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At 27°C, the vapour pressure of an aqueous solution of urea is equal to the osmotic pressure of 5 × 10–3 M aqueous solution of glucose. If vapour pressure of pure water at 27°C is 114 torr then the mole fraction of urea in its solution is - [Given : R = 0.08 L-atm/mol-K]a)4/5b)0.25c)3/4d)0.2Correct answer is option 'D'. 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 At 27°C, the vapour pressure of an aqueous solution of urea is equal to the osmotic pressure of 5 × 10–3 M aqueous solution of glucose. If vapour pressure of pure water at 27°C is 114 torr then the mole fraction of urea in its solution is - [Given : R = 0.08 L-atm/mol-K]a)4/5b)0.25c)3/4d)0.2Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 27°C, the vapour pressure of an aqueous solution of urea is equal to the osmotic pressure of 5 × 10–3 M aqueous solution of glucose. If vapour pressure of pure water at 27°C is 114 torr then the mole fraction of urea in its solution is - [Given : R = 0.08 L-atm/mol-K]a)4/5b)0.25c)3/4d)0.2Correct answer is option 'D'. Can you explain this answer?.

Solutions for At 27°C, the vapour pressure of an aqueous solution of urea is equal to the osmotic pressure of 5 × 10–3 M aqueous solution of glucose. If vapour pressure of pure water at 27°C is 114 torr then the mole fraction of urea in its solution is - [Given : R = 0.08 L-atm/mol-K]a)4/5b)0.25c)3/4d)0.2Correct answer is option 'D'. 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 At 27°C, the vapour pressure of an aqueous solution of urea is equal to the osmotic pressure of 5 × 10–3 M aqueous solution of glucose. If vapour pressure of pure water at 27°C is 114 torr then the mole fraction of urea in its solution is - [Given : R = 0.08 L-atm/mol-K]a)4/5b)0.25c)3/4d)0.2Correct answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of At 27°C, the vapour pressure of an aqueous solution of urea is equal to the osmotic pressure of 5 × 10–3 M aqueous solution of glucose. If vapour pressure of pure water at 27°C is 114 torr then the mole fraction of urea in its solution is - [Given : R = 0.08 L-atm/mol-K]a)4/5b)0.25c)3/4d)0.2Correct answer is option 'D'. Can you explain this answer?, a detailed solution for At 27°C, the vapour pressure of an aqueous solution of urea is equal to the osmotic pressure of 5 × 10–3 M aqueous solution of glucose. If vapour pressure of pure water at 27°C is 114 torr then the mole fraction of urea in its solution is - [Given : R = 0.08 L-atm/mol-K]a)4/5b)0.25c)3/4d)0.2Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of At 27°C, the vapour pressure of an aqueous solution of urea is equal to the osmotic pressure of 5 × 10–3 M aqueous solution of glucose. If vapour pressure of pure water at 27°C is 114 torr then the mole fraction of urea in its solution is - [Given : R = 0.08 L-atm/mol-K]a)4/5b)0.25c)3/4d)0.2Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice At 27°C, the vapour pressure of an aqueous solution of urea is equal to the osmotic pressure of 5 × 10–3 M aqueous solution of glucose. If vapour pressure of pure water at 27°C is 114 torr then the mole fraction of urea in its solution is - [Given : R = 0.08 L-atm/mol-K]a)4/5b)0.25c)3/4d)0.2Correct answer is option 'D'. Can you explain this answer? tests, examples and also practice JEE tests.