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A liquid which is immiscible in water was steam distilled at 95.2°C at a total pressure of 0.983 atmosphere. What is the mass of the liquid present per gram of water in the distillate?
Molar mass of the liquid is 134.3 gm/mol and vapour pressure of water is 0.84 atm.
(Round off up to 2 decimal places)
Molar mass of the liquid is 134.3 gm/mol and vapour pressure of water is 0.84 atm.
(Round off up to 2 decimal places)
Correct answer is '1.27'. Can you explain this answer?
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Most Upvoted Answer
A liquid which is immiscible in water was steam distilled at 95.2°...
PT = 0.983 atm
Pwater = 0.84 atm
Pliquid = 0.143 atm
Therefore, mass of the liquid present per gram of water in the distillate is 1.27 g.
Pwater = 0.84 atm
Pliquid = 0.143 atm
Therefore, mass of the liquid present per gram of water in the distillate is 1.27 g.
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A liquid which is immiscible in water was steam distilled at 95.2°...
The steam distillation process involves heating a mixture of two immiscible liquids to vaporize the more volatile component, which is then condensed and collected separately. In this case, the liquid that is immiscible in water was heated and vaporized at a temperature of 95.2 degrees Celsius.
During steam distillation, water is typically used as the solvent and is heated to its boiling point to produce steam. The steam carries the volatile component of the immiscible liquid along with it, while leaving behind the less volatile component. The steam is then condensed back into liquid form, resulting in the separation of the two components.
It is important to note that the boiling point of the immiscible liquid is crucial in determining the appropriate temperature for steam distillation. In this case, the immiscible liquid vaporized at a temperature of 95.2 degrees Celsius, indicating that it has a lower boiling point than water.
By using steam distillation, the immiscible liquid can be effectively separated from water, allowing for the isolation and collection of the volatile component.
During steam distillation, water is typically used as the solvent and is heated to its boiling point to produce steam. The steam carries the volatile component of the immiscible liquid along with it, while leaving behind the less volatile component. The steam is then condensed back into liquid form, resulting in the separation of the two components.
It is important to note that the boiling point of the immiscible liquid is crucial in determining the appropriate temperature for steam distillation. In this case, the immiscible liquid vaporized at a temperature of 95.2 degrees Celsius, indicating that it has a lower boiling point than water.
By using steam distillation, the immiscible liquid can be effectively separated from water, allowing for the isolation and collection of the volatile component.
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A liquid which is immiscible in water was steam distilled at 95.2°C at a total pressure of 0.983 atmosphere. What is the mass of the liquid present per gram of water in the distillate?Molar mass of the liquid is 134.3 gm/mol and vapour pressure of water is 0.84 atm.(Round off up to 2 decimal places)Correct answer is '1.27'. Can you explain this answer?
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A liquid which is immiscible in water was steam distilled at 95.2°C at a total pressure of 0.983 atmosphere. What is the mass of the liquid present per gram of water in the distillate?Molar mass of the liquid is 134.3 gm/mol and vapour pressure of water is 0.84 atm.(Round off up to 2 decimal places)Correct answer is '1.27'. 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 A liquid which is immiscible in water was steam distilled at 95.2°C at a total pressure of 0.983 atmosphere. What is the mass of the liquid present per gram of water in the distillate?Molar mass of the liquid is 134.3 gm/mol and vapour pressure of water is 0.84 atm.(Round off up to 2 decimal places)Correct answer is '1.27'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A liquid which is immiscible in water was steam distilled at 95.2°C at a total pressure of 0.983 atmosphere. What is the mass of the liquid present per gram of water in the distillate?Molar mass of the liquid is 134.3 gm/mol and vapour pressure of water is 0.84 atm.(Round off up to 2 decimal places)Correct answer is '1.27'. Can you explain this answer?.
A liquid which is immiscible in water was steam distilled at 95.2°C at a total pressure of 0.983 atmosphere. What is the mass of the liquid present per gram of water in the distillate?Molar mass of the liquid is 134.3 gm/mol and vapour pressure of water is 0.84 atm.(Round off up to 2 decimal places)Correct answer is '1.27'. 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 A liquid which is immiscible in water was steam distilled at 95.2°C at a total pressure of 0.983 atmosphere. What is the mass of the liquid present per gram of water in the distillate?Molar mass of the liquid is 134.3 gm/mol and vapour pressure of water is 0.84 atm.(Round off up to 2 decimal places)Correct answer is '1.27'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A liquid which is immiscible in water was steam distilled at 95.2°C at a total pressure of 0.983 atmosphere. What is the mass of the liquid present per gram of water in the distillate?Molar mass of the liquid is 134.3 gm/mol and vapour pressure of water is 0.84 atm.(Round off up to 2 decimal places)Correct answer is '1.27'. Can you explain this answer?.
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