Question Description
In an experiment, 1 g of a non-volatile solute was dissolved in 100 g of acetone (mol.weight=58) at K. The vapour pressure of the solution was found to be 192.5 mm of Hg. The molecular weight of the solute is (vapour pressure of acetone = 195 mm of Hg)a)25.24b)35.24c)45.24d)55.24Correct answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about In an experiment, 1 g of a non-volatile solute was dissolved in 100 g of acetone (mol.weight=58) at K. The vapour pressure of the solution was found to be 192.5 mm of Hg. The molecular weight of the solute is (vapour pressure of acetone = 195 mm of Hg)a)25.24b)35.24c)45.24d)55.24Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In an experiment, 1 g of a non-volatile solute was dissolved in 100 g of acetone (mol.weight=58) at K. The vapour pressure of the solution was found to be 192.5 mm of Hg. The molecular weight of the solute is (vapour pressure of acetone = 195 mm of Hg)a)25.24b)35.24c)45.24d)55.24Correct answer is option 'C'. Can you explain this answer?.

Solutions for In an experiment, 1 g of a non-volatile solute was dissolved in 100 g of acetone (mol.weight=58) at K. The vapour pressure of the solution was found to be 192.5 mm of Hg. The molecular weight of the solute is (vapour pressure of acetone = 195 mm of Hg)a)25.24b)35.24c)45.24d)55.24Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for NEET. Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free.

Here you can find the meaning of In an experiment, 1 g of a non-volatile solute was dissolved in 100 g of acetone (mol.weight=58) at K. The vapour pressure of the solution was found to be 192.5 mm of Hg. The molecular weight of the solute is (vapour pressure of acetone = 195 mm of Hg)a)25.24b)35.24c)45.24d)55.24Correct answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of In an experiment, 1 g of a non-volatile solute was dissolved in 100 g of acetone (mol.weight=58) at K. The vapour pressure of the solution was found to be 192.5 mm of Hg. The molecular weight of the solute is (vapour pressure of acetone = 195 mm of Hg)a)25.24b)35.24c)45.24d)55.24Correct answer is option 'C'. Can you explain this answer?, a detailed solution for In an experiment, 1 g of a non-volatile solute was dissolved in 100 g of acetone (mol.weight=58) at K. The vapour pressure of the solution was found to be 192.5 mm of Hg. The molecular weight of the solute is (vapour pressure of acetone = 195 mm of Hg)a)25.24b)35.24c)45.24d)55.24Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of In an experiment, 1 g of a non-volatile solute was dissolved in 100 g of acetone (mol.weight=58) at K. The vapour pressure of the solution was found to be 192.5 mm of Hg. The molecular weight of the solute is (vapour pressure of acetone = 195 mm of Hg)a)25.24b)35.24c)45.24d)55.24Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice In an experiment, 1 g of a non-volatile solute was dissolved in 100 g of acetone (mol.weight=58) at K. The vapour pressure of the solution was found to be 192.5 mm of Hg. The molecular weight of the solute is (vapour pressure of acetone = 195 mm of Hg)a)25.24b)35.24c)45.24d)55.24Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice NEET tests.