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Total vapour pressure of mixture of 1 mole of volatile component and 3 mole of volatile component B is 90 mm Hg. For such case:a)There is negativedeviation from Raoult’s law.b)Boiling point has been lowered.c)Force of attraction between A and B is smaller than that between A and A or between B and B.d)There is positive deviation from Raoult’s law.Correct answer is option 'D'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Total vapour pressure of mixture of 1 mole of volatile component and 3 mole of volatile component B is 90 mm Hg. For such case:a)There is negativedeviation from Raoult’s law.b)Boiling point has been lowered.c)Force of attraction between A and B is smaller than that between A and A or between B and B.d)There is positive deviation from Raoult’s law.Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Total vapour pressure of mixture of 1 mole of volatile component and 3 mole of volatile component B is 90 mm Hg. For such case:a)There is negativedeviation from Raoult’s law.b)Boiling point has been lowered.c)Force of attraction between A and B is smaller than that between A and A or between B and B.d)There is positive deviation from Raoult’s law.Correct answer is option 'D'. Can you explain this answer?.

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Here you can find the meaning of Total vapour pressure of mixture of 1 mole of volatile component and 3 mole of volatile component B is 90 mm Hg. For such case:a)There is negativedeviation from Raoult’s law.b)Boiling point has been lowered.c)Force of attraction between A and B is smaller than that between A and A or between B and B.d)There is positive deviation from Raoult’s law.Correct answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Total vapour pressure of mixture of 1 mole of volatile component and 3 mole of volatile component B is 90 mm Hg. For such case:a)There is negativedeviation from Raoult’s law.b)Boiling point has been lowered.c)Force of attraction between A and B is smaller than that between A and A or between B and B.d)There is positive deviation from Raoult’s law.Correct answer is option 'D'. Can you explain this answer?, a detailed solution for Total vapour pressure of mixture of 1 mole of volatile component and 3 mole of volatile component B is 90 mm Hg. For such case:a)There is negativedeviation from Raoult’s law.b)Boiling point has been lowered.c)Force of attraction between A and B is smaller than that between A and A or between B and B.d)There is positive deviation from Raoult’s law.Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of Total vapour pressure of mixture of 1 mole of volatile component and 3 mole of volatile component B is 90 mm Hg. For such case:a)There is negativedeviation from Raoult’s law.b)Boiling point has been lowered.c)Force of attraction between A and B is smaller than that between A and A or between B and B.d)There is positive deviation from Raoult’s law.Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Total vapour pressure of mixture of 1 mole of volatile component and 3 mole of volatile component B is 90 mm Hg. For such case:a)There is negativedeviation from Raoult’s law.b)Boiling point has been lowered.c)Force of attraction between A and B is smaller than that between A and A or between B and B.d)There is positive deviation from Raoult’s law.Correct answer is option 'D'. Can you explain this answer? tests, examples and also practice Chemistry tests.