Proust’s Law

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Prout's hypothesis

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The hypothesis put forward by the British chemist William Prout (1785–1850) in 1815 that all atomic weights are integer multiples of the atomic weight of hydrogen and hence that all atoms are made out of hydrogen. Subsequent work on atomic weights in the 19th century showed that this hypothesis is incorrect (with chlorine having an atomic weight of 35.5 being a glaring example of this). The understanding of atomic structure that emerged in the 20th century, with atomic number being the number of protons in an atom and non-integer atomic weights being due to mixtures of isotopes, has vindicated the spirit of Prout's hypothesis.

http://web.lemoyne.edu/~giunta/prout.html William Prout's original paper

From:   Prout's hypothesis   in  A Dictionary of Chemistry »

Subjects: Science and technology — Chemistry

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Prout’s hypothesis

Learn about this topic in these articles:, aid in chemical research.

…atomic weight of hydrogen (Prout’s hypothesis). This theory proved highly fruitful for later investigations of atomic weights, atomic theory, and the classification of the elements. Prout’s theory concerning the relative densities and weights of gases was in agreement with Avogadro’s law (1811), which was not generally accepted until the…

Joseph Proust and the Law of Constant Composition

Joseph Proust (1754-1826)

On September 26, 1754 ,  French chemist Joseph Louis Proust was born. He was best known for his discovery of the law of constant composition in 1799 , stating that in chemical reactions matter is neither created nor destroyed.

A Young Chemist

In 1780   Proust returned to Paris , where he taught chemistry at the Musée, a private teaching institution founded by scientific impresario and aeronaut Jean-François Pilâtre de Rozier . Part of this association involved  Proust with aerostatic experiments, which culminated in a balloon ascent with  Pilâtre on June 23, 1784 , at Versailles , in the presence of the king and queen of France, the king of Sweden , and the French court. In 1786   Proust returned to Spain to teach chemistry , first at Madrid and then in 1788 at the Royal Artillery School in Segovia . Founded in 1764 , this school was part of the program of the government of Charles III to bring Spain abreast of the northern European countries regarding military training. Because of Spain ’s scientific backwardness, expert instructors had to be sought abroad and  Proust had been recommended by no less than the great French chemist Antoine-Laurent de Lavoisier .

Hydrogen Sulfide

Proust’s law.

His second achievement derived from a controversy with C.L. Berthollet on the law of definite proportions, which is sometimes also known as Proust’s Law .  Proust studied copper carbonate , the two tin oxides, and the two iron sulfides to prove this law. He did this by making artificial copper carbonate and comparing it to natural copper carbonate . With this he showed that each had the same proportion of weights between the three elements involved . Between the two types of the other compounds,  Proust showed that no intermediate compounds exist between them.  Proust published this paper in 1794 , and his famous opponent Berthollet did not believe that substances always combine in constant and definite proportions. Moreover, Bethollet claimed that that the products of a reaction depend on the ratio of reactants.  Proust’s law was not accepted until 1812 , when the  Swedish chemist Jöns Jacob Berzelius gave him credit for it.

Although  Proust was correct in his observations, the reason why reagents behave in the way he described did not become clear until  English chemist John Dalton formulated his atomic theory in 1803 . According to Dalton , a fixed number of  atoms of one substance always combined with a fixed number of atoms of another substance in forming a compound. Dalton realized that substances must combine in the same proportions by weight as the weight proportions of their atoms . Other chemists had already observed that pure substances do combine in fixed proportions. They called that finding the law of definite (or constant) proportions. Dalton’s theory explained the law.

Proust  also performed a series of researches to characterize different types of sugars , present in vegetable products. After the death of his wife in 1817 ,  Proust moved to Angers , where in 1820 he took over the pharmacy of his brother Joachim. In 1819 he became a chevalier of the Legion of Honor , and in 1820 he was granted a pension by Louis XVIII . [2] On July 5, 1826 he died in Angers, France .

References and Further Reading:

  • [1] Joseph Proust at Encyclopedia Britannica
  • [2] Joseph Proust at Encyclopedia.com
  • [3] Modern Chemistry started with Lavosier
  • [4] Jöns Jacob Berzelius – One of the Founders of Modern Chemistry
  • [5] John Dalton and the Atomic Theory
  • [6] August Kekulé and the Carbon Ring Structure
  • [7] A Life of Discoveries – The Great Michael Faraday
  • [8] Joseph Proust at Wikidata
  • [9]  GCSE Chemistry – History of the Model of the Atom #7 ,  Cognito  @ youtube
  • [10]  Chisholm, Hugh, ed. (1911).   “Proust, Joseph Louis”  .   Encyclopædia Britannica . Vol. 22 (11th ed.). Cambridge University Press.
  • [11] Timeline of French Chemists via DBpedia and Wikidata

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  • CBSE Notes For Class 11
  • Chemistry Notes Class 11 Chapter 1

CBSE Class 11 Notes Chapter 1 - Some Basic Concepts of Chemistry

Chemistry is referred to as the “Central Science” as it interconnects geology, biology, environmental science, and physics to each other. Ancient Indians had knowledge of various concepts from chemistry even before it emerged as a discipline. In chemistry, one of the core fundamental concepts is matter – it is defined as a substance that occupies physical space and has inertia. There are three states of matter viz solid, gas, liquid. Matter can also be classified into compounds, mixtures or elements.

Laws of Chemical Combination

There are five basic laws that govern the chemical combination of elements. They are as follows:

Law of Conservation of Mass

It states that during a chemical reaction, the mass of the products and reactants will always be equal.

Explore more:   Law of Conservation of Mass

Law of Definite Proportions

It states that every chemical compound will contain a fixed and constant proportion by mass, of its constituent elements. Joseph Proust proposed this law.

Read more:  Law of Constant Proportions

Law of Multiple Proportions

It is a rule of stoichiometry formulated by John Dalton.

Also Read:  Stoichiometric Calculations

Gay Lussac’s Law of Gaseous Volumes

Gay Lussac gave this law in the year 1808. This law was properly explained by Avogadro.

Read more: Gay Lussac’s Law

Avogadro Law

It states that when the temperature and pressure conditions are the same, gases of equal volumes contain the same number of molecules.

Read more:   Avogadro Law

The above-described laws led to Dalton’s atomic theory.

For solving problems on Stoichiometry and Redox Reaction, watch the below video

proust hypothesis class 11

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John dalton's atomic theory: class-11 chemistry notes for iit-jee and neet, dalton’s atomic theory: a detailed exploration, introduction.

  • Pre-Daltonian Understanding of Matter

Revival of Atomic Theory

  • Indivisibility of Atoms
  • Identical Atoms of an Element
  • Distinct Atoms of Different Elements
  • Simple Whole-Number Ratios
  • Conservation in Chemical Reactions

Law of Conservation of Mass

Law of definite proportions, law of multiple proportions, chemical formulas and equations, atomic weights and the periodic table, predicting chemical behavior, discovery of subatomic particles, isotopes and atomic mass, modern atomic theory, frequently asked questions (faqs) about dalton’s atomic theory.

