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Mass Transfer

In a liquid-liquid extraction, 10 kg of a solution containing 2 kg of solute C and 8 kg of solvent A is brought into contact with 10 kg of solvent B. Solvent A and B are completely immiscible in each other whereas solute C is soluble in both the solvents. The extraction process attains equilibrium. The equilibrium relationship between the two phases is Y* = 0.9X, where Y* is the kg of C/kg of B and X is kg of C/kg of A. Choose the correct answer.

Question: In a liquid-liquid extraction, 10 kg of a solution containing 2 kg of solute C and 8 kg of solvent A is brought into contact with 10 kg of solvent B. Solvent A and B are completely immiscible in each other whereas solute C is soluble in both the solvents. The extraction process attains equilibrium. The equilibrium relationship between the two phases is Y* = 0.9X, where Y* is the kg of C/kg of B and X is kg of C/kg of A. Choose the correct answer.
[A].

The entire amount of C is transferred to solvent B.

[B].

Less than 2 kg but more than 1 kg of C is transferred to solvent B.

[C].

Less than 1 kg of C is transferred to B.

[D].

No amount of C is tranferred to B.

Answer: Option A

Explanation:

No answer description available for this question.

In a liquid-liquid extraction, 10 kg of a solution containing 2 kg of solute C and 8 kg of solvent A is brought into contact with 10 kg of solvent B. Solvent A and B are completely immiscible in each other whereas solute C is soluble in both the solvents. The extraction process attains equilibrium. The equilibrium relationship between the two phases is Y* = 0.9X, where Y* is the kg of C/kg of B and X is kg of C/kg of A. Choose the correct answer. Read More »

CHEMICAL ENGINEERING, Mass Transfer

Acetone is to be removed from air in an isothermal dilute absorber using pure water as solvent. The incoming air contains 5 mole% of acetone (yin = 0.05). The design equation to be used for obtaining the number of trays (N) of the absorber is, N+2 = 6 log (yin/yout).For 98% recovery of acetone, the number of trays required is/are

Question: Acetone is to be removed from air in an isothermal dilute absorber using pure water as solvent. The incoming air contains 5 mole% of acetone (yin = 0.05). The design equation to be used for obtaining the number of trays (N) of the absorber is, N+2 = 6 log (yin/yout).For 98% recovery of acetone, the number of trays required is/are
[A].

1

[B].

8

[C].

9

[D].

10

Answer: Option C

Explanation:

No answer description available for this question.

Acetone is to be removed from air in an isothermal dilute absorber using pure water as solvent. The incoming air contains 5 mole% of acetone (yin = 0.05). The design equation to be used for obtaining the number of trays (N) of the absorber is, N+2 = 6 log (yin/yout).For 98% recovery of acetone, the number of trays required is/are Read More »

CHEMICAL ENGINEERING, Mass Transfer

An air-water vapour mixture has a dry bulb temperature of 60°C and a dew point temperature of 40°C. The total pressure is 101.3 kPa and the vapour pressure of water at 40°C and 60°C are 7.30 kPa and 19.91 kPa respectively.The humidity of air sample expressed as kg of water vapour/kg of dry air is

Question: An air-water vapour mixture has a dry bulb temperature of 60°C and a dew point temperature of 40°C. The total pressure is 101.3 kPa and the vapour pressure of water at 40°C and 60°C are 7.30 kPa and 19.91 kPa respectively.The humidity of air sample expressed as kg of water vapour/kg of dry air is
[A].

0.048

[B].

0.079

[C].

0.122

[D].

0.152

Answer: Option D

Explanation:

No answer description available for this question.

An air-water vapour mixture has a dry bulb temperature of 60°C and a dew point temperature of 40°C. The total pressure is 101.3 kPa and the vapour pressure of water at 40°C and 60°C are 7.30 kPa and 19.91 kPa respectively.The humidity of air sample expressed as kg of water vapour/kg of dry air is Read More »

CHEMICAL ENGINEERING, Mass Transfer

A distillation column separates 10000 kg/hr of a benzene-toluene mixture as shown in the figure below :In the figure xF, xD and xW represent the weight fraction of benzene in the feed, distillate and residue respectively. The reflux ratio is

Question: A distillation column separates 10000 kg/hr of a benzene-toluene mixture as shown in the figure below :In the figure xF, xD and xW represent the weight fraction of benzene in the feed, distillate and residue respectively. The reflux ratio is
[A].

0.5

[B].

0.6

[C].

1.0

[D].

2.0

Answer: Option B

Explanation:

No answer description available for this question.

A distillation column separates 10000 kg/hr of a benzene-toluene mixture as shown in the figure below :In the figure xF, xD and xW represent the weight fraction of benzene in the feed, distillate and residue respectively. The reflux ratio is Read More »

CHEMICAL ENGINEERING, Mass Transfer

Compound A is extracted from a solution of A + B into a pure solvent S. A Co-current unit is used for the liquid-liquid extraction. The inlet rate of the solution containing A is 200 moles of B/hr.m2 and the solvent flow, rate is 400 moles of S/m2. hr. The equilibrium data is represented by Y = 3X2 , where Y is in moles of a A/moles of B and X is in moles A/moles of S. The maximum percentage extraction achieved in the unit is

Question: Compound A is extracted from a solution of A + B into a pure solvent S. A Co-current unit is used for the liquid-liquid extraction. The inlet rate of the solution containing A is 200 moles of B/hr.m2 and the solvent flow, rate is 400 moles of S/m2. hr. The equilibrium data is represented by Y = 3X2 , where Y is in moles of a A/moles of B and X is in moles A/moles of S. The maximum percentage extraction achieved in the unit is
[A].

25%

[B].

50%

[C].

70%

[D].

