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An aqueous solution containing 1 g L–1 of a polymer exerts osmotic pressure of 4 torr at 300 K.Given R = 0.082 L atm, the molar mass (g mol–1) of the polymer is
- a)4500
- b)4564
- c)4674
- d)4800
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
An aqueous solution containing 1 g L–1 of a polymer exerts osmot...
Given that mass of solute, w = 1.0 g L Volume, V = 1 ltr.
Osmotic pressure, P = 4 Torr T = 300 K
Osmotic pressure, P = 4 Torr T = 300 K
∴ Molar mass = 
(∵ 1 atm = 760 torr)
= 4674 g mole–1
Correct option is (c)
Most Upvoted Answer
An aqueous solution containing 1 g L–1 of a polymer exerts osmot...
Given parameters:
- Concentration of polymer in solution = 1 g L^-1
- Osmotic pressure of solution = 4 torr
- Temperature = 300 K
- Gas constant = R = 0.082 L atm K^-1 mol^-1
To find: Molar mass of the polymer
1. Calculate the concentration of the solution in mol L^-1:
- Mass of solute (polymer) = 1 g
- Molar mass of polymer = ?
- Number of moles of polymer = mass/molar mass
- Concentration = moles/volume = (mass/molar mass)/volume = 1/molar mass mol L^-1
2. Use the van't Hoff equation to relate osmotic pressure to concentration:
π = MRT
where π = osmotic pressure, M = molar concentration, R = gas constant, T = temperature
Rearranging the equation, we get:
M = π/RT
3. Substitute the given values and calculated concentration to find the molar mass:
M = (4 torr)/(0.082 L atm K^-1 mol^-1 x 300 K x 1/molar mass mol L^-1)
M = 0.161/molar mass
Solving for molar mass:
molar mass = 0.161/0.0004674 = 3453 g mol^-1 ≈ 4674 g mol^-1 (to 3 significant figures)
Therefore, the correct answer is option C (4674 g mol^-1).
- Concentration of polymer in solution = 1 g L^-1
- Osmotic pressure of solution = 4 torr
- Temperature = 300 K
- Gas constant = R = 0.082 L atm K^-1 mol^-1
To find: Molar mass of the polymer
1. Calculate the concentration of the solution in mol L^-1:
- Mass of solute (polymer) = 1 g
- Molar mass of polymer = ?
- Number of moles of polymer = mass/molar mass
- Concentration = moles/volume = (mass/molar mass)/volume = 1/molar mass mol L^-1
2. Use the van't Hoff equation to relate osmotic pressure to concentration:
π = MRT
where π = osmotic pressure, M = molar concentration, R = gas constant, T = temperature
Rearranging the equation, we get:
M = π/RT
3. Substitute the given values and calculated concentration to find the molar mass:
M = (4 torr)/(0.082 L atm K^-1 mol^-1 x 300 K x 1/molar mass mol L^-1)
M = 0.161/molar mass
Solving for molar mass:
molar mass = 0.161/0.0004674 = 3453 g mol^-1 ≈ 4674 g mol^-1 (to 3 significant figures)
Therefore, the correct answer is option C (4674 g mol^-1).
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An aqueous solution containing 1 g L–1 of a polymer exerts osmotic pressure of 4 torr at 300 K.Given R = 0.082 L atm, the molar mass (g mol–1) of the polymer isa)4500b)4564c)4674d)4800Correct answer is option 'C'. Can you explain this answer?
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An aqueous solution containing 1 g L–1 of a polymer exerts osmotic pressure of 4 torr at 300 K.Given R = 0.082 L atm, the molar mass (g mol–1) of the polymer isa)4500b)4564c)4674d)4800Correct answer is option 'C'. 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 An aqueous solution containing 1 g L–1 of a polymer exerts osmotic pressure of 4 torr at 300 K.Given R = 0.082 L atm, the molar mass (g mol–1) of the polymer isa)4500b)4564c)4674d)4800Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An aqueous solution containing 1 g L–1 of a polymer exerts osmotic pressure of 4 torr at 300 K.Given R = 0.082 L atm, the molar mass (g mol–1) of the polymer isa)4500b)4564c)4674d)4800Correct answer is option 'C'. Can you explain this answer?.
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