{"id":865993,"date":"2025-07-31T15:41:34","date_gmt":"2025-07-31T10:11:34","guid":{"rendered":"https:\/\/leverageedu.com\/discover\/?p=865993"},"modified":"2025-07-31T15:41:34","modified_gmt":"2025-07-31T10:11:34","slug":"ncert-notes-class-11-chemistry-part-i-chapter-1-some-basic-concepts-of-chemistry","status":"publish","type":"post","link":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-class-11-chemistry-part-i-chapter-1-some-basic-concepts-of-chemistry\/","title":{"rendered":"NCERT Notes Class 11 Chemistry (Part-I) Chapter 1: Some Basic Concepts of Chemistry (Free PDF)"},"content":{"rendered":"\n<p>Chemistry is an important branch of natural science that deals with the study of matter, energy, and the interactions between them. In this unit on some basic concepts of chemistry, we discuss the importance and development of chemistry, its branches, the nature and scope of the subject, the states and classification of matter, as well as key concepts such as mass, volume, density, atomic and molecular masses, mole concept, and stoichiometry.<\/p>\n\n\n\n\n\n\n<p><strong>Explore Notes of Class 11 Chemistry<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-background has-fixed-layout\" style=\"background-color:#f9d0b2\"><tbody><tr><td><strong>Chapter 2<\/strong><\/td><td><strong>Chapter 3<\/strong><\/td><td><strong>Chapter 4<\/strong><\/td><td><strong>Chapter 5<\/strong><\/td><td><strong>Chapter 6<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-background has-fixed-layout\" style=\"background-color:#9ed9fb\"><tbody><tr><td><strong><a href=\"https:\/\/drive.google.com\/file\/d\/1nw49jFRds5lRuzeEHZ0qrQek9cMsUjK4\/view?usp=drive_link\"><strong>Download PDF of NCERT Notes Class 11 Chemistry Chapter 1: Some Basic Concepts of Chemistry<\/strong><\/a><\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-introduction\">Introduction<\/h2>\n\n\n\n<p>Science aims to systematically understand and describe nature. Chemistry is the branch of science dealing with the composition, structure, properties, and reactions of material substances. Everyday changes such as curd formation, vinegar formation, and rusting of iron are chemical in nature. Chemistry evolved through practices like alchemy, iatrochemistry, and later modern chemistry.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-development-of-chemistry\">Development of Chemistry<\/h2>\n\n\n\n<p>In this section, we have presented a sneak peek at the historical development of chemistry.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-alchemical-traditions-in-india\">Alchemical Traditions in India<\/h3>\n\n\n\n<p>The alchemical traditions in India are briefly mentioned below.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Ancient India had well-developed chemical knowledge, including:\n<ul class=\"wp-block-list\">\n<li>Metallurgy, medicine, dyeing, glass and ceramic production, cosmetics, etc.<\/li>\n\n\n\n<li>Found in texts like Rasayan Shastra, Rasvidya, and Rastantra.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Excavations at Mohenjo-Daro and Harappa show the use of:\n<ul class=\"wp-block-list\">\n<li>Baked bricks, glazed pottery, and gypsum cement.<\/li>\n\n\n\n<li>Faience: a glass-like material used for ornaments.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-contributions-to-chemistry\">Contributions to Chemistry<\/h3>\n\n\n\n<p>Some of the notable Indian teachers who contributed to the field of Chemistry in India are briefly mentioned below.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Charaka Samhita<\/strong>: Describes sulphuric acid, nitric acid, and metal oxides.<\/li>\n\n\n\n<li><strong>Sushruta Samhita<\/strong>: Highlights the use of alkalies in treatment.<\/li>\n\n\n\n<li><strong>Kautilya\u2019s Arthashastra<\/strong>: Discusses salt production, dyeing, and fermentation.<\/li>\n\n\n\n<li><strong>Bhasma<\/strong>: Finely powdered metal used in Ayurvedic treatments (now considered to contain nanoparticles).<\/li>\n\n\n\n<li><strong>Acharya Kanda (600 BCE)<\/strong>: Proposed atomic theory using Param\u0101\u1e47u \u2014 indivisible, eternal, in motion.<\/li>\n\n\n\n<li><strong>Nagarjuna<\/strong>: Described mercury compounds and metal extraction.<\/li>\n\n\n\n<li><strong>Chakrapani<\/strong>: Discovered mercury sulphide (HgS), and made early soap using alkalies and mustard oil.<\/li>\n<\/ul>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-importance-of-chemistry\">Importance of Chemistry<\/h2>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Chemistry is central to all sciences and is used in:\n<ul class=\"wp-block-list\">\n<li>Weather prediction<\/li>\n\n\n\n<li>Functioning of computers<\/li>\n\n\n\n<li>Brain processes<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Chemistry helps in the production of:\n<ul class=\"wp-block-list\">\n<li>Fertilisers, dyes, soaps, drugs, detergents, alloys, metals, and synthetic materials.<\/li>\n\n\n\n<li>Healthcare: Drugs like cisplatin (for cancer) and AZT (for AIDS) were developed using chemical processes.<\/li>\n\n\n\n<li>Environment: Safer CFC substitutes have been created using chemical knowledge.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Chemistry contributes to the economy and employment generation through industries.