  • Main Postulates of Dalton’s Atomic Theory
  • Contribution to the Law of Conservation of Mass
  • Limitations Identified by Later Discoveries
  • Influence on the Development of the Periodic Table
  • Revolutionary Impact of Dalton’s Atomic Theory

Dalton's Atomic Theory is founded on five key postulates

In the early 19th century, John Dalton, an English chemist, meteorologist, and physicist, proposed one of the most significant theories in the history of science: Dalton’s Atomic Theory. This theory fundamentally transformed our understanding of matter and laid the groundwork for modern chemistry. In this comprehensive article, we will delve into Dalton’s Atomic Theory, examining its historical context, core postulates, and the scientific principles underlying it. We will also explore its implications, limitations, and how it set the stage for future developments in atomic theory.

Historical Context

The pre-daltonian understanding of matter.

Before Dalton, the concept of atoms had been speculated by ancient Greek philosophers such as Democritus and Leucippus, who proposed that matter was composed of indivisible particles called atoms. However, these early ideas were largely philosophical and lacked empirical support. Over the centuries, the notion of atoms faded, and the dominant theory of matter was based on the classical elements—earth, water, air, and fire—propounded by Aristotle.

The late 18th and early 19th centuries witnessed significant advancements in the field of chemistry. Scientists like Antoine Lavoisier, Joseph Proust, and Humphry Davy conducted experiments that challenged the existing paradigms. Lavoisier’s law of conservation of mass and Proust’s law of definite proportions provided crucial empirical evidence that matter was composed of discrete units.

John Dalton, influenced by these scientific developments and his own work on gases, revived and formalized the concept of atoms. His work culminated in the publication of “A New System of Chemical Philosophy” in 1808, where he articulated his atomic theory.

Core Postulates of Dalton’s Atomic Theory

Dalton’s Atomic Theory is founded on five key postulates:

  • All matter is composed of indivisible atoms.
  • Atoms of a given element are identical in mass and properties.
  • Atoms of different elements have distinct masses and properties.
  • Atoms combine in simple, fixed, whole-number ratios to form compounds.
  • Chemical reactions involve the rearrangement of atoms, without their creation or destruction.

Let’s explore each of these postulates in detail.

Postulate 1: Indivisibility of Atoms

Dalton asserted that atoms are the fundamental building blocks of matter and cannot be subdivided, created, or destroyed. This idea was revolutionary because it provided a concrete explanation for the conservation of mass observed in chemical reactions. Dalton envisioned atoms as solid, indestructible spheres, much like tiny billiard balls.

Postulate 2: Identical Atoms of an Element

According to Dalton, all atoms of a particular element are identical in mass and chemical properties. This postulate explained why a given element always exhibits consistent chemical behavior. For example, all atoms of hydrogen have the same mass and react similarly with other elements.

Postulate 3: Distinct Atoms of Different Elements

Dalton proposed that atoms of different elements vary in mass and properties. This differentiation explained the diversity of chemical substances. For instance, carbon atoms have different masses and properties compared to oxygen atoms, leading to the formation of different compounds like carbon dioxide (CO₂) and water (H₂O).

Postulate 4: Simple Whole-Number Ratios

Dalton observed that atoms combine in simple, fixed, whole-number ratios to form compounds. This postulate was crucial in understanding chemical stoichiometry and predicting the outcomes of chemical reactions. For example, water (H₂O) always consists of two hydrogen atoms for every oxygen atom, and carbon dioxide (CO₂) always has one carbon atom for every two oxygen atoms.

Postulate 5: Conservation in Chemical Reactions

Dalton’s final postulate stated that chemical reactions involve the rearrangement of atoms, but the atoms themselves are neither created nor destroyed. This idea was consistent with Lavoisier’s law of conservation of mass, providing a solid foundation for modern chemistry.

Scientific Principles Underlying Dalton’s Atomic Theory

The law of conservation of mass, formulated by Antoine Lavoisier, states that mass is neither created nor destroyed in a chemical reaction. Dalton’s atomic theory provided a microscopic explanation for this law, as the rearrangement of indestructible atoms during a reaction ensures that the total mass remains constant.

Joseph Proust’s law of definite proportions states that a chemical compound always contains the same elements in the same proportion by mass. Dalton’s theory explained this observation by positing that compounds are composed of atoms in fixed ratios. For instance, water always has a mass ratio of approximately 8:1 (oxygen to hydrogen) because it is composed of two hydrogen atoms for every oxygen atom.

Dalton himself formulated the law of multiple proportions, which states that if two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in simple whole-number ratios. This law was crucial in distinguishing between different compounds composed of the same elements. For example, carbon and oxygen can form both carbon monoxide (CO) and carbon dioxide (CO₂). If we compare the mass of oxygen that combines with a fixed mass of carbon in these two compounds, the ratio is 1:2.

Implications of Dalton’s Atomic Theory

Dalton’s theory provided a framework for understanding chemical formulas and equations. By representing compounds as combinations of atoms, chemists could write chemical equations that accurately depicted the conservation of mass and the stoichiometric relationships between reactants and products.

Dalton’s idea that atoms of different elements have distinct masses led to the concept of atomic weights (or atomic masses). By determining the relative masses of different atoms, scientists could begin to categorize elements and understand their relationships. This work eventually contributed to the development of the periodic table by Dmitri Mendeleev.

Dalton’s theory allowed chemists to predict the behavior of substances in chemical reactions. Understanding that atoms combine in fixed ratios helped in anticipating the quantities of products formed and reactants required in a reaction.

Limitations and Evolution of Dalton’s Atomic Theory

One of the major limitations of Dalton’s theory was the idea that atoms are indivisible. In the late 19th and early 20th centuries, experiments by scientists like J.J. Thomson, Ernest Rutherford, and James Chadwick revealed the existence of subatomic particles—electrons, protons, and neutrons. These discoveries showed that atoms are not indivisible but have an internal structure.

Dalton’s assertion that all atoms of a given element are identical was challenged by the discovery of isotopes. Isotopes are atoms of the same element that have different numbers of neutrons, resulting in different atomic masses. For example, carbon has isotopes like carbon-12 and carbon-14, which have different masses but are chemically identical.

Modern atomic theory has evolved significantly since Dalton’s time. The development of quantum mechanics and the understanding of atomic structure have provided a more detailed and accurate description of atoms. Today, we know that atoms consist of a nucleus containing protons and neutrons, surrounded by electrons in defined energy levels or orbitals.

John Dalton’s Atomic Theory was a groundbreaking milestone in the history of science. Despite its limitations and the subsequent advancements in atomic theory, Dalton’s work laid the essential foundation for modern chemistry. His theory provided a coherent framework to understand the nature of matter, chemical reactions, and the principles governing the behavior of elements and compounds. The core ideas of Dalton’s Atomic Theory—such as the existence of atoms, the conservation of mass in reactions, and the fixed ratios of atoms in compounds—remain integral to the study of chemistry today. As we continue to explore the complexities of atomic and molecular structures, the legacy of Dalton’s pioneering work endures.

