The Molecular Weight Of A Hydrogen Molecule Is M=2.016 ... - Vedantu

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AnswerQuestion Answers for Class 12Class 12 BiologyClass 12 ChemistryClass 12 EnglishClass 12 MathsClass 12 PhysicsClass 12 Social ScienceClass 12 Business StudiesClass 12 EconomicsQuestion Answers for Class 11Class 11 EconomicsClass 11 Computer ScienceClass 11 BiologyClass 11 ChemistryClass 11 EnglishClass 11 MathsClass 11 PhysicsClass 11 Social ScienceClass 11 AccountancyClass 11 Business StudiesQuestion Answers for Class 10Class 10 ScienceClass 10 EnglishClass 10 MathsClass 10 Social ScienceClass 10 General KnowledgeQuestion Answers for Class 9Class 9 General KnowledgeClass 9 ScienceClass 9 EnglishClass 9 MathsClass 9 Social ScienceQuestion Answers for Class 8Class 8 ScienceClass 8 EnglishClass 8 MathsClass 8 Social ScienceQuestion Answers for Class 7Class 7 ScienceClass 7 EnglishClass 7 MathsClass 7 Social ScienceQuestion Answers for Class 6Class 6 ScienceClass 6 EnglishClass 6 MathsClass 6 Social ScienceQuestion Answers for Class 5Class 5 ScienceClass 5 EnglishClass 5 MathsClass 5 Social ScienceQuestion Answers for Class 4Class 4 ScienceClass 4 EnglishClass 4 MathsThe molecular weight of a hydrogen molecule is M=2.016×${{10}^{-3}}$ kg/mol. If the root-mean-square speed of hydrogen molecules (H$_{2}$ ​) at 373.15K(100${}^{\circ }C$) is ${v_{rms}}$​=\[\dfrac{x+210}{100}km/s\] Find x.AnswerVerified599.7k+ viewsHint: At first, we need to calculate the mass of hydrogen from the molecular weight of hydrogen that is given. Then compare the equation to the formula for kinetic energy. After getting the $V_{rms}$, put it in the given equation and solve it for the value of x.Complete step by step answer:The mass of hydrogen molecule can be calculated from molecular weight,$m=\dfrac{M}{{{N}_{0}}}=\dfrac{2.016\times {{10}^{-3}}kg/mol}{6.02\times {{10}^{23}}mo{{l}^{-1}}}=3.35\times {{10}^{-27}}kg$ Then,$\dfrac{1}{2}mv_{rms}^{2}=\dfrac{3}{2}kT$ On solving,$\Rightarrow {{v}_{rms}}=\sqrt{\dfrac{3kT}{m}}$ $\Rightarrow {{v}_{rms}}=\sqrt{\dfrac{3\times 1.38\times {{10}^{-23}}\times 373.15}{3.35\times {{10}^{-27}}}}$$\Rightarrow {{v}_{rms}}=2.15Km/s$Now,${{V}_{rms}}=\dfrac{x+210}{100}km/s$2.15=$\dfrac{x+210}{100}$,which on solving, we get the value of x as 5 km/sAdditional Information:The sum of the atomic masses of all atoms in a molecule, based on a scale in which the atomic masses of different elements is known as Molecular weight. For example, the molecular weight of water, which has two parts of hydrogen and one part of oxygen, is 18 (i.e., 2 + 16).The ${{V}_{rms}}$ is defined as the square root of the mean of the squares for one time period of the sine wave.A molecule of hydrogen is the simplest possible molecule. It consists of two protons and two electrons that are held together by forces. Like atomic hydrogen, the assemblage can exist in one or more energy levels.The root mean square (RMS or rms) is defined as the square root of the mean square. The RMS is also known as the quadratic mean and is a particular case of where the generalized mean is with exponent 2.Note:Students must calculate the mass of the hydrogen molecule from molecular weight properly. Relation between rms value and kinetic energy must be clear. Always remember that hydrogen is a diatomic molecule.Recently Updated PagesMaster Class 11 Business Studies: Engaging Questions & Answers for SuccessMaster Class 11 Computer Science: Engaging Questions & Answers for SuccessMaster Class 11 Economics: Engaging Questions & Answers for SuccessMaster Class 11 Social Science: Engaging Questions & Answers for SuccessMaster Class 11 English: Engaging Questions & Answers for SuccessMaster Class 11 Chemistry: Engaging Questions & Answers for SuccessMaster Class 11 Business Studies: Engaging Questions & Answers for SuccessMaster Class 11 Computer Science: Engaging Questions & Answers for SuccessMaster Class 11 Economics: Engaging Questions & Answers for SuccessMaster Class 11 Social Science: Engaging Questions & Answers for SuccessMaster Class 11 English: Engaging Questions & Answers for SuccessMaster Class 11 Chemistry: Engaging Questions & Answers for Success

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