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Two liquids A and B from ideal solutions. At 300 K, the vapour pressure of solution containing 1 mole of A and 3 mole of B is 550 mm Hg. At the same temperature, if one more mole of B is added this solution, the vapour pressure of the solution increases by 10 mm Hg. Determine the vapour pressure of A and B in their pure states (in mm Hg):
- a)400, 600
- b)500, 500
- c)600, 400
- d)None of these.
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Two liquids A and B from ideal solutions. At 300 K, the vapour pressur...
Since P=xAP0A+xBP0B
⇒(11+3)P0A+(31+3)P∘B=550mm Hg
⇒(11+4)P0A+(41+4)P∘B=560mm Hg
That is,
0.25P0A+0.75P0B=550mm Hg
0.20P0A+0.8P0B=560mm Hg
Solving for P0A and P0B we get,
P0A=400 mm Hg
P0B=600 mm Hg
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Two liquids A and B from ideal solutions. At 300 K, the vapour pressur...
To solve this problem, we can use Raoult's law, which states that the vapor pressure of a component in an ideal solution is directly proportional to its mole fraction in the solution.
Let's assume that the vapor pressures of pure A and B are P°A and P°B, respectively.
According to the problem, at 300 K, the vapor pressure of a solution containing 1 mole of A and 3 moles of B is 550 mm Hg.
We can write the vapor pressure of the solution as follows:
Ptotal = PA + PB
Given that Ptotal = 550 mm Hg and the mole fraction of A in the solution (XA) is 1/4, we can write:
550 = (1/4)P°A + (3/4)P°B
Now, let's consider what happens when one more mole of B is added to the solution.
The mole fraction of A remains the same (1/4), but the mole fraction of B changes to (4/5) since there are now 4 moles of B in a total of 5 moles of solution.
The new vapor pressure of the solution is increased by 10 mm Hg, so we can write:
(Ptotal + 10) = (1/4)P°A + (4/5)P°B
Now we have a system of two equations with two variables:
550 = (1/4)P°A + (3/4)P°B --(1)
(Ptotal + 10) = (1/4)P°A + (4/5)P°B --(2)
We can solve these equations simultaneously to find the values of P°A and P°B.
To solve for P°A, we can multiply equation (1) by 5:
2750 = 5(1/4)P°A + 5(3/4)P°B
2750 = (5/4)P°A + (15/4)P°B
To solve for P°B, we can multiply equation (2) by 4:
4(Ptotal + 10) = 4(1/4)P°A + 4(4/5)P°B
4Ptotal + 40 = P°A + (16/5)P°B
Now we have the following system of equations:
2750 = (5/4)P°A + (15/4)P°B --(3)
4Ptotal + 40 = P°A + (16/5)P°B --(4)
By solving equations (3) and (4), we find that P°A = 400 mm Hg and P°B = 600 mm Hg.
Therefore, the vapor pressure of A and B in their pure states is 400 mm Hg and 600 mm Hg, respectively.
Hence, the correct answer is option A) 400, 600.
Let's assume that the vapor pressures of pure A and B are P°A and P°B, respectively.
According to the problem, at 300 K, the vapor pressure of a solution containing 1 mole of A and 3 moles of B is 550 mm Hg.
We can write the vapor pressure of the solution as follows:
Ptotal = PA + PB
Given that Ptotal = 550 mm Hg and the mole fraction of A in the solution (XA) is 1/4, we can write:
550 = (1/4)P°A + (3/4)P°B
Now, let's consider what happens when one more mole of B is added to the solution.
The mole fraction of A remains the same (1/4), but the mole fraction of B changes to (4/5) since there are now 4 moles of B in a total of 5 moles of solution.
The new vapor pressure of the solution is increased by 10 mm Hg, so we can write:
(Ptotal + 10) = (1/4)P°A + (4/5)P°B
Now we have a system of two equations with two variables:
550 = (1/4)P°A + (3/4)P°B --(1)
(Ptotal + 10) = (1/4)P°A + (4/5)P°B --(2)
We can solve these equations simultaneously to find the values of P°A and P°B.