1. What are the main postulates of Dalton’s Atomic Theory?

Answer: Dalton’s Atomic Theory is based on five key postulates:

2. How did Dalton’s Atomic Theory contribute to the law of conservation of mass?

Answer: Dalton’s Atomic Theory provided a microscopic explanation for the law of conservation of mass. It posited that atoms are indestructible and cannot be created or destroyed in chemical reactions. Thus, during a chemical reaction, the total mass of the atoms involved remains constant because the atoms are simply rearranged, not altered in their fundamental nature.

3. What limitations of Dalton’s Atomic Theory were identified by later scientific discoveries?

Answer: Several limitations of Dalton’s Atomic Theory were uncovered by later research:

  • Indivisibility of Atoms: Dalton’s idea that atoms are indivisible was challenged by the discovery of subatomic particles (electrons, protons, and neutrons).
  • Identical Atoms of an Element: The discovery of isotopes showed that atoms of the same element can have different masses due to varying numbers of neutrons.
  • Modern Atomic Theory: Advances in quantum mechanics and atomic structure provided a more detailed and accurate description of atoms, revealing their internal complexity and the behavior of electrons in defined energy levels.

4. How did Dalton’s Atomic Theory influence the development of the periodic table?

Answer: Dalton’s concept that atoms of different elements have distinct masses led to the development of atomic weights. By determining the relative masses of atoms, scientists were able to categorize elements and recognize patterns in their properties. This work laid the groundwork for Dmitri Mendeleev’s development of the periodic table, which organized elements based on atomic weight and recurring chemical properties.

5. Why was Dalton’s Atomic Theory considered revolutionary for its time?

Answer: Dalton’s Atomic Theory was revolutionary because it provided a systematic and empirical explanation for the nature of matter, which was previously based on philosophical speculation and classical elements. It introduced the idea of atoms as the fundamental building blocks of matter, explained the laws of chemical combination (such as the laws of conservation of mass, definite proportions, and multiple proportions), and offered a framework for understanding chemical reactions. This theory marked a significant shift from qualitative to quantitative scientific analysis in chemistry.

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Unit 3: Classification of elements & periodicity in properties

About this unit.

The periodic table is the everyday support for students, it suggests new avenues of research to professionals, and it provides a succinct organization of the whole of chemistry.

In this third unit of class 11, chemistry, we will study the historical development of the periodic table and the modern periodic law. We will also learn how the periodic classification follows as a logical consequence of the electronic configuration of atoms. Finally, we shall examine some of the periodic trends in the physical and chemical properties of the elements.

Modern periodic table

  • The periodic table (Opens a modal)

s, p, d, f subshells

  • Groups of the periodic table (Opens a modal)
  • Counting valence electrons for main group elements (Opens a modal)
  • The periodic table - transition metals (Opens a modal)
  • Electron configurations 7 questions Practice
  • Electronic configurations and position of elements 4 questions Practice

Periodic trends in physical properties of elements

  • Atomic radius trends on periodic table (Opens a modal)
  • Atomic and ionic radii (Opens a modal)
  • Mini-video on ion size (Opens a modal)
  • Ionization energy trends (Opens a modal)
  • Ionization energy: group trend (Opens a modal)
  • Ionization energy: period trend (Opens a modal)
  • First and second ionization energy (Opens a modal)
  • Electron affinity: period trend (Opens a modal)
  • Electronegativity (Opens a modal)
  • Electronegativity and bonding (Opens a modal)
  • Metallic nature (Opens a modal)
  • Periodic trends 7 questions Practice

Periodic trend in chemical properties of elements

  • Oxidation state trends in periodic table (Opens a modal)
  • Valence electrons and ionic compounds (Opens a modal)
  • Worked example: Finding the formula of an ionic compound (Opens a modal)
  • Valence electrons and ionic compounds 4 questions Practice

Class 11 Chemistry MCQ – Laws of Chemical Combination

This set of Class 11 Chemistry Chapter 1 Multiple Choice Questions & Answers (MCQs) focuses on “Laws of Chemical Combination”.

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Home » Class 11 » Periodic Table & Periodicity in Properties Notes: Class 11, JEE, NEET & AIIMS

Periodic Table & Periodicity in Properties Notes: Class 11, JEE, NEET & AIIMS

PERIODIC TABLE

CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES IN PERIODIC TABLE

Periodic Table is the most important concept in chemistry, both in principle and in practice. It is the everyday support for students, it suggests new avenues of research to professionals, and it provides a better organization of the whole of chemistry.

Earlier attempts to classify elements:

The atomic weights of all other elements are exact multiples of that of hydrogen and hence hydrogen is the primary substance from which the other elements have been formed. It is also called the unitary method.
The  atomic mass of the middle element of a Triad is the arithmetic mean of the atomic masses of the other two elements.

Ex..

He called this relationship as the Law of octaves. According to Newlands’ law of octaves when the elements are arranged in order of increasing atomic weights then every eighth element has properties similar to that of the first element. analogy with the intervals of the musical scale. Ex..

He classifies elements in the form of curve b/w atomic volume and atomic mass. The properties of the elements are the periodic function of their atomic volumes.

Mendeleev’s Periodic Table

  • He observed that the properties of elements, both physical and chemical, were periodically related to the atomic mass of the elements.
  • The Periodic Law (also referred to as Mendeleev’s Law), states that the chemical properties of elements are a periodic function of their atomic weights

 The advantages of Mendeleev’s Periodic table are:

  • The inclusion of these newly discovered elements did not disturb the periodic table. Examples include germanium, gallium, and scandium.
  •  It was used to correct the wrong atomic weights in use at that time.
  • A variance from atomic weight order was provided by Mendeleev’s table.

The limitations of Mendeleev’s Periodic table are:

  •  Hydrogen position was in the group of alkali metals but hydrogen also exhibited halogen like qualities.
  •  Isotopes were positioned differently since this type of classification of elements was done by considering the atomic weight of the element. Therefore – protium, deuterium, and tritium would occupy varying positions in Mendeleev’s table.
  • An anomalous positioning of a few elements showed that the atomic masses did not increase regularly from one element to the next. the atomic mass of 58.9) before nickel (atomic mass of 58.7).
  • He does not explain the electronic arrangement of elements.

Modern Periodic Table (By Moseley )

Modern periodic law:.

  •  According to the properties of elements all periodic function of their atomic numbers.
  • So, when elements are arranged according to increasing atomic no.’s there is periodicity in the electronic configuration that leads to periodicity in their chemical properties.

Introduction of the modern periodic table: It consists of:

horizontal rows (Periods) Vertical column (Groups)

There are 7 periods and 18 groups in this long form of the periodic table. The number of elements in each period:

  •  Ist period has 2 elements
  •  IInd period has 8 elements
  •  IIIrd period has 8 elements
  • IVth period has 18 elements
  •  Vth period has 18 elements
  • VIth period has 32 elements
  • VIIth period has rest of elements
  •  Number of valence electron in atom of elements decides which elements will be first in period and which will be last.