<\/li>\n<\/ul>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-nature-of-matter\">Nature of Matter<\/h2>\n\n\n\n<p>Matter is anything that has mass and occupies space. It exists in 3 forms mainly: solid, liquid, and gas. Below, we have tabulated the general differences between these matters.<\/p>\n\n\n\n<figure class=\"wp-block-table is-style-stripes\"><table class=\"has-fixed-layout\"><tbody><tr><td><strong>State<\/strong><\/td><td><strong>Volume<\/strong><\/td><td><strong>Shape<\/strong><\/td><td><strong>Particle Arrangement<\/strong><\/td><\/tr><tr><td>Solid<\/td><td>Definite<\/td><td>Definite<\/td><td>Closely packed<\/td><\/tr><tr><td>Liquid<\/td><td>Definite<\/td><td>Not Definite<\/td><td>Loosely packed<\/td><\/tr><tr><td>Gas<\/td><td>Not Definite<\/td><td>Not Definite<\/td><td>Widely spaced<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><strong>Interconversion<\/strong>: Matter can be converted from one state to another by changing temperature or pressure<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-classification-of-matter\">Classification of Matter<\/h2>\n\n\n\n<p>Below, we have provided a tabular description of the classification of matter as a mixture.<\/p>\n\n\n\n<figure class=\"wp-block-table is-style-stripes\"><table class=\"has-fixed-layout\"><tbody><tr><td><strong>Type<\/strong><\/td><td><strong>Details<\/strong><\/td><td><strong>Examples<\/strong><\/td><\/tr><tr><td>Mixture<\/td><td>Contains 2+ substances mixed physically<\/td><td>Air, tea, salt water<\/td><\/tr><tr><td>Homegeneous<\/td><td>Uniform composition<\/td><td>Sugar solution<\/td><\/tr><tr><td>Hetreogeneous<\/td><td>Non-Uniform composition<\/td><td>Salt and sugar, oil and water<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>Below, we have provided a tabular description of the classification of matter as pure substance.<\/p>\n\n\n\n<figure class=\"wp-block-table is-style-stripes\"><table class=\"has-fixed-layout\"><tbody><tr><td><strong>Type<\/strong><\/td><td><strong>Details<\/strong><\/td><td><strong>Examples<\/strong><\/td><\/tr><tr><td>Pure Substance<\/td><td>Same type of particles with fixed composition<\/td><td>Water, copper, glucose<\/td><\/tr><tr><td>Elements<\/td><td>Made of identical atoms<\/td><td>O\u2082, Na, H\u2082<\/td><\/tr><tr><td>Compounds<\/td><td>Atoms of different elements in a fixed ratio; chemically combined<\/td><td>H\u2082O, CO\u2082, NH\u2083<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><strong>Also Read: <\/strong><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-english-snapshots-chapter-5-the-tale-of-melon-city\/\"><strong>NCERT Solutions Class 11 English Snapshots Chapter 5: The Tale of Melon City (Free PDF)<\/strong><\/a><\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-properties-of-matter-amp-their-measurement\">Properties of Matter & their Measurement<\/h2>\n\n\n\n<p>Every substance has unique characteristics or properties. These properties can be classified into two categories \u2014 physical properties, such as colour, odour, melting point, boiling point, density, etc., and chemical properties, like composition, combustibility, reactivity with acids and bases, etc. Physical properties can be measured or observed without changing the identity or the composition of the substance. The measurement or observation of chemical properties requires a chemical change to occur. The examples of chemical properties are characteristic reactions of different substances; these include acidity or basicity, combustibility, etc.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-measurement-of-physical-properties\">Measurement of Physical Properties<\/h3>\n\n\n\n<p>Scientific investigation requires quantitative measurements. Quantities are expressed as a number followed by its unit. Example:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Length = 6 m (where 6 is the number, and m is the unit of metre)<\/li>\n<\/ul>\n\n\n\n<p>Earlier, two systems of measurement were used:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>English system<\/strong><\/li>\n\n\n\n<li><strong>Metric system<\/strong><\/li>\n<\/ul>\n\n\n\n<p>The metric system was preferred because it\u2019s based on the decimal system. In 1960, an internationally accepted standard system of units was established, called the SI system (International System of Units).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-the-international-system-of-units-si\">The International System of Units (SI)<\/h3>\n\n\n\n<p>Based on 7 fundamental physical quantities, each with its own unit.<\/p>\n\n\n\n<figure class=\"wp-block-table is-style-stripes\"><table class=\"has-fixed-layout\"><tbody><tr><td><strong>Physical Quantity<\/strong><\/td><td><strong>Symbol<\/strong><\/td><td><strong>SI Unit<\/strong><\/td><td><strong>Expressed as<\/strong><\/td><\/tr><tr><td>Length<\/td><td>l<\/td><td>metre<\/td><td>m<\/td><\/tr><tr><td>Mass<\/td><td>m<\/td><td>kilogram<\/td><td>kg<\/td><\/tr><tr><td>Time<\/td><td>t<\/td><td>second<\/td><td>s<\/td><\/tr><tr><td>Electric Current<\/td><td>I<\/td><td>ampere<\/td><td>A<\/td><\/tr><tr><td>Thermodynamic Temperature<\/td><td>T<\/td><td>kelvin<\/td><td>K<\/td><\/tr><tr><td>Amount of Substance<\/td><td>n<\/td><td>mole<\/td><td>mol<\/td><\/tr><tr><td>Luminous Intensity<\/td><td>lv<\/td><td>candela<\/td><td>cd<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>In India, the National Physical Laboratory (NPL), New Delhi, maintains the national standards of measurements. Units are redefined when newer, more accurate methods are developed.