To solve for P°A, we can multiply equation (1) by 5:
2750 = 5(1/4)P°A + 5(3/4)P°B
2750 = (5/4)P°A + (15/4)P°B
To solve for P°B, we can multiply equation (2) by 4:
4(Ptotal + 10) = 4(1/4)P°A + 4(4/5)P°B
4Ptotal + 40 = P°A + (16/5)P°B
Now we have the following system of equations:
2750 = (5/4)P°A + (15/4)P°B --(3)
4Ptotal + 40 = P°A + (16/5)P°B --(4)
By solving equations (3) and (4), we find that P°A = 400 mm Hg and P°B = 600 mm Hg.
Therefore, the vapor pressure of A and B in their pure states is 400 mm Hg and 600 mm Hg, respectively.
Hence, the correct answer is option A) 400, 600.
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Two liquids A and B from ideal solutions. At 300 K, the vapour pressure of solution containing 1 mole of A and 3 mole of B is 550 mm Hg. At the same temperature, if one more mole of B is added this solution, the vapour pressure of the solution increases by 10 mm Hg. Determine the vapour pressure of A and B in their pure states (in mm Hg):a)400, 600b)500, 500c)600, 400d)None of these.Correct answer is option 'A'. Can you explain this answer?
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Two liquids A and B from ideal solutions. At 300 K, the vapour pressure of solution containing 1 mole of A and 3 mole of B is 550 mm Hg. At the same temperature, if one more mole of B is added this solution, the vapour pressure of the solution increases by 10 mm Hg. Determine the vapour pressure of A and B in their pure states (in mm Hg):a)400, 600b)500, 500c)600, 400d)None of these.Correct answer is option 'A'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Two liquids A and B from ideal solutions. At 300 K, the vapour pressure of solution containing 1 mole of A and 3 mole of B is 550 mm Hg. At the same temperature, if one more mole of B is added this solution, the vapour pressure of the solution increases by 10 mm Hg. Determine the vapour pressure of A and B in their pure states (in mm Hg):a)400, 600b)500, 500c)600, 400d)None of these.Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two liquids A and B from ideal solutions. At 300 K, the vapour pressure of solution containing 1 mole of A and 3 mole of B is 550 mm Hg. At the same temperature, if one more mole of B is added this solution, the vapour pressure of the solution increases by 10 mm Hg. Determine the vapour pressure of A and B in their pure states (in mm Hg):a)400, 600b)500, 500c)600, 400d)None of these.Correct answer is option 'A'. Can you explain this answer?.
Two liquids A and B from ideal solutions. At 300 K, the vapour pressure of solution containing 1 mole of A and 3 mole of B is 550 mm Hg. At the same temperature, if one more mole of B is added this solution, the vapour pressure of the solution increases by 10 mm Hg. Determine the vapour pressure of A and B in their pure states (in mm Hg):a)400, 600b)500, 500c)600, 400d)None of these.Correct answer is option 'A'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about Two liquids A and B from ideal solutions. At 300 K, the vapour pressure of solution containing 1 mole of A and 3 mole of B is 550 mm Hg. At the same temperature, if one more mole of B is added this solution, the vapour pressure of the solution increases by 10 mm Hg. Determine the vapour pressure of A and B in their pure states (in mm Hg):a)400, 600b)500, 500c)600, 400d)None of these.Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two liquids A and B from ideal solutions. At 300 K, the vapour pressure of solution containing 1 mole of A and 3 mole of B is 550 mm Hg. At the same temperature, if one more mole of B is added this solution, the vapour pressure of the solution increases by 10 mm Hg. Determine the vapour pressure of A and B in their pure states (in mm Hg):a)400, 600b)500, 500c)600, 400d)None of these.Correct answer is option 'A'. Can you explain this answer?.
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