Families of different elements

  • 1 to 2 group and 13 to 17 contain normal or representative elements
  • 3 to 12 group –  transition elements.
  • 57 to 71   lanthanides
  •  89 to 103  Actinides.Left hand side – metals. Right hand side – non-metals

Valence electrons of different groups:

  • Hydrogen element has been placed at top of Ist group. Electronic configuration of H is similar to alkali metal both have 1 valence electron.
Valence electron of group Element
1 1
2 2
13 3
14 4
15 5
16 6
17 7
18 8

IUPAC  Nomenclature of elements with atomic numbers more than 100

Digit Root Abbreviation
0 Nil N
1 Un U
2 Bi B
3 Tri T
4 Quad Q
5 Pent P
6  Hex H
7 Sept S
8  Oct O

Recommended and official names of elements with Z more than 100

Z Recommended Symbol IUPAC official  name IUPAC Symbol
101 Unnilunium Unu Mendelevium Md
102 Unnilbium Unb Nobelium No
103 Unniltium Unt Lawrencium Lr
104 Unnilquadium Unq Rutherfordium Rf
105 Unnilpentium unp Dubnium Db

Element in s,p, d and f block and their electronic configuration in Periodic Table

  s-block elements:.

  •  Elements in which the last electron enters the s-orbitals.
  • General outer shell electronic configuration of s-block elements – ns 1-2 where n = 2-7

General characteristics of s-block elements in Periodic Table:

They are soft metals with low melting and boiling points,

  •  low ionization enthalpies (energies) and are highly electropositive.
  •   s -block elements lose the valence (outermost) electrons readily to form +1 (in case of alkali metals) and +2 ions (in case of alkaline earth metals)
  •  very reactive metals. The metallic character and the reactivity increase as we move down the group. Because of high reactivity, they are never found pure in nature.
  • The compounds of s-block elements with the exception of those of beryllium are predominantly ionic.

    P-Block Elements:

  • Elements in which the last electron enters any one of three p-orbitals of their respective outermost shells are called P-Block elements.

General characteristics of p-block elements:

  • P-block elements include both metal and non-metals but the number of non-metals is much higher than that of metals. Further, the metallic character increases from top to bottom within a group and non-metallic character increase from left to right along a period in this block.
  •  Their ionization enthalpies are relatively higher as compared to those s-block elements.
  •  They mostly form covalent compounds.
  •  Some of the show more than one (variable) oxidation states in their compounds.
  •  Their oxidizing character increase from left to right in a period and reducing character increase from top to bottom in a group.

D-block Elements:

  • Elements in which the last electron enters any one of the five d-orbitals of their respective penultimate shells are called d-block elements.

General Characteristics of d-block elements :

They are hard, malleable (i.e. can be converted into sheets) and ductile (i.e. can be drawn into wires) metals with high melting and boiling points,,  a good conductor of heat and electricity., ionization enthalpies are between s- and p- block elements.,  show variables oxidation states., form both ionic and covalent compounds., f-block elements:.

  • Elements in which the last electron entre any one of the seven f-orbitals of their respective ante-penultimate shells are called f block elements.
  • General outer shells electronic configuration of f-block elements – (n – 2)f 1-14 (n – 1)d 0-2 ns 2

General Characteristics of f-block elements:

  • They have generally high melting and boiling points,
  • show variable oxidation states
  •  Their compounds are generally colored,
  •  have heavy metals,
  •  a high tendency to form complexes.

Atomic Properties in Periodic Table:

i:
Distance an atom.from the centre of the nucleus to outermost shell of The distance between the centre of nuclei of atoms and mean position of shared pair of electrons between the bonded atoms one half the intern nuclear distances between two neighboring metals in lattice it is the measure of distance between cation and anion in ionic crystals. A  half of internuclear distance between the adjacent atoms of substance belonging to two different molecules.
For example:

The internuclear distance between the two hydrogen atoms in an H2 molecule is measured to be 74 pm. Therefore, the atomic radius of a hydrogen atom is 742=37 pm 74 2 = 37 pm.

For example :

H2 molecule the bond length is 74 pm, therefore, covalent radii is 37 pm

For example:

Cu metal, the bond length is 256pm therefore radii are128pm

Radius of Fe2+ is 76 pm, while that of Fe3+ is 65 pm. Hydrogen= 1.2A

Carbon= 1.7 A

Nitrogen= 1.55A

TRENDS IN ATOMIC PROPERTIES in Periodic Table:

Trends in atomic properties in Periodic Table

Atomic size decreases on moving from left to right in a period

Due to an increase in effective nuclear charge.Increases on moving from top to bottom in a group.

Due to addition of new subshells.

As we move towards right the atomic no. increase so as no. of protons increase, attraction increase. So the order is that it keeps on decreasing

Along period the size keeps on decreasing due to an increase in nuclear charge. Nuclear charge is the attraction of positively charged protons towards electron.

For example, in 3rdperiod: the order is

As we move down in the periodic table metallic radius increases due to the addition of shell along the group.

It is because every time a new shell is added so the effect of the addition of shell is more than the pronounced nuclear charge, therefore, the order as we discussed is increasing down the group.

For example: In the first group the order is

 

As you move from left to right across an element period (row) the ionic radius decreases. Even though the size of the atomic nucleus increases with larger atomic numbers moving across a period, the ionic and atomic radius decreases.

This is because the effective positive force of the nucleus also increases,

As you move from top to bottom down an element group (column) ionic radius increases.

This is because a new electron shell is added as you move down the periodic table. This increases the overall size of the atom.

 

Isoelectronic species:

  • They are those that have the same number of electron
  • The species with a higher positive charge is smaller and the species with a lesser positive charge is bigger similarly the species with a higher negative charge is bigger than the species with a lesser negative charge.
  • For example: O 2- , F – , Na + etc. are isoelectronic

Atomic radii:

Atomic radii in Periodic Table

Metallic radii:

Metallic radii in Periodic Table

Ionic radii:

Ionic radii

  • Cation is always smaller than parent atom as it is formed by removal of the electron due to which the same nuclear charge acts on a lesser number of electrons, therefore, the nuclear charge increases and the size decreases
  • Anion is always bigger than parent atom as it is formed by gaining electrons due to which the same nuclear charge acts on more number of electrons and moreover the magnitude of repulsions increase, therefore the nuclear charge decrease and size increase

Covalent radii:

Covalent radii in Periodic Table

Ionization energy in Periodic Table:

  •  It is the amount of energy required to remove electron from valence shell of isolated gaseous atom.
  • The word required is used because it means ionization energy is positive that is it means it is always given from outside to remove electron.

Successive ionization energies:

It is the amount of energy required to remove the second electron from the ion.

For example: If we compare ionization energy 1, 2 and 3 we found that: M (g) +∆ i H 1àM+(g) + e- (g)

M 1 (g) +∆ i H 2 àM 2+ (g) + e – (g)

M 2+ (g) +∆ i H 3 aM 3+ (g) + e – (g)

There are certain factors on which ionization energy depends:

  • Size of atom: If the size is small, nuclear charge will be high so it will attract electron more effectively, therefore, ionization energy will be high. This means ionization energy is inversely proportional to the size of an atom.
  • C harge on nucleus:

If nuclear charge is more than attraction for electron will be more, therefore, ionization energy will be high or vice versa.