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-mass-and-weight\">Mass and Weight<\/h3>\n\n\n\n<p>Mass and weight are discussed briefly below.<\/p>\n\n\n\n<p><strong>Mass<\/strong>:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass is defined as the amount of matter present in a substance.<\/li>\n\n\n\n<li>It is an intrinsic property of matter.<\/li>\n\n\n\n<li>The mass of a substance remains constant and does not vary with location or gravitational pull.<\/li>\n\n\n\n<li>It is measured using an analytical balance in laboratories.<\/li>\n\n\n\n<li>The SI unit of mass is the kilogram (kg).<\/li>\n\n\n\n<li>In laboratories, smaller units like grams (g) are commonly used.<\/li>\n<\/ul>\n\n\n\n<p><strong>Weight<\/strong>:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Weight is the force with which an object is attracted towards the Earth.<\/li>\n<\/ul>\n\n\n\n<ul class=\"wp-block-list\">\n<li>It depends on mass and gravitational acceleration (g).\n<ul class=\"wp-block-list\">\n<li>Weight = Mass \u00d7 g<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Unlike mass, weight is not constant. It varies from place to place depending on the gravity.\n<ul class=\"wp-block-list\">\n<li>Example: A body weighs less on the moon than on Earth.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>The SI unit of weight is also newton (N) (being a force), but it is commonly expressed in kg-force in everyday life.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-volume\">Volume<\/h3>\n\n\n\n<p>The volume is briefly described below.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume refers to the space occupied by matter.<\/li>\n\n\n\n<li>It is an extensive property and varies with the amount of substance.<\/li>\n\n\n\n<li>SI unit of volume: cubic metre (m\u00b3)<\/li>\n\n\n\n<li>Commonly used units in the laboratory:\n<ul class=\"wp-block-list\">\n<li>1 litre (L) = 1000 millilitres (mL)<\/li>\n\n\n\n<li>1 L = 1000 cm\u00b3 = 1 dm\u00b3<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Instruments used to measure volume:\n<ul class=\"wp-block-list\">\n<li>Graduated cylinders<\/li>\n\n\n\n<li>Pippttes<\/li>\n\n\n\n<li>Burettes<\/li>\n\n\n\n<li>Volumetric flasks<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-density\">Density<\/h3>\n\n\n\n<p>The density is briefly described below.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Density is defined as the mass per unit volume of a substance.<\/li>\n\n\n\n<li>It is represented by the formula: Density = Mass \/ Volume<\/li>\n\n\n\n<li>SI unit: kg\/m\u00b3<\/li>\n\n\n\n<li>Common unit in laboratories: g\/cm\u00b3 or g\/mL<\/li>\n\n\n\n<li>It helps in identifying substances and determining purity.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-temperature\">Temperature<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Temperature is a measure of the degree of hotness or coldness of a body.<\/li>\n\n\n\n<li>It determines the direction of heat flow.<\/li>\n\n\n\n<li>Heat flows from a body at a higher temperature to a lower temperature.<\/li>\n\n\n\n<li>Common temperature scales:\n<ol class=\"wp-block-list\">\n<li>Celsius scale (\u00b0C)<\/li>\n\n\n\n<li>Fahrenheit scale (\u00b0F)<\/li>\n\n\n\n<li>Kelvin scale (K) \u2014 the SI unit of temperature<\/li>\n<\/ol>\n<\/li>\n<\/ul>\n\n\n\n<p><strong>Conversion formulas:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>K = \u00b0C + 273.15<\/li>\n\n\n\n<li>\u00b0F = (\u00b0C \u00d7 9\/5) + 32<\/li>\n\n\n\n<li>The Kelvin scale begins at absolute zero (0 K), the lowest possible temperature where the motion of particles ceases.<\/li>\n\n\n\n<li>It does not have negative values.<\/li>\n<\/ul>\n\n\n\n<p class=\"has-pale-ocean-gradient-background has-background\"><strong>Also Read: <\/strong><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-class-11-english-snapshots-chapter-5-the-tale-of-melon-city\/\"><strong>NCERT Notes Class 11 English Snapshots Chapter 5: The Tale of Melon City (Free PDF)<\/strong><\/a><\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-uncertainty-in-measurement\">Uncertainty in Measurement<\/h2>\n\n\n\n<p>In scientific experiments, uncertainty may arise due to limitations of instruments and human errors.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-scientific-notation\">Scientific Notation<\/h3>\n\n\n\n<p>It is used to express very large or very small numbers in compact form. General form:<br>N \u00d7 10\u207f, where 1 \u2264 N < 10 and n is an integer.<\/p>\n\n\n\n<p><strong>Examples:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>0.00016 = 1.6 \u00d7 10\u207b\u2074<\/li>\n\n\n\n<li>232.508 = 2.32508 \u00d7 10\u00b2<\/li>\n\n\n\n<li>6,02,200,000,000,000,000,000,000 = 6.022 \u00d7 10\u00b2\u00b3 (Avogadro\u2019s number)<\/li>\n\n\n\n<li>Useful for representing:\n<ul class=\"wp-block-list\">\n<li>Atomic and molecular sizes<\/li>\n\n\n\n<li>Number of particles in a mole<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-significant-figures\">Significant Figures<\/h3>\n\n\n\n<p>The number of meaningful digits in a measured quantity is called significant figures. These include all digits that are known reliably, plus one uncertain digit.<\/p>\n\n\n\n<p><strong>Rules for determining significant figures:<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>All non-zero digits are significant.