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 |Read what you need to know about our industry portal chemeurope.com.
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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
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
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.
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.
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
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?
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
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 )
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.
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.
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.
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.
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
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).
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.
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
Long Answer Type Questions
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
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.
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.
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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 ...
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.
️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
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...
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 ...
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.
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 ...
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 ...
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.
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 ...
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 ...
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.
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 ...
CBSE NCERT Class 10: Chapter Periodic classification of Elements: Proust hypothesis - tutorial by Aliya Gupta. CBSE Exam, class 10.
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.
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 ...
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.
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.
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.
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 ...
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 ...
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.