Screening effect: Due to filling of inner orbital’s the nuclear charge is somehow reduced for outermost electron. As a result, the outermost electron will be loosely bounded and therefore ionization energy decreases.

  •   Penetration effect: The orbital’s which are closer to the nucleus will experience high nuclear charge so the penetrating effect will be more. The order it follows is:

S>P>d>f

Variation along period:

  • Along period it increases It is because along period the size decreases, nuclear charge increase, therefore, ionization energy increases but the anomalous behavior is seen let us see that: Li<Be>B<C<N<O<F<Ne

In this Be has high ionization energy than B because of fully filled shell N has high ionization energy than O because N is half-filled, more stable therefore ionization energy is high.

Variation along group:

  • Ionization energy decreases along the group as the size increases, nuclear charge decreases, as a result, less ionization energy is required to remove the electron.
  • For example in 1st group the order is:
  • Li>Na>K>Rb>Cs out of these cesium has the largest size and least ionization  Energy.

Electron gain enthalpy in Periodic Table:

  •  It is the amount of energy released when the electron is added to the outermost shell of an atom.
  • This energy release depends upon the extent of attraction for an incoming electron.
  • In the case of inert gases, they have almost no attraction Because of stable configuration.

Therefore it is negative for inert gases this means energy should be supplied in order to make them accept an electron. This energy can be mathematically shown as: M + e – à M 1 (negatively charged ion) Successive electron gain energies: Factors affecting electron gain enthalpies:

Atomic radius :

More is the size, less is the nuclear charge, therefore, less attraction for incoming electron therefore electron gain enthalpy is less negative for those atoms or vice versa

 Nuclear charge :

less is the nuclear charge, therefore, less attraction for incoming electron, therefore, electron gain enthalpy is less negative for those atoms or vice versa

Electronic configuration:

atoms with stable electronic configurations have less negative electron gain enthalpies.

  •  Electron gain enthalpy becomes less negative as we move down the group
  • As we down the group, the size increases, and nuclear charge decreases due to which the attraction for incoming electron decreases so as electron gain enthalpy decreases.
  • But certain anomalies are seen like out of chlorine and fluorine, fluorine has less negative electron enthalpy because of its extremely small size the incoming electron suffers interelectronic repulsions, therefore, electron gain enthalpy is less negative as compared to chlorine

Variation along period

  • Along period it increases the reason being the size decreases and nuclear charge increases.
  • But still, the exceptional behavior is seen like Electron gain enthalpy of inert gases is positive because of their stable electronic configuration.

Electronegativity in Periodic Table:

Fluorine daily use

  • It is the tendency of an atom to attract a shared pair of electrons more towards itself. This property depends upon the size and the nuclear charge.
  • If the size is less then more will be the nuclear charge hence more attraction for shared pair of electrons.

Fluorine ( most electronegative element )

Other factors on which electronegativity depends:

  • State of hybridization: The order is sp>sp 2 >sp 3

Oxidation state of element:

Along period it increases as size decreases and nuclear charge increases, therefore, the attraction for a shared pair of electrons increase

Applications of electronegativity: It tells us about the metallic and non metallic character of atom more is the electronegativity lesser is the metallic character and more is the non metallic character or vice versa. Polar or non-polar : If the difference in the electronegativity between the two bonded atoms is more than the bond is polar that is it has lost but if electronegativity difference is less or zero than the bonded non polar. Metallic and non-metallic character

Metallic character is a tendency to lose an electron.

Non-metallic character: It is the tendency to gain an electron

Along group

 along periods.

Metallic character decreases and non-metallic character increases. As we move the Size decreases along period due to which Nuclear Charge increases, therefore, tendency to gain electron increases and to lose electron decreases.

Na      Mg      Al      Si      P        S          Cl

  • Metallic character decreases

                  Nature of oxides:-

Along group:, basic oxides increase as we come across more metallic character down the group., along period:.

  • Basic nature of oxides decreases and acidic nature of oxides increases because size decrease and tendency to lose electron decreases, therefore, oxides are acidic

Chemical reactivity in Periodic Table:

  • It decreases (in case of non-metals)
  • It increases (in case of metals)
  • F                                     Li Cl      Decrease               Na Br                                   K I                                      Rb       Increases Cs Fr ALONG PERIOD It decreases and then increases.

Diagonal relationship in Periodic Table

  • It is the anomalous behavior of second group elements. It is defined as the similarity of some elements of second group with elements of third group present diagonally.” Like few diagonal relationships are given:

DIAGONAL RELATIONSHIP

  • Elements of 3rd period called as bridge elements .

Important Questions With Answer of Periodic table:

1 Considering the atomic number and position in the periodic table, arrange the following elements in the increasing order of metallic character : Si, Be, Mg, Na, P.

Metallic character increases down a group and decreases along a period as we move from left to right. Hence the order of increasing metallic character is       P < Si < Be < Mg < Na.

2 Which of the following species will have the largest and the smallest size? Mg, Mg 2+ , Al, Al 3+

Atomic radii decrease across a period. Cations are smaller than their parent atoms. Among isoelectronic species, the one with the larger positive nuclear charge will have a smaller radius. Hence the largest species is Mg;  the smallest one is Al 3+

3 Are the oxidation state and covalency of Al in [AlCl(H 2 O 5 ) 2+ ] same?

No. The oxidation state of Al is +3 and the covalency is 6.

4 Show by a chemical reaction with water that Na 2 O is a basic oxide and Cl 2 O 7 is an acidic oxide.

Na 2 O with water forms a strong base whereas Cl 2 O 7 forms strong acid. Na 2 O + H 2 O → 2NaOH Cl 2 O7 + H 2 O → 2HClO 4 Their basic or acidic nature can be qualitatively tested with litmus paper.

Published in Chemistry and Class 11

  • bridge elements
  • Covalent radius
  • Covalent radius trend
  • ionic radius
  • Ionization energy
  • Isoelectronic species
  • Periodic Table trends
  • Thermal energy examples
  • Hydrogen and its Compounds Important Questions And Answers

Hydrogen

  • Thermodynamics Important Questions And Answers
  • Environmental Chemistry Important Questions And Answers
  • Chemical Equilibrium Important Questions And Answers

PIYUSH K.S GAMMER

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NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12

NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements

June 20, 2022 by Sastry CBSE

NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements are part of NCERT Exemplar Class 11 Physics . Here we have given NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements.