<\/strong>\n<ul class=\"wp-block-list\">\n<li>Example: 237 \u2192 3 significant figures<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Zeros between non-zero digits are significant.<\/strong>\n<ul class=\"wp-block-list\">\n<li>Example: 1005 \u2192 4 significant figures<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Leading zeros are not significant.<\/strong>\n<ul class=\"wp-block-list\">\n<li>Example: 0.0025 \u2192 2 significant figures<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Trailing zeros in a number with a decimal point are significant.<\/strong>\n<ul class=\"wp-block-list\">\n<li>Example: 3.00 \u2192 3 significant figures<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Exact numbers (e.g., 2 pens, 12 eggs)<\/strong> have <strong>infinite significant figures<\/strong>.<\/li>\n<\/ol>\n\n\n\n<p><strong>Rounding off significant figures:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>If the digit to be dropped is less than 5, the preceding digit is not changed.<\/li>\n\n\n\n<li>If the digit to be dropped is more than 5, the preceding digit is increased by one.<\/li>\n\n\n\n<li>If the digit to be dropped is exactly 5, round to make the preceding digit even.<\/li>\n<\/ul>\n\n\n\n<p><strong>Rules for operations<\/strong>:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Addition\/Subtraction<\/strong>:\n<ul class=\"wp-block-list\">\n<li>Final answer should have the same number of decimal places as the number with the fewest decimal places.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Multiplication\/Division<\/strong>:\n<ul class=\"wp-block-list\">\n<li>Final result should have the same number of significant figures as the quantity with the least number of significant figures.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-dimensional-analysis\">Dimensional Analysis<\/h3>\n\n\n\n<p>It is also called the factor label method or unit factor method. Used to convert units from one system to another using conversion factors.<\/p>\n\n\n\n<p><strong>Examples:<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Convert 3 inches to cm<\/strong>:<br>1 inch = 2.54 cm<br>\u2192 3 \u00d7 2.54 = <strong>7.62 cm<\/strong><strong><br><\/strong><\/li>\n\n\n\n<li><strong>Convert 2 L to m\u00b3<\/strong>:<br>1 L = 1 dm\u00b3 = 10\u207b\u00b3 m\u00b3<br>\u2192 2 L = 2 \u00d7 10\u207b\u00b3 m\u00b3<br><\/li>\n\n\n\n<li><strong>Convert 2 days into seconds<\/strong>:<br>1 day = 24 \u00d7 60 \u00d7 60 = 86400 seconds<br>\u2192 2 days = <strong>172800 seconds<\/strong><\/li>\n<\/ol>\n\n\n\n<p class=\"has-pale-ocean-gradient-background has-background\"><strong>Also Read: <\/strong><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-class-11-psychology-chapter-5-learning\/\"><strong>NCERT Notes Class 11 Psychology Chapter 5: Learning (Free PDF)<\/strong><\/a><\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-laws-of-chemical-combinations\">Laws of Chemical Combinations<\/h2>\n\n\n\n<p>These laws were developed after experimental observations regarding how elements combine to form compounds.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-law-of-conservation-of-mass\">Law of Conservation of Mass<\/h3>\n\n\n\n<p>It was proposed by Antoine Lavoisier in 1789. It states that:<br><br>\u201cMass can neither be created nor destroyed during a chemical reaction.\u201d<\/p>\n\n\n\n<p><strong>Example<\/strong>: When 10 g of calcium reacts with 17 g of chlorine, 27 g of calcium chloride is formed.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-law-of-definite-proportions\">Law of Definite Proportions<\/h3>\n\n\n\n<p>It was proposed by Joseph Proust. It states that:<br><br>\u201cA given compound always contains the same elements in the same proportion by mass.\u201d<br><br>Example: Water from any source contains 88.89% oxygen and 11.11% hydrogen by mass.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-law-of-multiple-proportions\">Law of Multiple Proportions<\/h3>\n\n\n\n<p>It was proposed by John Dalton in 1803. It states that:<br><br>\u201cIf two elements combine to form two or more compounds, the masses of one element that combine with a fixed mass of the other are in the ratio of small whole numbers.\u201d<strong><br><\/strong><strong><br><\/strong>Example: CO and CO\u2082:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>12 g of carbon combines with 16 g (in CO) and 32 g (in CO\u2082) of oxygen<\/li>\n\n\n\n<li>Ratio = 16:32 = 1:2<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-gay-lussac-s-law-of-gaseous-volumes\">Gay Lussac\u2019s Law of Gaseous Volumes<\/h3>\n\n\n\n<p>It was proposed by Gay Lussac in 1808. It states that:<br><br>\u201cWhen gases combine or are produced in a chemical reaction, they do so in simple whole number ratios by volume, provided all gases are measured at the same temperature and pressure.\u201d<\/p>\n\n\n\n<p>Example: H\u2082 + O\u2082 \u2192 H\u2082O<br>2 volumes of hydrogen + 1 volume of oxygen \u2192 2 volumes of water vapor<\/p>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-avogadro-s-law\">Avogadro\u2019s Law<\/h3>\n\n\n\n<p>It was proposed by Amedeo Avogadro in 1811. It states that:<br><br>\u201cEqual volumes of all gases, at the same temperature and pressure, contain equal number of molecules.