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At room temperature, the mole fraction of a solute is 0.25 and the vapour pressure of pure solvent is 0.80 atm. The vapour pressure (in atm) is lowered by
(Round off up to 1 decimal place)
Correct answer is '0.2'. Can you explain this answer?
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At room temperature, the mole fraction of a solute is 0.25 and the vap...
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At room temperature, the mole fraction of a solute is 0.25 and the vap...
Understanding the Problem
To determine the lowering of vapor pressure due to a solute, we apply Raoult’s Law. This law states that the vapor pressure of a solvent in a solution is directly proportional to the mole fraction of the solvent.
Given Data
- Mole fraction of solute (Xsolute) = 0.25
- Mole fraction of solvent (Xsolvent) = 1 - Xsolute = 1 - 0.25 = 0.75
- Vapor pressure of pure solvent (P°solvent) = 0.80 atm
Applying Raoult’s Law
According to Raoult’s Law:
Psolution = Xsolvent * P°solvent
Where:
- Psolution = vapor pressure of the solution
- Xsolvent = mole fraction of the solvent
- P°solvent = vapor pressure of pure solvent
Calculating Vapor Pressure of the Solution
- Psolution = 0.75 * 0.80 atm
Calculating this gives:
Psolution = 0.60 atm
Calculating the Change in Vapor Pressure
To find the lowering of vapor pressure:
Lowering = P°solvent - Psolution
Substituting the values:
Lowering = 0.80 atm - 0.60 atm = 0.20 atm
Final Answer
The vapor pressure is lowered by 0.2 atm when rounding to one decimal place. This demonstrates the effect of solute on the vapor pressure of a solvent, confirming the application of Raoult’s Law effectively.
To determine the lowering of vapor pressure due to a solute, we apply Raoult’s Law. This law states that the vapor pressure of a solvent in a solution is directly proportional to the mole fraction of the solvent.
Given Data
- Mole fraction of solute (Xsolute) = 0.25
- Mole fraction of solvent (Xsolvent) = 1 - Xsolute = 1 - 0.25 = 0.75
- Vapor pressure of pure solvent (P°solvent) = 0.80 atm
Applying Raoult’s Law
According to Raoult’s Law:
Psolution = Xsolvent * P°solvent
Where:
- Psolution = vapor pressure of the solution
- Xsolvent = mole fraction of the solvent
- P°solvent = vapor pressure of pure solvent
Calculating Vapor Pressure of the Solution
- Psolution = 0.75 * 0.80 atm
Calculating this gives:
Psolution = 0.60 atm
Calculating the Change in Vapor Pressure
To find the lowering of vapor pressure:
Lowering = P°solvent - Psolution
Substituting the values:
Lowering = 0.80 atm - 0.60 atm = 0.20 atm
Final Answer
The vapor pressure is lowered by 0.2 atm when rounding to one decimal place. This demonstrates the effect of solute on the vapor pressure of a solvent, confirming the application of Raoult’s Law effectively.
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At room temperature, the mole fraction of a solute is 0.25 and the vapour pressure of pure solvent is 0.80 atm. The vapour pressure (in atm) is lowered by(Round off up to 1 decimal place)Correct answer is '0.2'. Can you explain this answer?
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At room temperature, the mole fraction of a solute is 0.25 and the vapour pressure of pure solvent is 0.80 atm. The vapour pressure (in atm) is lowered by(Round off up to 1 decimal place)Correct answer is '0.2'. 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 At room temperature, the mole fraction of a solute is 0.25 and the vapour pressure of pure solvent is 0.80 atm. The vapour pressure (in atm) is lowered by(Round off up to 1 decimal place)Correct answer is '0.2'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At room temperature, the mole fraction of a solute is 0.25 and the vapour pressure of pure solvent is 0.80 atm. The vapour pressure (in atm) is lowered by(Round off up to 1 decimal place)Correct answer is '0.2'. Can you explain this answer?.
At room temperature, the mole fraction of a solute is 0.25 and the vapour pressure of pure solvent is 0.80 atm. The vapour pressure (in atm) is lowered by(Round off up to 1 decimal place)Correct answer is '0.2'. 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 At room temperature, the mole fraction of a solute is 0.25 and the vapour pressure of pure solvent is 0.80 atm. The vapour pressure (in atm) is lowered by(Round off up to 1 decimal place)Correct answer is '0.2'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At room temperature, the mole fraction of a solute is 0.25 and the vapour pressure of pure solvent is 0.80 atm. The vapour pressure (in atm) is lowered by(Round off up to 1 decimal place)Correct answer is '0.2'. Can you explain this answer?.
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