Advanced Mass Transfer Questions and Answers

This set of Advanced Mass Transfer Questions & Answers focuses on “Design of Extractor for Single Stage Operations”.

1. Stages in the extractor are _______________
a) Equilibrium stages
b) Theoretical stages
c) Equilibrium or theoretical stages
d) None of the mentioned
View Answer

Answer: c
Explanation: The extractor stages are maintained in equilibrium to get the effect mass transfer.

2. Extraction includes mixing alone.
a) True
b) False
View Answer

Answer: b
Explanation: Extraction needs mixing and separation.

3. Every stage of extractor act as mixer and settler in the extraction process.
a) True
b) False
View Answer

Answer: a
Explanation: Every extracting process needs mixing and separating unit.

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4. If the pure solvent is used find the value of “ys” by analysing the below single stage extractor.

Where, F,S are feed and solvent
a) 0
b) 1
c) 0-1
d) None of the mentioned
View Answer

Answer: a
Explanation: Pure solvent in the sense no solute present initially so ys= 0.

5. Find the rate of extract in mol/hr if

	Rate(mol/hr)	Composition
Feed	100	0.5
Solvent	50	0
Residue	25	0.25

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a) 100
b) 125
c) 150
d) 50
View Answer

Answer: b
Explanation: Feed + solvent = Extract + residue
100+50-25 = 125 mol/hr.

6. Is the extraction is possible if a gas solvent is used for extracting solute from liquid mixture.
a) True
b) False
View Answer

Answer: b
Explanation: The extraction is possible only if liquid-liquid equilibrium occurs. If gas solvent is used such equilibrium cannot be achieve.

7. Find the process.

a) Single stage batch process
b) Single stage continuous process
c) Multistage continuous process
d) None of the mentioned
View Answer

Answer: b
Explanation: Here stage used is one and there is a continuous input and output so the process is a single stage continuous process.

8. Find the sum of R and E if the sum of F and S is 80 kmol.

a) 20 kmol
b) 40 kmol
c) 60 kmol
d) 80 kmol
View Answer

Answer: d
Explanation: M= F+M= R+S =80.

9. Find the pure solvent rate if the feed ( F= 100 kmol/hr) composition is 0.65 and mixture solute composition is 0.5.
a) 10 kmol/hr
b) 20 kmol/hr
c) 30 kmol/hr
d) 40 kmol/hr
View Answer

Answer: c
Explanation: S/F = 0.65- 0.5/0.5 -0
= 0.3
= 100 x 0.3 = 30 kmol/hr.

10. Find the mixture composition of feed and pure solvent if the ratio of solvent to feed rate is 0.5 and the feed composition is 0.5.
a) 0.5
b) 0.25
c) 0.33
d) 0.4
View Answer

Answer: c
Explanation: S/F= 0.5-x/x-0
0.5x+x = 0.5 then x= 0.33.

Sanfoundry Global Education & Learning Series – Mass Transfer.

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