\u201d<br><br>Avogadro\u2019s number (N\u2090): 6.022 \u00d7 10\u00b2\u00b3 particles\/mol<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-dalton-s-atomic-theory\">Dalton\u2019s Atomic Theory<\/h2>\n\n\n\n<p>Dalton\u2019s Atomic Theory was proposed by John Dalton in 1808.<\/p>\n\n\n\n<p><strong>Postulates of Dalton\u2019s Atomic Theory:<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Matter is made up of tiny indivisible particles called atoms.<\/li>\n\n\n\n<li>Atoms of the same element are identical in all respects.<\/li>\n\n\n\n<li>Atoms of different elements have different properties and masses.<\/li>\n\n\n\n<li>Compounds are formed when atoms combine in simple whole-number ratios.<\/li>\n\n\n\n<li>Atoms can neither be created nor destroyed in chemical reactions.<\/li>\n<\/ol>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-atomic-and-molecular-masses\">Atomic and Molecular Masses<\/h2>\n\n\n\n<p>In this section, we have provided a summary of the concept of atomic and molecular masses.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-atomic-mass\">Atomic Mass<\/h3>\n\n\n\n<p>Atoms are extremely small, so their masses are expressed relative to a standard. The standard used is the carbon-12 isotope. 1 atomic mass unit (1 amu or 1 u) = 1\/12 the mass of one atom of carbon-12.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>In grams:<br>1 u = 1.66056 \u00d7 10\u207b\u00b2\u2074 g<\/li>\n\n\n\n<li>Example:\n<ul class=\"wp-block-list\">\n<li>Mass of one hydrogen atom = 1.6736 \u00d7 10\u207b\u00b2\u2074 g<br>\u2192 in atomic mass units = 1.0078 u<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-average-atomic-mass\">Average Atomic Mass<\/h3>\n\n\n\n<p>In our universe, many elements exist as isotopes. The atomic mass of such elements is the weighted average of the masses of the isotopes based on their natural abundance.<\/p>\n\n\n\n<p><strong>Formula<\/strong>:<br>Average atomic mass = \u03a3 (isotopic mass \u00d7 % abundance) \/ 100<\/p>\n\n\n\n<p><strong>Example \u2013 Chlorine<\/strong>:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u00b3\u2075Cl (75.77%, mass = 34.9689 u)<\/li>\n\n\n\n<li>\u00b3\u2077Cl (24.23%, mass = 36.9659 u)<\/li>\n<\/ul>\n\n\n\n<p>Average atomic mass =<br>(75.77 \u00d7 34.9689 + 24.23 \u00d7 36.9659)\/100 = 35.5 u<\/p>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-molecular-mass\">Molecular Mass<\/h3>\n\n\n\n<p>The molecular mass is the sum of the atomic masses of all atoms present in a molecule.<\/p>\n\n\n\n<p>Formula:<br>Molecular mass = \u03a3 (atomic masses of constituent atoms)<\/p>\n\n\n\n<p><strong>Example \u2013 CH\u2084 (Methane)<\/strong>:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>C = 12.01 u, H = 1.008 u<\/li>\n\n\n\n<li>Molecular mass = 12.01 + 4 \u00d7 1.008 = 16.043 u<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-formula-mass\">Formula Mass<\/h3>\n\n\n\n<p>For ionic compounds (which do not exist as discrete molecules), we use the term formula mass instead of molecular mass. It is calculated the same way: by adding the atomic masses of the ions in the empirical formula.<\/p>\n\n\n\n<p><strong>Example \u2013 NaCl<\/strong>:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Na = 22.99 u, Cl = 35.45 u<\/li>\n\n\n\n<li>Formula mass = 22.99 + 35.45 = 58.44 u<\/li>\n<\/ul>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-mole-concept-and-molar-masses\">Mole Concept and Molar Masses<\/h2>\n\n\n\n<p>The mole concept and molar masses are discussed below.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-mole-concept\">Mole Concept<\/h3>\n\n\n\n<p>1 mole is defined as the amount of a substance that contains as many entities (atoms, molecules, ions, etc.) as there are in 12 g of carbon-12. 1 mole = 6.022 \u00d7 10\u00b2\u00b3 particles (Avogadro\u2019s number, N\u2090). So:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>1 mole of atoms = 6.022 \u00d7 10\u00b2\u00b3 atoms<\/li>\n\n\n\n<li>1 mole of molecules = 6.022 \u00d7 10\u00b2\u00b3 molecules<\/li>\n\n\n\n<li>1 mole of electrons = 6.022 \u00d7 10\u00b2\u00b3 electrons<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-molar-mass\">Molar Mass<\/h3>\n\n\n\n<p>Molar mass = Mass of one mole of a substance. It is numerically equal to the atomic\/molecular\/formula mass, but expressed in grams per mole (g\/mol)<\/p>\n\n\n\n<p><strong>Examples<\/strong>:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>H\u2082O: Molecular mass = 18.02 u \u2192 Molar mass = 18.02 g\/mol<\/li>\n\n\n\n<li>NaCl: Formula mass = 58.44 u \u2192 Molar mass = 58.44 g\/mol<\/li>\n<\/ul>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-percentage-composition\">Percentage Composition<\/h2>\n\n\n\n<p>It refers to the mass per cent of each element present in a compound.<\/p>\n\n\n\n<p>Formula:<br>Mass % of element = (Mass of element in 1 mol of compound \/ Molar mass of compound) \u00d7 100<\/p>\n\n\n\n<p><strong>Example \u2013 H\u2082O<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass = 2 \u00d7 1.008 + 16.00 = 18.016 g\/mol<\/li>\n\n\n\n<li>Mass % of H = (2.016 \/ 18.016) \u00d7 100 = 11.19%<\/li>\n\n\n\n<li>Mass % of O = (16.00 \/ 18.016) \u00d7 100 = 88.81%<\/li>\n<\/ul>\n\n\n\n<p><strong>Example \u2013 C\u2082H\u2086 (Ethane)<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>C = 12.01, H = 1.008<\/li>\n\n\n\n<li>Molar mass = 2 \u00d7 12.01 + 6 \u00d7 1.008 = 30.07 g\/mol<\/li>\n\n\n\n<li>Mass % of C = (24.02 \/ 30.07) \u00d7 100 = 79.89%<\/li>\n\n\n\n<li>Mass % of H = (6.048 \/ 30.07) \u00d7 100 = 20.11%<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-empirical-formula-and-molecular-formula\">Empirical Formula and Molecular Formula<\/h3>\n\n\n\n<p>In this section, we have discussed the empirical formula and molecular formula.