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In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.From the given data, it can be interpreted that .a)If degree of dissociation is more, then vapour density is more for PCI5b)More the degree of dissociation, less is the vapour density of PCI5c)Degree of dissociation and vapour density cannot be correlated.d)a is independent of temperature.Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.From the given data, it can be interpreted that .a)If degree of dissociation is more, then vapour density is more for PCI5b)More the degree of dissociation, less is the vapour density of PCI5c)Degree of dissociation and vapour density cannot be correlated.d)a is independent of temperature.Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.From the given data, it can be interpreted that .a)If degree of dissociation is more, then vapour density is more for PCI5b)More the degree of dissociation, less is the vapour density of PCI5c)Degree of dissociation and vapour density cannot be correlated.d)a is independent of temperature.Correct answer is option 'B'. Can you explain this answer?.

Solutions for In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.From the given data, it can be interpreted that .a)If degree of dissociation is more, then vapour density is more for PCI5b)More the degree of dissociation, less is the vapour density of PCI5c)Degree of dissociation and vapour density cannot be correlated.d)a is independent of temperature.Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

Here you can find the meaning of In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.From the given data, it can be interpreted that .a)If degree of dissociation is more, then vapour density is more for PCI5b)More the degree of dissociation, less is the vapour density of PCI5c)Degree of dissociation and vapour density cannot be correlated.d)a is independent of temperature.Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.From the given data, it can be interpreted that .a)If degree of dissociation is more, then vapour density is more for PCI5b)More the degree of dissociation, less is the vapour density of PCI5c)Degree of dissociation and vapour density cannot be correlated.d)a is independent of temperature.Correct answer is option 'B'. Can you explain this answer?, a detailed solution for In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.From the given data, it can be interpreted that .a)If degree of dissociation is more, then vapour density is more for PCI5b)More the degree of dissociation, less is the vapour density of PCI5c)Degree of dissociation and vapour density cannot be correlated.d)a is independent of temperature.Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.From the given data, it can be interpreted that .a)If degree of dissociation is more, then vapour density is more for PCI5b)More the degree of dissociation, less is the vapour density of PCI5c)Degree of dissociation and vapour density cannot be correlated.d)a is independent of temperature.Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.From the given data, it can be interpreted that .a)If degree of dissociation is more, then vapour density is more for PCI5b)More the degree of dissociation, less is the vapour density of PCI5c)Degree of dissociation and vapour density cannot be correlated.d)a is independent of temperature.Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.