Practice important
Membrane Osmometry
MCQ questions and answers.
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MCQ Questions
1. A dilute polymer solution at 300 K develops an osmotic pressure of 0.246 kPa across a semipermeable membrane. If its concentration is 2.00 g L⁻¹ and the solution behaves ideally, what is the number-average molar mass of the polymer? Use R = 8.314 kPa L mol⁻¹ K⁻¹.
A.
2.03 × 10³ g mol⁻¹
B.
2.03 × 10⁴ g mol⁻¹
C.
6.08 × 10⁴ g mol⁻¹
D.
1.01 × 10⁵ g mol⁻¹
Correct Answer:
2.03 × 10⁴ g mol⁻¹
Explanation:
For membrane osmometry under ideal dilute conditions, π = cRT/Mn. Thus, Mn = cRT/π = (2.00 × 8.314 × 300)/0.246 = 2.03 × 10⁴ g mol⁻¹. Osmotic pressure gives the number-average molar mass.
2. For a polymer solution, the measured values of π/c are 0.620 and 0.700 kPa L g⁻¹ at concentrations 0.020 and 0.060 g L⁻¹, respectively. Assuming π/c = RT/Mn + Bc and T = 298 K, what is the number-average molar mass?
A.
2.13 × 10³ g mol⁻¹
B.
3.57 × 10³ g mol⁻¹
C.
4.27 × 10³ g mol⁻¹
D.
7.11 × 10³ g mol⁻¹
Correct Answer:
4.27 × 10³ g mol⁻¹
Explanation:
The intercept of the π/c versus c plot is RT/Mn. From the two data points, the intercept is 0.620 − [(0.700 − 0.620)/(0.060 − 0.020)](0.020) = 0.580 kPa L g⁻¹. Therefore, Mn = RT/intercept = (8.314 × 298)/0.580 = 4.27 × 10³ g mol⁻¹.
3. Why is a polymer solution commonly dialysed before determining its molar mass by membrane osmometry?
A.
To remove polymer molecules that would otherwise pass through the membrane
B.
To remove low-molar-mass solutes whose osmotic contribution would be measured along with that of the polymer
C.
To increase the viscosity and stabilize the polymer colloid
D.
To convert the polymer into a monodisperse sample
Correct Answer:
To remove low-molar-mass solutes whose osmotic contribution would be measured along with that of the polymer
Explanation:
The membrane retains macromolecules but generally allows solvent and small ions or molecules to pass. Dialysis removes these diffusible solutes, preventing their osmotic pressure from being incorrectly attributed to the polymer.
4. A membrane osmometer is used on two polymer samples at identical temperature and mass concentration. Sample X produces a larger osmotic pressure than sample Y. Which conclusion is most appropriate, assuming ideal dilute behavior and identical chemical nature?
A.
Sample X has the larger number-average molar mass
B.
Sample X has the smaller number-average molar mass
C.
Both samples necessarily have the same number-average molar mass
D.
Osmotic pressure cannot provide any comparison of their molar masses
Correct Answer:
Sample X has the smaller number-average molar mass
Explanation:
For a fixed mass concentration, π = cRT/Mn. Therefore, a larger osmotic pressure corresponds to a smaller number-average molar mass because the solution contains more polymer molecules per unit mass.
5. In membrane osmometry, the quantity π/c is plotted against polymer concentration c rather than using a single measured value of π/c. What is obtained from extrapolation to c = 0?
A.
The weight-average molar mass directly
B.
The z-average molar mass directly
C.
RT/Mn, from which the number-average molar mass is calculated
D.
The polymer density independent of temperature
Correct Answer:
RT/Mn, from which the number-average molar mass is calculated
Explanation:
For dilute polymer solutions, π/c = RT/Mn + Bc + higher-order terms. Extrapolating π/c to zero concentration removes concentration-dependent nonideal effects, leaving RT/Mn.
6. A polymer solution is measured by membrane osmometry at constant concentration in the dilute ideal limit. If the absolute temperature is increased from 300 K to 330 K, what happens to the osmotic pressure?
A.
It decreases to 300/330 of its original value
B.
It remains unchanged because the polymer cannot cross the membrane
C.
It increases by a factor of 1.10
D.
It increases by a factor of 1.21
Correct Answer:
It increases by a factor of 1.10
Explanation:
For an ideal polymer solution, π = cRT/Mn. At constant c and Mn, osmotic pressure is directly proportional to absolute temperature. Therefore, π₂/π₁ = 330/300 = 1.10.
7. A polymer solution contains macromolecules that reversibly associate in pairs in solution. If the molar mass of an isolated polymer chain is M, what molar mass would be inferred from ideal membrane osmometry when all chains are present as dimers?
A.
M/2
B.
M
C.
2M
D.
4M
Correct Answer:
2M
Explanation:
Dimerization halves the number of independently osmotically active particles at a given mass concentration. Since π is halved, the apparent molar mass obtained from Mn = cRT/π becomes twice the monomer-chain molar mass.
8. A polymer solution produces an equilibrium height difference of 25.4 cm in a membrane osmometer containing water of density 1000 kg m⁻³. Taking g = 9.8 m s⁻², T = 298 K, and polymer concentration as 1.00 g L⁻¹, what is the approximate number-average molar mass? Use R = 8.314 kPa L mol⁻¹ K⁻¹.
A.
9.95 × 10² g mol⁻¹
B.
2.48 × 10³ g mol⁻¹
C.
9.95 × 10³ g mol⁻¹
D.
2.48 × 10⁴ g mol⁻¹
Correct Answer:
9.95 × 10² g mol⁻¹
Explanation:
The osmotic pressure is π = ρgh = 1000 × 9.8 × 0.254 Pa = 2.49 kPa. Using Mn = cRT/π gives Mn = (1.00 × 8.314 × 298)/2.49 ≈ 9.95 × 10² g mol⁻¹.
9. A small fraction of polymer molecules leaks through the membrane during a membrane-osmometry experiment, while the leaked molecules are not included in the measured polymer concentration. What is the most likely effect on the calculated molar mass?
A.
The calculated molar mass is lower because the measured osmotic pressure is too high
B.
The calculated molar mass is higher because the measured osmotic pressure is too low for the stated concentration
C.
The calculated molar mass is unchanged because osmotic pressure is independent of particle number
D.
The calculated molar mass becomes zero because the membrane loses selectivity
Correct Answer:
The calculated molar mass is higher because the measured osmotic pressure is too low for the stated concentration
Explanation:
If polymer escapes, the actual number of retained polymer particles is smaller than assumed from the stated concentration, reducing π. Since Mn = cRT/π, using the underestimated pressure yields an artificially high molar mass.
10. A polymer sample contains species with molar masses M₁ and M₂ in amounts n₁ and n₂. Which expression is consistent with the molar mass determined by membrane osmometry for an ideal dilute solution?
A.
(n₁M₁ + n₂M₂)/(n₁ + n₂)
B.
(n₁M₁² + n₂M₂²)/(n₁M₁ + n₂M₂)
C.
(n₁M₁ + n₂M₂)/(n₁/M₁ + n₂/M₂)
D.
(n₁ + n₂)/(n₁/M₁ + n₂/M₂)
Correct Answer:
(n₁M₁ + n₂M₂)/(n₁ + n₂)
Explanation:
Membrane osmotic pressure depends on the total number of macromolecular particles, π = (n₁ + n₂)RT/V, while total mass is n₁M₁ + n₂M₂. Thus, the inferred molar mass is total mass divided by total number of molecules, giving the number-average molar mass.
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