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1 mole of volatile liquid A is mixed with 3 mole of volatile liquid B to form an ideal solution. If half of the liquid is vapourised by heating then calculate the new mole fraction of A in liquid solution at equilibrium (pA=400mm, pB=200 mm, (33)^(1/2)=5.74)?
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1 mole of volatile liquid A is mixed with 3 mole of volatile liquid B ...
Solution:
Given:
Number of moles of liquid A = 1
Number of moles of liquid B = 3
To find:
The new mole fraction of A in the liquid solution at equilibrium.
Since the solution is ideal, the vapor pressure of each component in the solution can be calculated using Raoult's law. According to Raoult's law, the vapor pressure of a component in an ideal solution is directly proportional to its mole fraction.
Let's calculate the mole fraction of A and B in the solution before vaporization.
Mole fraction of A (xA) = Moles of A / Total moles of solution
= 1 / (1 + 3) = 1/4
Mole fraction of B (xB) = Moles of B / Total moles of solution
= 3 / (1 + 3) = 3/4
Now, let's calculate the vapor pressure of A and B using Raoult's law.
Vapor pressure of A (pA) = xA * p°A
= (1/4) * 400 mm = 100 mm
Vapor pressure of B (pB) = xB * p°B
= (3/4) * 200 mm = 150 mm
At equilibrium, half of the liquid is vaporized. This means that the mole fraction of A in the vapor phase will be equal to the mole fraction of A in the liquid phase.
Let the mole fraction of A in the vapor phase be yA.
Now, let's calculate the new mole fraction of A in the liquid solution at equilibrium.
Mole fraction of A in the liquid phase = (Initial moles of A - Moles of A vaporized) / Total moles of solution
= (1 - 0.5) / (1 + 3) = 0.125
Mole fraction of B in the liquid phase = (Initial moles of B - Moles of B vaporized) / Total moles of solution
= (3 - 0.5) / (1 + 3) = 0.875
Therefore, the new mole fraction of A in the liquid solution at equilibrium is 0.125.
Given:
Number of moles of liquid A = 1
Number of moles of liquid B = 3
To find:
The new mole fraction of A in the liquid solution at equilibrium.
Since the solution is ideal, the vapor pressure of each component in the solution can be calculated using Raoult's law. According to Raoult's law, the vapor pressure of a component in an ideal solution is directly proportional to its mole fraction.
Let's calculate the mole fraction of A and B in the solution before vaporization.
Mole fraction of A (xA) = Moles of A / Total moles of solution
= 1 / (1 + 3) = 1/4
Mole fraction of B (xB) = Moles of B / Total moles of solution
= 3 / (1 + 3) = 3/4
Now, let's calculate the vapor pressure of A and B using Raoult's law.
Vapor pressure of A (pA) = xA * p°A
= (1/4) * 400 mm = 100 mm
Vapor pressure of B (pB) = xB * p°B
= (3/4) * 200 mm = 150 mm
At equilibrium, half of the liquid is vaporized. This means that the mole fraction of A in the vapor phase will be equal to the mole fraction of A in the liquid phase.
Let the mole fraction of A in the vapor phase be yA.
Now, let's calculate the new mole fraction of A in the liquid solution at equilibrium.
Mole fraction of A in the liquid phase = (Initial moles of A - Moles of A vaporized) / Total moles of solution
= (1 - 0.5) / (1 + 3) = 0.125
Mole fraction of B in the liquid phase = (Initial moles of B - Moles of B vaporized) / Total moles of solution
= (3 - 0.5) / (1 + 3) = 0.875
Therefore, the new mole fraction of A in the liquid solution at equilibrium is 0.125.
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1 mole of volatile liquid A is mixed with 3 mole of volatile liquid B to form an ideal solution. If half of the liquid is vapourised by heating then calculate the new mole fraction of A in liquid solution at equilibrium (pA=400mm, pB=200 mm, (33)^(1/2)=5.74)?
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1 mole of volatile liquid A is mixed with 3 mole of volatile liquid B to form an ideal solution. If half of the liquid is vapourised by heating then calculate the new mole fraction of A in liquid solution at equilibrium (pA=400mm, pB=200 mm, (33)^(1/2)=5.74)? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 1 mole of volatile liquid A is mixed with 3 mole of volatile liquid B to form an ideal solution. If half of the liquid is vapourised by heating then calculate the new mole fraction of A in liquid solution at equilibrium (pA=400mm, pB=200 mm, (33)^(1/2)=5.74)? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1 mole of volatile liquid A is mixed with 3 mole of volatile liquid B to form an ideal solution. If half of the liquid is vapourised by heating then calculate the new mole fraction of A in liquid solution at equilibrium (pA=400mm, pB=200 mm, (33)^(1/2)=5.74)?.
1 mole of volatile liquid A is mixed with 3 mole of volatile liquid B to form an ideal solution. If half of the liquid is vapourised by heating then calculate the new mole fraction of A in liquid solution at equilibrium (pA=400mm, pB=200 mm, (33)^(1/2)=5.74)? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 1 mole of volatile liquid A is mixed with 3 mole of volatile liquid B to form an ideal solution. If half of the liquid is vapourised by heating then calculate the new mole fraction of A in liquid solution at equilibrium (pA=400mm, pB=200 mm, (33)^(1/2)=5.74)? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1 mole of volatile liquid A is mixed with 3 mole of volatile liquid B to form an ideal solution. If half of the liquid is vapourised by heating then calculate the new mole fraction of A in liquid solution at equilibrium (pA=400mm, pB=200 mm, (33)^(1/2)=5.74)?.
Solutions for 1 mole of volatile liquid A is mixed with 3 mole of volatile liquid B to form an ideal solution. If half of the liquid is vapourised by heating then calculate the new mole fraction of A in liquid solution at equilibrium (pA=400mm, pB=200 mm, (33)^(1/2)=5.74)? in English & in Hindi are available as part of our courses for JEE.
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