Single Correct Answer Type Q1. The number of significant figures in 0.06900 is (a) 5 (b) 4 (c) 2 (d) 3 Sol: (b) Key concept: Significant figures in the measured value of a physical quantity tell the number of digits in which we have confidence. Larger the number of significant figures obtained in a measurement, greater is the accuracy of the measurement. The reverse is also true. The following rules are observed in counting the number of significant figures in a given measured quantity. 1. All non-zero digits are significant. 2. A zero becomes significant figure if it appears between two non¬zero digits. 3. Leading zeros or the zeros placed to the left of the number are never significant. 4. Trailing zeros or the zeros placed to the right of the number are significant. 5. In exponential notation, the numerical portion gives the number of significant figures. Leading zeros or the zeros placed to the left of the number are never

NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements Img 1

Q3. The mass and volume of a body are 4.237 g and 2.5 cm 3 , respectively. The density of the material of the body in correct significant figures is (a) 1. 6048 g cm -3 (b) 1.69 g cm -3 (c) 1.7 g cm 3                                             (d) 1.695 g cm -3

NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements Img 3

After rounding off the number, we get density =1.7

Q4. The numbers 2.745 and 2.735 on rounding off to 3 significant figures will give (a) 2.75 and 2.74 (b) 2.74 and 2.73 (c) 2.75 and 2.73 (d) 2.74 and 2.74 Sol: (d) Key concept: While rounding off measurements, we use the following rules by convention: 1. If the digit to be dropped is less than 5, then the preceding digit is left unchanged. 2. If the digit to be dropped is more than 5, then the preceding digit is raised by one. 3. If the digit to be dropped is 5 followed by digits other than zero, then the preceding digit is raised by one. 4. If digit to be dropped is 5 or 5 followed by zeros, then preceding digit is left unchanged, if it is even. 5. If digit to be dropped is 5 or 5 followed by zeros, then the preceding digit is raised by one, if it is odd. Units and Measurements Let us round off 2.745 to 3 significant figures. Here the digit to be dropped is 5, then preceding digit is left unchanged, if it is even. Hence on rounding off 2.745, it would be 2.74. Now consider 2.737, here also the digit to be dropped is 5, then the preceding digit is raised by one, if it is odd. Hence on rounding off 2.735 to 3 significant figures, it would be 2.74.

NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements Img 4

Q9. Which of the following measurements is most precise? (a) 5.00 mm (b) 5.00 cm (c) 5.00 m (d) 5.00 km Sol: (a) Key concept: Precision is the degree to which several measurements provide answers very close to each other. It is an indicator of the scatter in the data. The lesser the scatter, higher the precision. Let us first check the units. In all the options magnitude is same but units of measurement are different. As here 5.00 mm has the smallest unit. All given measurements are correct upto two decimal places. However, the absolute error in (a) is 0.01 mm which is least of all the four. So it is most precise.

NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements Img 9

Q11. Young’s modulus of steel is 1.9 x 10 11 N/m 2 . When expressed in CGS units of dyne/cm 2 , it will be equal to (1 N = 10 5 dyne, 1 m 2 = 10 4 cm 2 )                     . (a) 1.9 xlO 10                                              (b) 1.9×10 12 (c) 1.9 xlO 12                                             (d) 1.9 xlO 13

NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements Img 10

More Than One Correct Answer Type Q13. On the basis of dimensions, decide which of the following relations for the displacement of a particle undergoing simple harmonic motion is not correct?

NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements Img 12

Q14. If P, Q, R are physical quantities, having different dimensions, which of the following combinations can never be a meaningful quantity? (a) (P-Q)/R           (b) PQ-R (c) PQ/R                                   (d) (PR-Q 2 )/R (e)(R + Q)/P Sol: (a, e) Key concept: Principle of Homogeneity of dimensions: It states that in a correct equation, the dimensions of each term added or subtracted must be same. Every correct equation must have same dimensions on both sides of the equation. According to the problem P, Q and R are having different dimensions, since, sum and difference of physical dimensions, are meaningless, i.e., (P – Q) and (R + Q) are not meaningful. So in option (b) and (c), PQ may have the same dimensions as those of R and in option (d) PR and Q 2 may have same dimensions as those of R. Hence, they cannot be added or subtracted, so we can say that (a) and (e) are not meaningful.

Q15. Photon is a quantum of radiation with energy E = hv , where v is frequency and h is Planck’s constant. The dimensions of h are the same as that of (a) Linear impulse (b) Angular impulse (c) Linear momentum                           (d) Angular momentum

NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements Img 13

Q16. If Planck’s constant (h) and speed of light in vacuum (c) are taken as two fundamental quantities, which one of the following can, in addition, be taken to express length, mass and time in terms of the three chosen fundamental quantities? (a) Mass of electron (m e )              (b) Universal gravitational constant (G) (c) Charge of electron (e)              (d) Mass of proton (m p )

NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements Img 14

Q17. Which of the following ratios express pressure? (a) Force/Area                                        (b) Energy/Volume (c) Energy/Area                                      (d) Force/Volume

Sol:  (a, b) Let us first express the relation of pressure with other physical quantities one by one with the help of dimensional analysis.

NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements Img 15

Very Short Answer Type Questions

Q19. Why do we have different units for the same physical quantity? Sol:  Magnitude of any given physical quantity may vary over a wide range, therefore, different units of same physical quantity are required. For example: 1.Mass ranges from 10 -30 kg (for an electron) to 10 53 kg (for the known universe). We need different units to measure them like miligram, gram, kilogram etc. 2.The length of a pen can be easily measured in cm, the height of a tree can be measured in metres, the distance between two cities can be measured in kilometres and distance between two heavenly bodies can be measured in light year.

NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements Img 16

Q21. Name the device used for measuring the mass of atoms and molecules. Sol: A mass spectrograph is a device which is used for measuring the mass of atoms and molecules.

Q22. Express unified atomic mass unit in kg. Sol: The unified atomic mass unit is the standard unit that is used for indicating mass on an atomic or molecular scale (atomic mass). One unified atomic mass unit is approximately the mass of one nucleon (either a single proton or neutron) and is numerically equivalent to 1 g/mol. It is defined as one- twelfth of the mass of an unbound neutral atom of carbon-12 in its nuclear and electronic ground state.

NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements Img 18

Q24. Why length, mass and time are chosen as base quantities in mechanics? Sol: Normally each physical quantity requires a unit or standard for its specification, so it appears that there must be as many units as there are physical quantities. However, it is not so. It has been found that if in mechanics we choose arbitrarily units of any three physical quantities we can express the units of all other physical quantities in mechanics in terms of these. So, length, mass and time are chosen as base quantities in mechanics because (i) Length, mass and time cannot be derived from one another, that is these quantities are independent. (ii) All other quantities in mechanics can be expressed in terms of length, mass and time.

Short Answer Type Questions 25. (a) The earth-moon distance is about 60 earth radius. What will be the . diameter of the earth (approximately in degrees) as seen from the moon? (b) Moon is seen to be of (1/2)° diameter from the earth. What must be the relative size compared to the earth? (c) From parallax measurement, the sun is found to be at a distance of about 400 times the earth-moon distance. Estimate the ratio of sun-earth diameters.

NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements Img 20

Q26. Which of the following time measuring devices is most precise? (a) A wallclock                                         (b) A stop watch (c) A digital watch                                   (d) An atomic clock Given reason for your answer. Sol: Option (d) is correct because a clock can measure time correctly up to one second. A stop watch can measure time correctly up to a fraction of a second. A digital watch can measure time up to a fraction of second whereas an atomic clock is the most accurate timekeeper and is based on characteristic frequencies of radiation emitted by certain atoms having precision of about 1 second in 300,000 years. So an atomic clock can measure time most precisely as precision of this clock is about 1 s in 10 13 s.