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-empirical-formula\">Empirical Formula<\/h3>\n\n\n\n<p>The simplest whole-number ratio of atoms of each element in a compound.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-molecular-formula\">Molecular Formula<\/h3>\n\n\n\n<p>It represents the actual number of atoms of each element present in one molecule. It is always a whole-number multiple of the empirical formula.<\/p>\n\n\n\n<p>Formula:<br>Molecular formula = (Empirical formula) \u00d7 n<\/p>\n\n\n\n<p>Where,<br>n = Molar mass \/ Empirical formula mass<\/p>\n\n\n\n<p><strong>Steps to Determine Empirical Formula<\/strong>:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Assume 100 g of the compound (mass % = grams).<\/li>\n\n\n\n<li>Convert masses to moles using atomic masses.<\/li>\n\n\n\n<li>Divide each mole value by the smallest mole number.<\/li>\n\n\n\n<li>Multiply to get a whole-number ratio (if needed).<\/li>\n\n\n\n<li>Write the empirical formula using these ratios.<\/li>\n<\/ol>\n\n\n\n<p><strong>Steps to Determine Molecular Formula:<\/strong><\/p>\n\n\n\n<p>Use the formula:<br>Molecular formula = Empirical formula \u00d7 (Molar mass \/ Empirical formula mass)<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-stoichiometry-and-stoichiometric-calculations\">Stoichiometry and Stoichiometric Calculations<\/h2>\n\n\n\n<p>Stoichiometry deals with quantitative relationships between reactants and products in a balanced chemical reaction.<\/p>\n\n\n\n<p><strong>Example Reaction:<\/strong><\/p>\n\n\n\n<p>CH\u2084 + 2O\u2082 \u2192 CO\u2082 + 2H\u2082O<\/p>\n\n\n\n<p>This means:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>1 mole of CH\u2084 reacts with 2 moles of O\u2082<\/li>\n\n\n\n<li>Produces 1 mole of CO\u2082 and 2 moles of H\u2082O<\/li>\n<\/ul>\n\n\n\n<p>So, if we have:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>16 g CH\u2084 reacts with 64 g O\u2082 \u2192 gives 44 g CO\u2082 + 36 g H\u2082O<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-limiting-reagent\">Limiting Reagent<\/h3>\n\n\n\n<p>In a chemical reaction, the limiting reagent is the reactant that is completely consumed first, limiting the amount of product formed. The other reactants are in excess.<\/p>\n\n\n\n<p><strong>Steps to Identify the Limiting Reagent:<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Calculate the moles of all reactants.<\/li>\n\n\n\n<li>Use mole ratios from the balanced equation to determine which reactant would produce the least amount of product.<\/li>\n\n\n\n<li>The reactant that forms the least amount of product is the limiting reagent.<\/li>\n<\/ol>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-class-11-chemistry-part-i-chapter-1-some-basic-concepts-of-chemistry-important-formulas-nbsp\">Class 11 Chemistry (Part-I) Chapter 1: Some Basic Concepts of Chemistry- Important Formulas\u00a0<\/h2>\n\n\n\n<p>Here are the important formulas and equations from NCERT Class 11 Chemistry Chapter Some Basic Concepts of Chemistry.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Number of atoms in 1 mole (Avogadro\u2019s number)<\/strong><strong><br><\/strong>N = 6.022 \u00d7 10\u00b2\u00b3<\/li>\n\n\n\n<li><strong>Molar mass<\/strong><strong><br><\/strong>Molar mass = Mass of 1 mole of a substance (in grams)<\/li>\n\n\n\n<li><strong>Mole formula<\/strong><strong><br><\/strong>Number of moles = Given mass \/ Molar mass<\/li>\n\n\n\n<li><strong>Number of particles (atoms, molecules, ions, etc.)<\/strong><strong><br><\/strong>Number of particles = Moles \u00d7 Avogadro\u2019s number<\/li>\n\n\n\n<li><strong>Mass from moles<\/strong><strong><br><\/strong>Mass = Number of moles \u00d7 Molar mass<\/li>\n\n\n\n<li><strong>Volume of gas at STP<\/strong><strong><br><\/strong>1 mole of gas at STP = 22.4 L<br>Volume = Moles \u00d7 22.4 L (at STP)<\/li>\n\n\n\n<li><strong>Percentage composition<\/strong><strong><br><\/strong>% of element = (Mass of element in 1 mole \/ Molar mass of compound) \u00d7 100<\/li>\n\n\n\n<li><strong>Empirical formula calculation<\/strong>\n<ul class=\"wp-block-list\">\n<li>Convert % composition to grams<\/li>\n\n\n\n<li>Divide by atomic masses to get moles<\/li>\n\n\n\n<li>Divide all mole values by the smallest mole to get the simplest ratio<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Molecular formula<\/strong><strong><br><\/strong>Molecular formula = (Empirical formula) \u00d7 (Molar mass \/ Empirical formula mass)<\/li>\n\n\n\n<li><strong>Limiting reagent<\/strong><strong><br><\/strong>The reagent that produces the lesser amount of product is the limiting reagent.<\/li>\n\n\n\n<li><strong>Concentration in molarity<\/strong><strong><br><\/strong>Molarity (M) = Moles of solute \/ Volume of solution in litres<\/li>\n\n\n\n<li><strong>Law of Conservation of Mass<\/strong><strong><br><\/strong>Total mass of reactants = Total mass of products<\/li>\n\n\n\n<li><strong>Law of Definite Proportions<\/strong><strong><br><\/strong>A given compound always contains the same elements in the same proportion by mass.