90%

Answer: Option B

Explanation:

No answer description available for this question.

Compound A is extracted from a solution of A + B into a pure solvent S. A Co-current unit is used for the liquid-liquid extraction. The inlet rate of the solution containing A is 200 moles of B/hr.m2 and the solvent flow, rate is 400 moles of S/m2. hr. The equilibrium data is represented by Y = 3X2 , where Y is in moles of a A/moles of B and X is in moles A/moles of S. The maximum percentage extraction achieved in the unit is Read More »

CHEMICAL ENGINEERING, Mass Transfer

A solid is being dried in the linear drying rate regime from moisture content Xo to XF. The drying rate is zero at X = 0 and the critical moisture content is the same as the initial moisture Xo. The drying time for M = (Ls/ARc) is (where, L = total mass of dry solid, A = total surface area for drying Rc = Constant maximum drying rate per unit area X = moisture content (in mass of water/mass of dry solids))

Question: A solid is being dried in the linear drying rate regime from moisture content Xo to XF. The drying rate is zero at X = 0 and the critical moisture content is the same as the initial moisture Xo. The drying time for M = (Ls/ARc) is (where, L = total mass of dry solid, A = total surface area for drying Rc = Constant maximum drying rate per unit area X = moisture content (in mass of water/mass of dry solids))
[A].

M(Xo – XF)

[B].

M(Xo/XF)

[C].

M ln(Xo/XF)

[D].

MXo ln(Xo/XF)

Answer: Option D

Explanation:

No answer description available for this question.

A solid is being dried in the linear drying rate regime from moisture content Xo to XF. The drying rate is zero at X = 0 and the critical moisture content is the same as the initial moisture Xo. The drying time for M = (Ls/ARc) is (where, L = total mass of dry solid, A = total surface area for drying Rc = Constant maximum drying rate per unit area X = moisture content (in mass of water/mass of dry solids)) Read More »

CHEMICAL ENGINEERING, Mass Transfer

A pure drug is administered as a sphere and as a cube. The amount of drug is the same in the two tablets. Assuming that the shape and size do not influence the mass transfer, the ratio of rate of dissolution in water at t = 0 for the cubic to spherical tablet is

Question: A pure drug is administered as a sphere and as a cube. The amount of drug is the same in the two tablets. Assuming that the shape and size do not influence the mass transfer, the ratio of rate of dissolution in water at t = 0 for the cubic to spherical tablet is
[A].

0.54

[B].

1.04

[C].

1.24

[D].

1.94

Answer: Option C

Explanation:

No answer description available for this question.

A pure drug is administered as a sphere and as a cube. The amount of drug is the same in the two tablets. Assuming that the shape and size do not influence the mass transfer, the ratio of rate of dissolution in water at t = 0 for the cubic to spherical tablet is Read More »

CHEMICAL ENGINEERING, Mass Transfer

H2S is being absorbed in a gas absorber unit. The height of the transfer unit based on the overall mass transfer coefficient on the gas side is 0.4 m. The equilibrium data is given by, y = 1.5 x. The bulk concentration of H2S has to be reduced from 0.05 to 0.001 mole fraction in the gas side. The height of the tower (in metres) corresponding to an operating line given by, y = 5x + 0.001 is

Question: H2S is being absorbed in a gas absorber unit. The height of the transfer unit based on the overall mass transfer coefficient on the gas side is 0.4 m. The equilibrium data is given by, y = 1.5 x. The bulk concentration of H2S has to be reduced from 0.05 to 0.001 mole fraction in the gas side. The height of the tower (in metres) corresponding to an operating line given by, y = 5x + 0.001 is
[A].

2.0

[B].

1.56

[C].

1.0

[D].

0.56

Answer: Option A

Explanation:

No answer description available for this question.

H2S is being absorbed in a gas absorber unit. The height of the transfer unit based on the overall mass transfer coefficient on the gas side is 0.4 m. The equilibrium data is given by, y = 1.5 x. The bulk concentration of H2S has to be reduced from 0.05 to 0.001 mole fraction in the gas side. The height of the tower (in metres) corresponding to an operating line given by, y = 5x + 0.001 is Read More »

CHEMICAL ENGINEERING, Mass Transfer

Experiments were conducted to determine the flux of a species A in a stagnant medium across a gas-liquid interface. The overall mass transfer co-efficient based on the liquid side for dilute systems for the above was estimated to be 4 x 10-3 kg mole/m2.s. The equilibrium data for the system is given as y = 2x. The flux across the interface (in kg mole/m2 .s) for bulk concentrations of A in gas phase and liquid phase as y = 0.4 and x = 0.01 respectively is

Question: Experiments were conducted to determine the flux of a species A in a stagnant medium across a gas-liquid interface. The overall mass transfer co-efficient based on the liquid side for dilute systems for the above was estimated to be 4 x 10-3 kg mole/m2.s. The equilibrium data for the system is given as y = 2x. The flux across the interface (in kg mole/m2 .s) for bulk concentrations of A in gas phase and liquid phase as y = 0.4 and x = 0.01 respectively is
[A].

5.6 x 10-4

[B].

8.5 x 10-4

[C].

5.6 x 10-3

[D].

8.5 x 10-3

Answer: Option B

Explanation:

No answer description available for this question.

Experiments were conducted to determine the flux of a species A in a stagnant medium across a gas-liquid interface. The overall mass transfer co-efficient based on the liquid side for dilute systems for the above was estimated to be 4 x 10-3 kg mole/m2.s. The equilibrium data for the system is given as y = 2x. The flux across the interface (in kg mole/m2 .s) for bulk concentrations of A in gas phase and liquid phase as y = 0.4 and x = 0.01 respectively is Read More »

CHEMICAL ENGINEERING, Mass Transfer