Q27. The distance of a galaxy is of the order of 10 25 Calculate the order of magnitude of time taken by light to reach us from the galaxy. Sol: According to the problem, distance of the galaxy = 10 25 m. Speed of light = 3 x 10 8 m/s Hence, time taken by light to reach us from galaxy is

NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements Img 22

Q29. During a total solar eclipse the moon almost entirely covers the sphere of the sun. Write the relation between the distances and sizes of the sun and moon. Sol: Key point: In geometry, a solid angle (symbol: Ω or w) is the two­dimensional angle in three-dimensional space that an object subtends at a point. It is a measure of how large the object appears to an observer looking from that point. In the International System of Units (SI), a solid angle is expressed in a dimensionless unit called a steradian (symbol: sr).

NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements Img 24

Q32. Calculate the length of the arc of a circle of radius 31.0 cm which subtends an angle of π/6 at the centre.

NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements Img 29

Q35. Time for 20 oscillations of a pendulum is measured as t 1 =39.6 s; t 2 = 39.9 s and t 3 = 39.5 s. What is the precision in the measurements? What is the accuracy of the measurement? Sol: According to the problem, time for 20 oscillations of a pendulum, t 1 = 39.6 s, t 2 = 39.9 s and t 3 = 39.5 s It is quite obvious from these observations that the least count of the watch is 0.1 s. As measurements have only one decimal place. Precision in the measurement = Least count of the measuring instrument= 0.1 s Precision in 20 oscillations = 0.1

NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements Img 32

Long Answer Type Questions

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Q40. If velocity of light c, Planck’s constant h and gravitational constant G are taken as fundamental quantities, then express mass, length and time in terms of dimensions of these quantities. Sol: We have to apply principle of homogeneity to solve this problem. Principle of homogeneity states that in a correct equation, the dimensions of each term added or subtracted must be same, i.e., dimensions of LHS and RHS should be equal, We know that, dimensions of

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Q42. In an experiment to estimate ‘the size of a molecule of oleic acid, 1 mL of oleic acid is dissolved in 19 mL of alcohol. Then 1 mL of this solution is diluted to 20 mL by adding alcohol. Now, 1 drop of this diluted solution is placed on water in a shallow trough. The solution spreads over the surface of water forming one molecule thick layer. Now, lycopodium powder is sprinkled evenly over the film and its diameter is measured. Knowing the volume of the drop and area of the film we can calculate the thickness of the film which will give us the size of oleic acid molecule.

Read the passage carefully and answer the following questions.

  • Why do we dissolve oleic acid in alcohol?
  • What is the role of lycopodium powder?
  • What would be the volume of oleic acid in each mL of solution prepared?
  • How will you calculate the volume of n drops of this solution of oleic
  • What will be the volume of oleic acid in one drop of this solution?

Sol: (a) Since Oleic acid does not dissolve in water, hence it is dissolved in alcohol.

(b)Lycopodium powder spreads on the entire surface of water when it is sprinkled evenly. When a drop of prepared solution of oleic acid and alcohol is dropped on water, oleic acid does not dissolve in water. Instead it spreads on the water surface pushing the lycopodium powder away to clear a circular area where the drop falls. We can thus be able to measure the area over which oleic acid spreads.

(c)Since 20 mL (1 mL oleic acid + 19 mL alcohol) contains 1 mL of oleic acid, oleic acid in each mL of the solution =1/20 mL. Further, as this 1 mL is diluted to 20 mL by adding alcohol. In each mL of solution prepared, volume of oleic acid = 1/20 mL x 1/20  = 1/400 mL

(d) Volume of n drops of this solution of oleic acid can be calculated by means of a burette (used to make solution in the form of countable drops) and measuring cylinder and measuring the number of drops.

(e) As 1 mL of solution contains n number of drops, then the volume of oleic acid in one drop will be = 1/(400)n mL

Q43. (a) How many astronomical units (AU) make 1 parsec? (b) Consider the sun like a star at a distance of 2 parsecs. When it is seen through a telescope with 100 magnification, what should be the angular size of the star? Sun appears to be (1/2)° from the earth. Due to atmospheric fluctuations, eye cannot resolve objects smaller than 1 arc minute. (c) Mars has approximately half of the earth’s diameter. When it is closest to the earth it is at about 1/2 AU from the earth. Calculate at what size it will appear when seen through the same telescope.

NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements Img 44

NCERT Exemplar Class 11 Physics Solutions

  • Chapter 1 Units and Measurements
  • Chapter 2 Motion in a Straight Line
  • Chapter 3 Motion in a Plane
  • Chapter 4 Laws of Motion
  • Chapter 5 Work, Energy and Power
  • Chapter 6 System of Particles and Rotational Motion
  • Chapter 7 Gravitation
  • Chapter 8 Mechanical Properties of Solids
  • Chapter 9 Mechanical Properties of Fluids
  • Chapter 10 Thermal Properties of Matter
  • Chapter 11 Thermodynamics
  • Chapter 12 Kinetic Theory
  • Chapter 13 Oscillations
  • Chapter 14 Waves

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was an early 19th century attempt to explain the existence of the various through a hypothesis regarding the internal structure of the . In 1815 and 1816, the English chemist published two papers in which he observed that the that had been measured for the elements known at that time appeared to be whole multiples of the atomic weight of . He then hypothesized that the hydrogen atom was the only truly fundamental object, and that the atoms of other elements were actually groupings of various numbers of hydrogen atoms.

Prout's hypothesis remained influential in chemistry throughout the 1820s. However, more careful measurements of the atomic weights, such as those compiled by in 1828 or Edward Turner in 1832, "disproved" the hypothesis. In particular the atomic weight of , which is 35.45 times that of , could not at the time be explained in terms of Prout's hypothesis. Some came up with the ad hoc claim that the basic unit was one-half of a hydrogen atom, but further discrepancies surfaced. This resulted in hypothesis that one-quarter of a hydrogen atom was the common unit. Although these turn out to be wrong, these conjectures catalyzed further measurement of atomic weights, a great benefit to chemistry.

The discrepancy in the was later understood to be the result of multiple of the same element. Although all elements are the product of of hydrogen into higher elemets, it is now understood that atoms consist of both (hydrogen nuclei) and . Neutrons in nuclei decay into protons by emission of beta radiation.

The modern version of Prout's rule is that the of a nucleus of and neutron number is equal to sum of the masses of its constituent protons and neutrons minus its binding energy.

William Prout (1815). On the relation between the specific gravities of bodies in their gaseous state and the weights of their atoms. , 6: 321–330. Online reprint William Prout (1816). Correction of a mistake in the essay on the relation between the specific gravities of bodies in their gaseous state and the weights of their atoms. , 7: 111–13. Online reprint |

. It uses material from the . A is available in Wikipedia.