<\/li>\n<\/ul>\n\n\n\n<p><strong>Explore Notes of Class 11 Chemistry<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-background has-fixed-layout\" style=\"background-color:#f7df9e\"><tbody><tr><td><strong>Chapter 2<\/strong><\/td><td><strong>Chapter 3<\/strong><\/td><td><strong>Chapter 4<\/strong><\/td><td><strong>Chapter 5<\/strong><\/td><td><strong>Chapter 6<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><strong>Download the Solutions of Other Chapters of Class 11 Chemistry<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-background has-fixed-layout\" style=\"background-color:#d5b5f3\"><tbody><tr><td><strong>Chapter 1<\/strong><\/td><td><strong>Chapter 2<\/strong><\/td><td><strong>Chapter 3<\/strong><\/td><td><strong>Chapter 4<\/strong><\/td><td><strong>Chapter 5<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><strong>Related Reads<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table is-style-stripes\"><table class=\"has-pale-ocean-gradient-background has-background has-fixed-layout\"><tbody><tr><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-psychology-chapter-7-thinking\/\"><strong>NCERT Solutions Class 11 Psychology Chapter 7: Thinking (Free PDF)<\/strong><\/a><\/td><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-class-11-psychology-chapter-7-thinking\/\"><strong>NCERT Notes Class 11 Psychology Chapter 7: Thinking (Free PDF)<\/strong><\/a><\/td><\/tr><tr><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-class-11-english-woven-words-chapter-1-the-lament\/\"><strong>NCERT Notes Class 11 English Woven Words Chapter 1: The Lament (Free PDF)<\/strong><\/a><\/td><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-english-woven-words-chapter-1-the-lament\/\"><strong>NCERT Solutions Class 11 English Woven Words Chapter 1: The Lament (Free PDF)<\/strong><\/a><\/td><\/tr><tr><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-class-11-psychology-chapter-6-memory\/\"><strong>NCERT Notes Class 11 Psychology Chapter 6: Memory (Free PDF)<\/strong><\/a><\/td><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-psychology-chapter-6-memory\/\"><strong>NCERT Solutions Class 11 Psychology Chapter 6: Memory (Free PDF)<\/strong><\/a><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<figure class=\"wp-block-embed is-type-video is-provider-youtube wp-block-embed-youtube wp-embed-aspect-16-9 wp-has-aspect-ratio\"><div class=\"wp-block-embed__wrapper\">\n<iframe loading=\"lazy\" title=\"Class 11th Chemistry | Some basic concepts of chemistry | Super One Shot with Questions by Ashu sir\" width=\"1200\" height=\"675\" src=\"https:\/\/www.youtube.com\/embed\/1AAdCJC4IzQ?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe>\n<\/div><figcaption class=\"wp-element-caption\"><strong><em>Source: Ashu Ghai 11 & 12<\/em><\/strong><\/figcaption><\/figure>\n\n\n\n<p><strong>Explore notes on other subjects in the NCERT Class 11<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-background has-fixed-layout\" style=\"background-color:#fbbbbb\"><tbody><tr><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-and-solutions-class-11-english\/\"><strong>English<\/strong><\/a><\/td><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-and-solutions-class-11-sociology\/\"><strong>Sociology<\/strong><\/a><\/td><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-and-solutions-class-11-geography\/\"><strong>Geography<\/strong><\/a><\/td><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-and-solutions-class-11-political-science\/\"><strong>Political Science<\/strong><\/a><\/td><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-and-solutions-class-11-psychology\/\"><strong>Psychology<\/strong><\/a><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-faqs\">FAQs<\/h2>\n\n\n\n<div class=\"schema-faq wp-block-yoast-faq-block\"><div class=\"schema-faq-section\" id=\"faq-question-1753955825957\"><strong class=\"schema-faq-question\">Q1. What is the difference between atomic mass and molar mass?<\/strong> <p class=\"schema-faq-answer\"><strong>Ans: <\/strong>Atomic mass is the relative mass of an atom compared to 1\/12th the mass of a carbon-12 atom and is expressed in unified mass units (u), while molar mass is the mass of one mole of a substance expressed in grams per mole (g\/mol). Numerically, both are equal.<\/p> <\/div> <div class=\"schema-faq-section\" id=\"faq-question-1753955850796\"><strong class=\"schema-faq-question\">Q2. Why is the mole concept important in chemistry?<\/strong> <p class=\"schema-faq-answer\"><strong>Ans: <\/strong>The mole concept allows chemists to count particles (atoms, molecules, ions) by weighing them and to relate mass, volume, and number of particles in chemical reactions using a standard quantity\u20146.022 \u00d7 10\u00b2\u00b3 particles per mole.<\/p> <\/div> <div class=\"schema-faq-section\" id=\"faq-question-1753955868773\"><strong class=\"schema-faq-question\">Q3. What is a limiting reagent, and how is it identified?<\/strong> <p class=\"schema-faq-answer\"><strong>Ans: <\/strong>A limiting reagent is the reactant that is completely consumed in a chemical reaction and limits the amount of product formed. It is identified by comparing the mole ratio of reactants used with the ratio in the balanced chemical equation.