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A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3. If vapour pressure of CH2Cl2 and CHCl3 at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl3 in vapour form is: (Molar mass of Cl=35.5 g mol−1)a)0.162b)0.675c)0.325d)0.486Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3. If vapour pressure of CH2Cl2 and CHCl3 at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl3 in vapour form is: (Molar mass of Cl=35.5 g mol−1)a)0.162b)0.675c)0.325d)0.486Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3. If vapour pressure of CH2Cl2 and CHCl3 at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl3 in vapour form is: (Molar mass of Cl=35.5 g mol−1)a)0.162b)0.675c)0.325d)0.486Correct answer is option 'A'. Can you explain this answer?.

Solutions for A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3. If vapour pressure of CH2Cl2 and CHCl3 at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl3 in vapour form is: (Molar mass of Cl=35.5 g mol−1)a)0.162b)0.675c)0.325d)0.486Correct answer is option 'A'. 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 A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3. If vapour pressure of CH2Cl2 and CHCl3 at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl3 in vapour form is: (Molar mass of Cl=35.5 g mol−1)a)0.162b)0.675c)0.325d)0.486Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3. If vapour pressure of CH2Cl2 and CHCl3 at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl3 in vapour form is: (Molar mass of Cl=35.5 g mol−1)a)0.162b)0.675c)0.325d)0.486Correct answer is option 'A'. Can you explain this answer?, a detailed solution for A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3. If vapour pressure of CH2Cl2 and CHCl3 at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl3 in vapour form is: (Molar mass of Cl=35.5 g mol−1)a)0.162b)0.675c)0.325d)0.486Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3. If vapour pressure of CH2Cl2 and CHCl3 at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl3 in vapour form is: (Molar mass of Cl=35.5 g mol−1)a)0.162b)0.675c)0.325d)0.486Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3. If vapour pressure of CH2Cl2 and CHCl3 at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl3 in vapour form is: (Molar mass of Cl=35.5 g mol−1)a)0.162b)0.675c)0.325d)0.486Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.