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  1. Law Of Constant Proportions

    NCERT Solutions Class 11 Commerce; NCERT Solutions For Class 10. ... The law of constant proportions is often referred to as Proust's law or as the law of definite proportions. ... The introduction of Dalton's atomic theory favoured this law and a relationship between these two concepts was established by the Swedish chemist Jacob Berzelius ...

  2. Proust's Law, Explained with examples

    Proust's law or law of definite proportions can be stated as follows: When two elements combine to give rise to another compound, they always do so in a constant mass ratio. In 1799, the French chemist Joseph Louis Proust (1754-1826) found in different samples of a compound the same elements in the same mass ratio.

  3. Periodic Table

    ️Hello Friends,In This Video I Have disscussed about Prouts Hypothesis of periodic table.♥️So Watch Full Video . For Notes Join Our Telegram Channelhttps://... CBSE Exam, class 12

  4. Prouts Unitary Theory

    Prouts Unitary Theory Video Lecture from Periodic table Chapter of Chemistry Class 11 for HSC, IIT JEE, CBSE & NEET.Watch Previous Videos of Chapter Periodic...

  5. Prout's hypothesis

    Prout's hypothesis. Prout's hypothesis was an early 19th-century attempt to explain the existence of the various chemical elements through a hypothesis regarding the internal structure of the atom. In 1815 [1] and 1816, [2] the English chemist William Prout published two papers in which he observed that the atomic weights that had been measured ...

  6. Prout's hypothesis

    Prout's hypothesis. William Prout advanced a hypothesis about atomic weights that had a lasting effect in chemistry far beyond the evidence on which it was based. Briefly, Prout noticed that a great many elements had atomic weights (combining weights, really) that were multiples of that of hydrogen, or very nearly so.

  7. Prout's hypothesis

    Prout's hypothesis. The hypothesis put forward by the British chemist William Prout (1785-1850) in 1815 that all atomic weights are integer multiples of the atomic weight of hydrogen and hence that all atoms are made out of hydrogen. Subsequent work on atomic weights in the 19th century showed that this hypothesis is incorrect (with chlorine ...

  8. Prout's hypothesis

    Prout's hypothesis was an early 19th century attempt to explain the existence of the various chemical elements through a hypothesis regarding the internal structure of the atom.In 1815 and 1816, the English chemist William Prout published two papers in which he observed that the atomic weights that had been measured for the elements known at that time appeared to be whole multiples of the ...

  9. Proust's Law of Definite Proportions

    Proust's Law of Definite Proportions. Proust's Law of Definite Proportions was created by Joseph Proust (a French chemist) in 1794 after realizing during his experiments that copper (II) carbonate is always in a fixed mass ratio of 5.3:1:4 (copper:carbon:oxygen) . Originally scientists believed that compounds could be composed of different mass ratios of elements and still be the same compound.

  10. Law of multiple proportions

    The discovery of this pattern led Dalton to develop the modern theory of atoms, ... One is a grey powder that Dalton referred to as "the protoxide of tin", which is 88.1% tin and 11.9% oxygen. The other is a white powder which Dalton referred to as "the deutoxide of tin", which is 78.7% tin and 21.3% oxygen. ... Joseph Louis Proust (1800 ...

  11. Prout's hypothesis

    Other articles where Prout's hypothesis is discussed: William Prout: …atomic weight of hydrogen (Prout's hypothesis). This theory proved highly fruitful for later investigations of atomic weights, atomic theory, and the classification of the elements. Prout's theory concerning the relative densities and weights of gases was in agreement with Avogadro's law (1811), which was not ...

  12. Joseph Proust and the Law of Constant Composition

    On September 26, 1754, French chemist Joseph Louis Proust was born. He was best known for his discovery of the law of constant composition in 1799, stating that in chemical reactions matter is neither created nor destroyed.. A Young Chemist. Joseph L. Proust was born on September 26, 1754 in Angers, France as the second son of Joseph Proust, an apothecary, and Rosalie Sartre.

  13. CBSE Class 11 Notes Chapter 1

    Laws of Chemical Combination Frequently Asked Questions on CBSE Class 11 Maths Notes Chapter 1 Some Basic Concepts of Chemistry. Chemistry is referred to as the "Central Science" as it interconnects geology, biology, environmental science, and physics to each other. Ancient Indians had knowledge of various concepts from chemistry even ...

  14. Proust hypothesis

    CBSE NCERT Class 10: Chapter Periodic classification of Elements: Proust hypothesis - tutorial by Aliya Gupta. CBSE Exam, class 10.

  15. John Dalton's Atomic Theory

    Dalton's Atomic Theory is founded on five key postulates: All matter is composed of indivisible atoms. Atoms of a given element are identical in mass and properties. Atoms of different elements have distinct masses and properties. Atoms combine in simple, fixed, whole-number ratios to form compounds.

  16. Classification of elements & periodicity in properties

    The periodic table is the everyday support for students, it suggests new avenues of research to professionals, and it provides a succinct organization of the whole of chemistry. In this third unit of class 11, chemistry, we will study the historical development of the periodic table and the modern periodic law. We will also learn how the periodic classification follows as a logical consequence ...

  17. Class 11 Chemistry MCQ

    This set of Class 11 Chemistry Chapter 1 Multiple Choice Questions & Answers (MCQs) focuses on "Laws of Chemical Combination". 1. How many basic laws are required to govern the combination of elements to form compounds? a) 6. b) 5.

  18. Joseph Proust

    Joseph-Louis Proust was born on 26 September 1754 in Angers, France.His father served as an apothecary in Angers. Joseph studied chemistry in his father's shop and later went to Paris where he gained the appointment of apothecary in chief to the Salpêtrière. [2] He also taught chemistry with Pilâtre de Rozier, a famous aeronaut. [2]Under Carlos IV's influence Proust went to Spain.

  19. Periodic Table & Periodicity in Properties:Class 11, JEE, NEET & AIIMS

    Periodic Table is the most important concept in chemistry, both in principle and in practice. It is the everyday support for students, it suggests new avenues of research to professionals, and it provides a better organization of the whole of chemistry. Earlier attempts to classify elements: Prout's hypothesis 1815.

  20. Flashcards for JEE Chemistry

    Document Description: Flashcards: Classification of Elements & Periodicity in Properties for NEET 2024 is part of Chemistry Class 11 preparation. The notes and questions for Flashcards: Classification of Elements & Periodicity in Properties have been prepared according to the NEET exam syllabus. Information about Flashcards: Classification of Elements & Periodicity in Properties covers topics ...

  21. Prout's_hypothesis

    Prout's hypothesis was an early 19th century attempt to explain the existence of the various chemical elements through a hypothesis regarding the internal structure of the atom.In 1815 and 1816, the English chemist William Prout published two papers in which he observed that the atomic weights that had been measured for the elements known at that time appeared to be whole multiples of the ...

  22. NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements

    NCERT Exemplar Class 11 Physics Chapter 1 Units and Measurements. Single Correct Answer Type. Q1. The number of significant figures in 0.06900 is. (a) 5 (b) 4 (c) 2 (d) 3. Sol: (b) Key concept: Significant figures in the measured value of a physical quantity tell the number of digits in which we have confidence.