<\/p> <\/div> <\/div>\n\n\n\n<p>For NCERT study material, follow NCERT Notes and Solutions Class 11 Chemistry by Leverage Edu now.\u00a0<\/p>\n\n\n\n<p><\/p>\n","protected":false},"excerpt":{"rendered":"Chemistry is an important branch of natural science that deals with the study of matter, energy, and the&hellip;\n","protected":false},"author":133,"featured_media":866002,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"editor_notices":[],"footnotes":""},"categories":[477,389],"tags":[],"class_list":{"0":"post-865993","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-ncert-study-material","8":"category-school-education"},"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v27.5 (Yoast SEO v27.5) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>NCERT Notes Class 11 Chemistry (Part-I) Chapter 1: Some Basic Concepts of Chemistry (Free PDF) - Leverage Edu Discover<\/title>\n<meta name=\"description\" content=\"Download free NCERT Class 11 Chemistry Chapter 1 notes (Some Basic Concepts of Chemistry) \u2013 concise, exam-ready, and easy to revise.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-class-11-chemistry-part-i-chapter-1-some-basic-concepts-of-chemistry\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Notes Class 11 Chemistry (Part-I) Chapter 1: Some Basic Concepts of Chemistry (Free PDF)\" \/>\n<meta property=\"og:description\" content=\"Download free NCERT Class 11 Chemistry Chapter 1 notes (Some Basic Concepts of Chemistry) \u2013 concise, exam-ready, and easy to revise.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-class-11-chemistry-part-i-chapter-1-some-basic-concepts-of-chemistry\/\" \/>\n<meta property=\"og:site_name\" content=\"Leverage Edu Discover\" \/>\n<meta property=\"article:published_time\" content=\"2025-07-31T10:11:34+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/blogassets.leverageedu.com\/media\/uploads\/sites\/9\/2025\/07\/21081747\/NCERT-Notes-Class-11-Chemistry-Part-I-Chapter-1-Some-Basic-Concepts-of-Chemistry-Free-PDF.webp\" \/>\n\t<meta property=\"og:image:width\" content=\"1024\" \/>\n\t<meta property=\"og:image:height\" content=\"640\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/webp\" \/>\n<meta name=\"author\" content=\"Devanshu Srivastava\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Devanshu Srivastava\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"14 minutes\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"NCERT Notes Class 11 Chemistry (Part-I) Chapter 1: Some Basic Concepts of Chemistry (Free PDF) - 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What is the difference between atomic mass and molar mass?","answerCount":1,"acceptedAnswer":{"@type":"Answer","text":"<strong>Ans: <\/strong>Atomic mass is the relative mass of an atom compared to 1\/12th the mass of a carbon-12 atom and is expressed in unified mass units (u), while molar mass is the mass of one mole of a substance expressed in grams per mole (g\/mol). Numerically, both are equal.","inLanguage":"en-US"},"inLanguage":"en-US"},{"@type":"Question","@id":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-class-11-chemistry-part-i-chapter-1-some-basic-concepts-of-chemistry\/#faq-question-1753955850796","position":2,"url":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-class-11-chemistry-part-i-chapter-1-some-basic-concepts-of-chemistry\/#faq-question-1753955850796","name":"Q2. Why is the mole concept important in chemistry?","answerCount":1,"acceptedAnswer":{"@type":"Answer","text":"<strong>Ans: <\/strong>The mole concept allows chemists to count particles (atoms, molecules, ions) by weighing them and to relate mass, volume, and number of particles in chemical reactions using a standard quantity\u20146.022 \u00d7 10\u00b2\u00b3 particles per mole.","inLanguage":"en-US"},"inLanguage":"en-US"},{"@type":"Question","@id":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-class-11-chemistry-part-i-chapter-1-some-basic-concepts-of-chemistry\/#faq-question-1753955868773","position":3,"url":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-notes-class-11-chemistry-part-i-chapter-1-some-basic-concepts-of-chemistry\/#faq-question-1753955868773","name":"Q3. What is a limiting reagent, and how is it identified?","answerCount":1,"acceptedAnswer":{"@type":"Answer","text":"<strong>Ans: <\/strong>A limiting reagent is the reactant that is completely consumed in a chemical reaction and limits the amount of product formed. It is identified by comparing the mole ratio of reactants used with the ratio in the balanced chemical equation.","inLanguage":"en-US"},"inLanguage":"en-US"}]}},"acf":[],"_links":{"self":[{"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/posts\/865993","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/users\/133"}],"replies":[{"embeddable":true,"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/comments?post=865993"}],"version-history":[{"count":0,"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/posts\/865993\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/media\/866002"}],"wp:attachment":[{"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/media?parent=865993"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/categories?post=865993"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/tags?post=865993"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}