${\mathrm{N}}_2$

$\begin{array}{rl}& =\frac{14}{28}\times (6.023\times {10}^{23})\\ & =0.5\times {(6.023\times {10}^{23})}_{}\end{array}$
Number of molecules of ${\mathrm{H}}_2$

$\begin{array}{rl}& =\frac{1}{2}\times 6.023\times {10}^{23}\\ & =0.5\times {(6.023\times {10}^{23})}_{}\end{array}$
Hence, the correct answer is option (3).

Approach to Solve Questions of Class 11 Chemistry Chapter 1

To solve the questions effectively, students have to focus on understanding the fundamental concepts along with practicing questions. Students should utilise the sources and content provided online to ensure a complete understanding of the subject. Here are a few tips that help students to solve the questions with a good approach

1. Understand the basic concepts
While solving questions, students are guided to first understand the basic concepts like,
SI Units & Measurements - significant figures, scientific notation and dimensional analysis.
Laws of Chemical Combination

2. Practice the formulas of the mole concept
The numericals from this chapter are common yet easy. You can solve them easily on the basis of formulas. Some important topics include

• Atomic Mass and number of moles
• Mole & Avogadro’s Number: 6.022 × 10²³
• Percentage Composition: Empirical Formula & Molecular Formula.
• Stoichiometry: finding the limiting reagents and balancing chemical equations

3. Calculations and practice.
As we all know, practice is key so do focus on topics like

• Conversions of mass into moles and moles into number of particles.
• Empirical and molecular formulas like finding % composition, moles of element and determining the simplest ratio, which is the empirical formula.

4. Solve Numerical Problems
Keep some formulas on your fingertips-

• Density & Molar Volume: For gases, use 1 mole = 22.4 L at STP.
• Molarity (M) = Moles of solute / Volume of solution (L)
• Molality (m) = Moles of solute / Mass of solvent (kg)
• Mole Fraction ($\chi$) = Moles of component / Total moles
• Percentage Purity: (Mass of pure substance / Total mass) × 100.

Apart from this, solve the in-text and end-of-exercise problems provided in the NCERT textbook. Questions from the NCERT books are asked directly in the NCERT boards and other competitive exams.

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