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Osmotic pressure of 40% (wt./vol.) urea solution is 1.64 atm and that of 3.42% (wt./vol.) cane sugar is 2.46 atm. When equal volumes of the above two solutions are mixed, the osmotic pressure of the resulting solution is -
- a)1.64 atm
- b)2.46 atm
- c)4.10 atm
- d)2.05 atm
Correct answer is option 'D'. Can you explain this answer?
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Solution:
Given,
Osmotic pressure of 40% (wt./vol.) urea solution = 1.64 atm
Osmotic pressure of 3.42% (wt./vol.) cane sugar solution = 2.46 atm
When equal volumes of the above two solutions are mixed, the osmotic pressure of the resulting solution is to be calculated.
Let us assume the volume of each solution to be V.
Then, the number of moles of urea present in V volume of 40% (wt./vol.) urea solution = 40/60.02 × V × 1000/60 = 0.666 V
Similarly, the number of moles of cane sugar present in V volume of 3.42% (wt./vol.) cane sugar solution = 3.42/342.3 × V × 1000/180 = 0.019 V
On mixing, the total volume of the solution = 2V
The total number of moles of solute in the mixture = 0.666 V + 0.019 V = 0.685 V
The concentration of the resulting solution = (0.685V/2V) × 100 = 34.25%
Using the formula, π = CRT, where π is the osmotic pressure, C is the concentration, R is the gas constant and T is the temperature, we can find the osmotic pressure of the resulting solution.
Let us assume the temperature to be constant and equal to room temperature, which is approximately 25°C or 298 K.
Therefore,
For the urea solution, π1 = 1.64 atm
For the cane sugar solution, π2 = 2.46 atm
For the resulting solution, C = 34.25%
R = 0.0821 L atm mol^-1 K^-1 (gas constant)
T = 298 K
π3 = CRT = (34.25/100) × 0.0821 × 298 = 2.05 atm
Therefore, the osmotic pressure of the resulting solution is 2.05 atm, which is option D.
Given,
Osmotic pressure of 40% (wt./vol.) urea solution = 1.64 atm
Osmotic pressure of 3.42% (wt./vol.) cane sugar solution = 2.46 atm
When equal volumes of the above two solutions are mixed, the osmotic pressure of the resulting solution is to be calculated.
Let us assume the volume of each solution to be V.
Then, the number of moles of urea present in V volume of 40% (wt./vol.) urea solution = 40/60.02 × V × 1000/60 = 0.666 V
Similarly, the number of moles of cane sugar present in V volume of 3.42% (wt./vol.) cane sugar solution = 3.42/342.3 × V × 1000/180 = 0.019 V
On mixing, the total volume of the solution = 2V
The total number of moles of solute in the mixture = 0.666 V + 0.019 V = 0.685 V
The concentration of the resulting solution = (0.685V/2V) × 100 = 34.25%
Using the formula, π = CRT, where π is the osmotic pressure, C is the concentration, R is the gas constant and T is the temperature, we can find the osmotic pressure of the resulting solution.
Let us assume the temperature to be constant and equal to room temperature, which is approximately 25°C or 298 K.
Therefore,
For the urea solution, π1 = 1.64 atm
For the cane sugar solution, π2 = 2.46 atm
For the resulting solution, C = 34.25%
R = 0.0821 L atm mol^-1 K^-1 (gas constant)
T = 298 K
π3 = CRT = (34.25/100) × 0.0821 × 298 = 2.05 atm
Therefore, the osmotic pressure of the resulting solution is 2.05 atm, which is option D.
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Osmotic pressure of 40% (wt./vol.) urea solution is 1.64 atm and that of 3.42% (wt./vol.) cane sugar is 2.46 atm. When equal volumes of the above two solutions are mixed, the osmotic pressure of the resulting solution is -a)1.64 atmb)2.46 atmc)4.10 atmd)2.05 atmCorrect answer is option 'D'. Can you explain this answer?
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Osmotic pressure of 40% (wt./vol.) urea solution is 1.64 atm and that of 3.42% (wt./vol.) cane sugar is 2.46 atm. When equal volumes of the above two solutions are mixed, the osmotic pressure of the resulting solution is -a)1.64 atmb)2.46 atmc)4.10 atmd)2.05 atmCorrect answer is option 'D'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Osmotic pressure of 40% (wt./vol.) urea solution is 1.64 atm and that of 3.42% (wt./vol.) cane sugar is 2.46 atm. When equal volumes of the above two solutions are mixed, the osmotic pressure of the resulting solution is -a)1.64 atmb)2.46 atmc)4.10 atmd)2.05 atmCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Osmotic pressure of 40% (wt./vol.) urea solution is 1.64 atm and that of 3.42% (wt./vol.) cane sugar is 2.46 atm. When equal volumes of the above two solutions are mixed, the osmotic pressure of the resulting solution is -a)1.64 atmb)2.46 atmc)4.10 atmd)2.05 atmCorrect answer is option 'D'. Can you explain this answer?.
Osmotic pressure of 40% (wt./vol.) urea solution is 1.64 atm and that of 3.42% (wt./vol.) cane sugar is 2.46 atm. When equal volumes of the above two solutions are mixed, the osmotic pressure of the resulting solution is -a)1.64 atmb)2.46 atmc)4.10 atmd)2.05 atmCorrect answer is option 'D'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Osmotic pressure of 40% (wt./vol.) urea solution is 1.64 atm and that of 3.42% (wt./vol.) cane sugar is 2.46 atm. When equal volumes of the above two solutions are mixed, the osmotic pressure of the resulting solution is -a)1.64 atmb)2.46 atmc)4.10 atmd)2.05 atmCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Osmotic pressure of 40% (wt./vol.) urea solution is 1.64 atm and that of 3.42% (wt./vol.) cane sugar is 2.46 atm. When equal volumes of the above two solutions are mixed, the osmotic pressure of the resulting solution is -a)1.64 atmb)2.46 atmc)4.10 atmd)2.05 atmCorrect answer is option 'D'. Can you explain this answer?.
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