This set of Bioseparation Technology Multiple Choice Questions & Answers (MCQs) focuses on “Mass Transfer – Diffusion of Solutes in Dense and Porous Solid”.
1. What is the example of diffusion of solutes in dense medium?
a) Diffusion of ions through dense membrane
b) Diffusion of ions in acetone
c) Diffusion of sodium chloride through microfiltration membrane
d) Diffusion of albumin through ultrafiltration membrane
View Answer
Explanation: The diffusion of ions through dense membrane is an example of diffusion of solutes in dense membrane. The molecules in the solute diffuse through the dense solid membrane once it gets dissolved in it. Therefore, the ions in the solute dissolve in the dense membrane and then diffuse through it showing the diffusion of solute in the dense solid.
2. Which parameter of the molecule of solid medium is not counter-diffused?
a) Solubility
b) Mobility
c) Diffusivity
d) Gravity
View Answer
Explanation: Mobility of the molecule of solid medium is not counter-diffused even if the parameter is limited. The mobility of the molecule and the diffusion of solutes, rate of diffusion depends on the structure of solid and the interaction of molecules with the solute.
3. Fick’s law is not applicable for describing the diffusion of solutes molecules in a solid medium.
a) False
b) True
View Answer
Explanation: Fick’s law is still used to describe the diffusion of solute molecule in a solid medium. Diffusion occurs depending on the Fick’s law and the rate of diffusion is proportional to the space available for the transfer of solute molecules and it is inversely proportional to the thickness of the membrane.
4. What will be the flux when the diffusivity is independent of the concentration and there is no bulk flow?
a) NA = DA (\(\frac{dC_A}{dZ}\))
b) NA = (\(\frac{dC_A}{dZ}\))
c) NA = -DA (\(\frac{dC_A}{dZ}\))
d) NA = -(\(\frac{dC_A}{dZ}\))
View Answer
Explanation: The molar flux (NA) in the direction of Z is given by Flux’s law which is NA = -DA (\(\frac{dC_A}{dZ}\)) where, DA is the diffusivity of A through the solid membrane, CA is the concentration of the solute and Z is the thickness of the solid membrane through which diffusion takes place.
5. Which process involves the use of diffusion of solutes?
a) Homogenous catalytic reaction
b) Homogenous non-catalytic reaction
c) Heterogeneous non-catalytic reaction
d) Heterogeneous catalytic reaction
View Answer
Explanation: Heterogeneous catalytic reaction involves the use of diffusion of solutes. In this reaction the reactants are adsorbed on the surface of catalyst and then the reactant diffuses on the surface of the catalyst and finally reaction takes place between the reactants and the products get desorbed from the surface of the solid catalyst.
6. Solutes molecules cannot diffuse through the pores present in porous solids.
a) True
b) False
View Answer
Explanation: Solute molecules can diffuse through the pores in the porous solids and it takes place by filling the pores with some liquid medium so that the there is no diffusion of solute molecules through the solid medium and it holds the liquid. Solid material can influence the diffusion within the liquid medium.
7. What will the response of the pore wall when the pores have dimension of same order of magnitude as the solute?
a) Hindrance to diffusion
b) Easy diffusion
c) Blockage in the pore wall
d) Change in the solute composition
View Answer
Explanation: The pore wall will cause hindrance to diffusion when the pores have dimensions of same order of magnitude as the solutes. Therefore, the pore size should have low depending on the liquid medium and the solute used for diffusion.
8. What is the example of diffusion of solutes in porous solid?
a) Diffusion of ions through dense membrane
b) Diffusion of ions in acetone
c) Diffusion of sodium chloride through microfiltration membrane
d) Diffusion of albumin through membrane
View Answer
Explanation: Diffusion of sodium chloride through microfiltration membrane is the example of diffusion of solutes in porous medium. The membrane has pores of micron size so that the pore walls should not cause hindrance to the diffusion of solutes through the membrane.
9. What is the steady state equation for un-hindered diffusion of solute from 1 point and 2 point slab of porous solid?
a) JA = –\(\frac{\varepsilon D(C_{A1} – C_{A2})}{\tau(x_1 – x_2)}\)
b) -JA = \(\frac{\varepsilon D(C_{A1} – C_{A2})}{\tau(x_1 – x_2)}\)
c) JA = \(\frac{\varepsilon D(C_{A1} – C_{A2})}{\tau(x_1 – x_2)}\)
d) JA = \(\frac{\varepsilon D(C_{A1} – C_{A2})}{\tau(x_1 – x_2)}\)
View Answer
Explanation: The steady state equation for unhindered diffusion of solute from 1 point and 2 within a slab of porous solid is JA = \(\frac{\varepsilon D(C_{A1} – C_{A2})}{\tau(x_1 – x_2)}\) where, D is the diffusivity of the solutes in the liquid with the pores ε is the porosity of the medium, τ is the tortuosity of the medium and the CA1 – CA2 is the concentration difference between two points.
10. Calculate the solute concentration in the other chamber after 50 minutes when the content of one of the chambers was replaced with 10 ml of 1 mg/ml human albumin solution at t = 0. The porosity of the membrane is 0.75; pore size is 1 micron, surface area is 2 cm2 and thickness is 0.1 mm of the membrane, volume of chamber is 10 ml.
a) C1 = 0.973 mg/ml and C2 = 0.027mg/ml
b) C1 = 0.85 mg/ml and C2 = 0.010mg/ml
c) C1 = 0.76 mg/ml and C2 = 0.027mg/ml
d) C1 = 0.973 mg/ml and C2 = 0.01mg/ml
View Answer
Explanation: It is assumed that the diffusion of albumin through the pores is un-hindered. The diameter of albumin is 7.2 x l0-3 microns. Therefore: D eff = D. The diffusivity of albumin is 5.94 x 10-11 m2/s. 5.94 × 10-11 = \(\frac{10 × 10^{-6} × 0.1
× 10^{-3}}{2 × 0.75 × 2 × 10^{-4} × 3000}\) ln (\(\frac{1}{C_1 – C_2}\)) the molar concentration with the mass concentration since the molecular weight cancels out between the numerator and denominator of the term within parenthesis. When, C10V = C1V + C2V i.e. 1 × 10 × 10-6 = (C1 + C2) × 10 × 10-6 and the concentrations are 0.093mg/ml and 0.027 mg/ml.
11. What will be the steady state flux of glucose between two points within the medium when the glucose is diffusing at 25°C in water within a porous medium and the porosity is 0.5, tortuosity is 1.8 and pore diameter is 8.6 x 10-3 micron and the points are separated by a distance of 1 mm and the concentrations are 1.50g/l and 1.51 g/l respectively?
a) 2.02 × 10-12 kg-moles/ m2s
b) 4.02 × 10-12 kg-moles/ m2s
c) 5.02 × 10-12 kg-moles/ m2s
d) 8.02 × 10-12 kg-moles/ m2s
View Answer
Explanation: The molecular weight of glucose is 180 kg/moles and the diffusivity is D = \(\frac{9.40 × 10^{-15} × 298}{0.001 × (180)^{1/3}}\) = 4.96 × 10-10 m2/s. and the diffusivity of glucose in the porous structure is D(eff) = 4.96 × 10-10 (1 – \(\frac{8.6 × 10^{-4}}{8.6 × 10^{-3}}\)) = 3.25 × 10-10 m2/s. Therefore, the flux of glucose is JA = \(\frac{3.25 × 10^{-10} × 0.5 × 0.01}{0.001 × 180 × 1.8}\) = 5.02 × 10-12 kg-moles/ m2s.
Sanfoundry Global Education & Learning Series – Bioseparation Technology.
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