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1 g of a protein polymer having molar mass 16×10^4 is present in 800ml solution at 300k then osmotic pressure in Pascal ?
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1 g of a protein polymer having molar mass 16×10^4 is present in 800ml...
Calculation of Osmotic Pressure of Protein Polymer Solution
Given Data:
- Mass of Polymer: 1 g
- Molar Mass of Polymer: 16×10^4 g/mol
- Volume of Solution: 800 mL = 0.8 L
- Temperature: 300 K
Formula Used:
Osmotic Pressure (π) = (n/V)RT
Where,
- n: Number of moles of solute
- V: Volume of solution
- R: Universal gas constant = 8.314 J/mol-K
- T: Temperature in Kelvin
Solution:
- Number of moles of solute:
n = Mass of solute/Molar mass of solute
= 1 g/16×10^4 g/mol
= 6.25×10^-5 mol - Osmotic Pressure:
π = (n/V)RT
= (6.25×10^-5 mol/0.8 L)×8.314 J/mol-K×300 K
= 1.55×10^3 Pa (Pascal)
Conclusion:
The osmotic pressure of the protein polymer solution at 300 K is 1.55×10^3 Pa. Osmotic pressure is a colligative property of a solution that depends on the number of solute particles in the solution but not on their size or nature. It is an important parameter in many biological and chemical processes, including osmosis, dialysis, and reverse osmosis.
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1 g of a protein polymer having molar mass 16×10^4 is present in 800ml...
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1 g of a protein polymer having molar mass 16×10^4 is present in 800ml solution at 300k then osmotic pressure in Pascal ?
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1 g of a protein polymer having molar mass 16×10^4 is present in 800ml solution at 300k then osmotic pressure in Pascal ? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 1 g of a protein polymer having molar mass 16×10^4 is present in 800ml solution at 300k then osmotic pressure in Pascal ? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1 g of a protein polymer having molar mass 16×10^4 is present in 800ml solution at 300k then osmotic pressure in Pascal ?.
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